Notes » Cost

Cost of amenable groups

jraimbau — Mon, 28 Nov 2016 10:29:04 +0000

The aim of these notes is to provide an introduction to Gaboriau’s paper « Sur le coût des relations d’équivalence et des groupes » and to gather in one place various arguments, occuring in diverse sources, to give a characterisation of amenable groups through their actions on probability spaces.

Cost

Setting

Let $ X, \mathcal B, \mu $ be a measured space (often we’ll omit $ \mathcal B $ from the notation). Usually $ \mu $ will be a probability measure (i.e. $ \mu(X) = 1 $). Let $ E \subset X \times X $ be an equivalence relation on $ X $, we will use the abbreviation $ xEy $ for $ (x, y) \in E $. We will suppose that:

$ E $ is measurable, meaning that it is a measurable subset of $ X \times X $; equivalently, if $ A \in \mathcal B $ then also $ \{ x : \exists y \in A : xEy \} \in \mathcal B $.
All equivalence classes for $ E $ are countable.
$ \mu $ is $ E $-invariant, that is for any partially defined Borel isomorphism $ \phi : A \to B $ (where $ A, B \in \mathcal B $) such that $ \forall x \in A: xE\phi(x) $ we have $ \phi_*\mu|_A = \mu|_B $.

The basic example is when a finitely generated group $ \Gamma $ acts on a space $ X, \mu $ by Borel automorphisms preserving the measure $ \mu $(if in addition $ \mu $ is a probability measure we’ll say that this is a p.m.p.—for probability measure preserving—action), and $ E $ is the orbit relation: $ xEy \Leftrightarrow \exists \gamma \in \Gamma: y =\gamma \cdot x $.

The group of Borel automorphisms whose graph is contained in $ E $ (the full group of $ E $) is denoted by $ [E] $. The pseudo group of partially defined Borel automorphisms with graph contained in $ E $ is denoted by $ [[E]] $.

Graphings and cost of relations

A graphing of $ E $ is an undirected graph $ \mathcal G $ on $ X $ such that:

$ \mathcal G $ is contained in $ E $ : that is, if $ x, y $ are neighbours in $ \mathcal G $ then $ xEy $ (connected components of $ \mathcal G $ are contained in $ E $-classes);
$ \mathcal G $ generates $ E $: that is, if $ xEy $ then $ x, y $ are in the same connected component of $ \mathcal G $.
$ \mathcal G $ is measurable in the sense that the subset of $ X \times X $ given by those $ (x, y) $ which are neighbours in $ \mathcal G $ is measurable.

Equivalently, a graphing is defined by a collection $ \Phi = (\phi_i : A_i \to B_i)_{i \in I} $ of partially defined Borel automorphisms, such that:
\[
xEy \Leftrightarrow \exists i_1, \ldots, i_r \in I : y = \phi_{i_r} \circ \cdots \circ \phi_{i_1} x.
\]
The cost of a graphing $ \mathcal G $ (with respect to the measure $ \mu$) is half the average vertex degree:
\[
\mathrm{cost}(\mathcal G, \mu) = \frac 1 2 \int_X d_{\mathcal G}(x) d\mu(x)
\]
(where $ d_{\mathcal G}(x) $ is the number of neighbours of $ x $ in $ \mathcal G $). If the graphing is defined by $ \Phi $ as above then the cost is equal to:
\[
\mathrm{cost}(\Phi, \mu) = \sum_{i \in I} \mu(A_i).
\]
The cost of a measured equivalence relation is the smallest possible cost for a graphing:
\[
\mathrm{cost}(E, \mu) = \inf_{\mathcal G} \mathrm{cost}(\mathcal G, \mu).
\]

Cost of groups and fixed price

The cost of a discrete group $ \Gamma $, denoted by $ \mathrm{cost}(\Gamma) $, is the infimum over all of its essentially free, p.m.p. actions of their cost. A group is said to have fixed price if the cost of any action is equal to the cost of the group. There are no known exemples of a group which does not have fixed price.

Computing cost

Basic properties

Let $ X, \mu $ a probability space and $ E $ a measured equivalence relation on $ X $ preserving $ \mu $, with infinite countable classes. Then we have the following facts due to Levitt:

$ \mathrm{cost}(E, \mu) \ge 1 $.
If there exists a graphing $ \mathcal G $ such that $ \mathrm{cost}(E, \mu) = \mathrm{cost}(\mathcal G, \mu) $ then almost all components of $ \mathcal G $ are trees; it is then referred to as a treeing.
If $ B \subset X $ is a positive subset (measurable with positive measure) and $ E $ has finite cost then
\[
\mathrm{cost}(E|_B, \mu|_B/ \mu(B)) – 1 = \mu(B)^{-1} (\mathrm{cost}(E, \mu) – 1)
\]
Moreover $ E|_B $ admits a treeing if and only if $ E $ does.

We give two proofs of 1.: the one by Levitt, and one by Gaboriau which deduces it immediately from 3 (they are essentially the same).

Levitt’s proof goes as follows: for a partial graphing $ \Phi $ of $ E $ (a Borel graph contained in $E$ but which does not necessarily generate it) let
\[
e(\Phi) = \int_X \frac 1 {|\langle \Phi \rangle x|} d\mu(x).
\]
This is 0 if $ \Phi $ is a graphing (all orbits are infinite), and in general it is equal to the infimum over the measures of all Borel sets meeting every orbit of $ \langle \Phi \rangle $. The claim is then that for all partial graphings we have
\[
(\ast) \qquad e(\Phi) + \mathrm{cost}(\Phi) \ge 1.
\]
This is proven by induction: for the empty graphing we have $ e = 1, \mathrm{cost} = 0 $. Now suppose that $ \Phi $ satisfies $ (\ast) $ and $ \psi = \Phi \cup \{\psi : U \to V \} $. Let $ A $ be any Borel subset meeting every orbit of $ \langle \Psi \rangle $: we want to find a subset $ B $ such that $ A \cup B $ meets every orbit of $ \langle \Phi \rangle $ and $ \mu(B) \le \mu(A) + \mu(U) $ (then we get $ \mathrm{cost}(\Phi) + \mu(U) \ge \mathrm{cost}(\Psi) $ and if, fixing an aritrarily small $ \varepsilon > 0 $ we choose $ B $ such that $ \mu(B) \le e(\Psi) + \varepsilon $ also
\[
e(\Phi) \le \mu(A) \le \mu(B) + \mu(U) \le e(\Psi) + \varepsilon + \mathrm{cost}(\Psi) – \mathrm{cost} \Phi
\]
which proves that $ \mathrm{cost}(\Psi) + e(\Psi) \ge \mathrm{cost}(\Phi) + e(\Phi) $).

For $ x \in X $ there exists $ a \in A $ and $ \phi_{i_k} \in \Phi, m_k \in \mathbb Z $, $ k = 1, \ldots, n $ such that $ x = \prod_{k=1}^n \phi_{i_k} \psi^{m_k} $. Let $ s(x) = \inf \sum_k |m_k| $ and define:
\[
B_1 = \{ x \in U : s(\psi x) < s(x) \}, \quad B_2 = \{ x \in V : s(\psi^{-1}x) < s(x) \}.
\]
Then $ \psi^{-1}B_2 \subset U $ and $ B_1 \cap \psi^{-1} B_2 = \emptyset $ so that $ \mu(B_1) + \mu(B_2) \le \mu(U) $. In addition it is clear that $ X = \langle \Phi \rangle B $, which finishes our proof in case $ \Phi $ is a finite graphing. If $ \Phi = ( \phi_i, i \in \mathbb N ), \Phi_n = (\phi_k, 1\le k \le n) $ then we can conclude bu using $ \mathrm{cost}(\Phi_n) + e(\Phi_n) \ge 1 $ and $ \mathrm{cost}(\Phi) = \lim\mathrm{cost}(\Phi_n) $ and $ \lim e(\Phi_n) = 0 $.

Gaboriau’s proof runs as follows: if $ \Phi $ is a graphing then we know that $ e(\Phi) = 0 $ and thus for every $ \varepsilon $ there is a subset $ X_\varepsilon $ which meets every orbit and has $ \mu(X_\varepsilon) = \varepsilon $. By 3. we have
\[
\mathrm{cost}(\Phi) – 1 = \mu(X_\varepsilon)(\mathrm{cost}(\Phi|_{X_\varepsilon}) – 1) \ge -\mu(X_\varepsilon) = -\varepsilon
\]
since cost is positive, and by taking $ \varepsilon \to 0 $ we conclude that $ \mathrm{cost}(\Phi) – 1 \ge 0 $.

The proof of 2. is as follows. Suppose that $ \Phi $ is a graphing of $ E $ which is not a treeing. Then there exists $ i_k, \varepsilon_k \in \{ \pm 1\}, k=1, \ldots, n $ and a nonzero set $ A $ such that $ \prod_{i=1}^n \phi_{i_k} x = x $ for all $ x \in A $. We may in addition assume that $ \prod_{i=1}^p \phi_{i_k}^{\varepsilon_k} x \not \in A $ for all $ p \le n-1 $ and that $ \varepsilon_k = 1 $. Then the graphing which has $ \phi_{i_k} : A_{i_k} \to B_{i_k} $ by $ \phi_{i_k} : A_{i_k}\setminus A \to B_{i_k}\setminus \phi_{i_k}A $ has the same orbits as $ \Phi $ and cost equal to $ \mathrm{cost}(\Phi) – \mu(A) $, so that $ \mathrm{cost}(\Phi) > \mathrm{cost}(E) $.

When is cost realised by a single graphing?

The main theorem in Gaboriau’s paper is the converse of 2. An immediate corollary is that any essentially free action of a free group on $ r $ generators has cost $ r $, hence the group has fixed price $ r $.

Theorem : If $ \mathcal G $ is a treeing of $ E $ then $ \mathrm{cost}(E, \mu) = \mathrm{cost}(\mathcal G, \mu) $.

A more precise special case is the following, essentially due to Ornstein–Weiss in the hard direction and appearing in Levitt’s paper.

Theorem : Let $ E $ be the relation induced by an essentially free p.m.p. discrete group action. Then there exists a graphing $ \mathcal G $ with $ \mathrm{cost}(E) = \mathrm{cost}(\Phi) = 1 $ if and only if $ \Gamma $ is amenable.

Modulo an easy argument (to go from « heving a graphing with cost 1 » to « being generated by a single automorphism ») this means that:

If the orbits of $ \Gamma $ on $ X $ are equal to the orbits of a single Borel automorphism $ T : X \to X $ then $ \Gamma $ is amenable;
Conversely, if $ \Gamma $ is amenable there exists a $ T $ whose orbits are exactly that of $ \Gamma $.

The proof of the reciprocal direction 1 goes as follows: the orbit equivalence between $ \Gamma $ and $ \langle T \rangle \cong \mathbb Z$ gives a measure equivalence between these two groups, i.e. a measure space $ \widehat X, \widehat \mu $ with an action of $ \Gamma \times \mathbb Z $ such that both factors admit a finite-volume fundamental set. (This can be obtained as $ \widehat X = X \times \mathbb Z $ with $ \mathbb Z $ acting on the right by translations in the $ \mathbb Z $ factor and $ \Gamma $ acting by $ \gamma\cdot(x, n) = (\gamma x, n+m ) $ where $ \gamma x = T^m x $).

Now this measure equivalence gives a map $ \pi \mapsto \tilde \pi $ from unitary representations of $ \mathbb Z $ to those of $ \Gamma $. If $ \pi $ is defined on a Hilbert space $ \mathcal H $ then the representation $ \tilde \pi $ is defined on the space:
\[
\tilde{\mathcal H} := \left\{ f: \widehat X \to \mathcal H: \forall m \in \mathbb Z, x \in \widehat X : f(x\cdot m) = \pi(m)f(x) \right\}
\]
on which it acts by $ \tilde \pi(\gamma)f(x) = f(\gamma^{-1} \cdot x) $. Then we have the following facts:

If $ \pi $ is the left-regular representation on $ L^2(\Lambda) $ then $ \tilde \pi $ is equivalent to the representation on $ L^2(\widehat X) $ (which is equivalent to the Hilbert sum $ w_n \in \overline \bigoplus_{\mathbb Z} L^2(\Gamma) $ of countably many copies of the left-regular representation on $ L^2(\Gamma) $).
If $ \pi $ has almost-invariant vectors then $ \tilde \pi $ also does.

Thus if $ \Lambda $ is amenable, we get a sequence of quasi-$ \Gamma $-invariant vectors $ w_n \in \overline \bigoplus_{\mathbb Z} L^2(\Gamma) $. Projecting them onto the factors gives a sequence of quasi-invariant vectors in $ L^2(\Gamma) $, hence it is amenable.

Proof of (a) and (b): For (a) we have the map from $ \tilde H $ to $ L^2(\widehat X) $ given by $ f \mapsto \langle f, \delta_{1_\Lambda} \rangle_{L^2(\Lambda)} $ which is easily checked to be an isometry and $ \Gamma $-equivariant. For (b), we choose a mesurable fundamental domain $ U \subset \widehat X $ for $ \Lambda $. Given a sequence $ v_n \in \mathcal H, \| v_n \|_{\mathcal H} = 1 $ of quasi -invariant vectors (meaning that $ \langle \pi(g)v_n, v_n \rangle_{\mathcal H} \to 1 $ for every $ g \in \Lambda $) we define $ f_n \in \tilde H $ by $ f_n(x) = v_n $ for $ x \in U $. Then $ f_n $ is a sequence of quasi-invariant vectors in $ L^2(\Gamma) $.

Rokhlin lemma and cost of amenable groups

Now we want to prove that if $ \Gamma $ is amenable then the orbit relation of every p.m.p. $ \Gamma $-action is generated by a single Borel automorphism. For this we will use the following characterisation: an equivalence relation is said to be hyperfinite if there exists a nested sequence of finite, measurable subrelations $ F_1 \subset \cdots \subset F_n \subset E $ such that $ E = \bigcup_n F_n $.

Lemma : A measurable equivalence relation is hyperfinite if and only if it is induced by a single Borel automorphism.

The construction of an automorphism from a sequence of nested finite relations is illustarted by the following pictures.

Thus we want to prove that an action of an amenable group gives rise to an hyperfinite equivalence relation. The result that will enable us to do so is the following theorem of Ornstein and Weiss, often called « Rokhlin’s lemma » (Rokhlin proved the case $ \Gamma = \mathbb Z $). Say that a finite subset $ T \subset \Gamma $ is a tile if there exists $ C \subset \Gamma $ such that $ \Gamma = CT $ and $ cT \cap c’T = \emptyset $ if $ c \not= c’ \in C $?

Theorem : Suppose that $ \Gamma $ is an amenable group and that $ T \subset \Gamma $ is a tile. Let $ \Gamma $ acting on (X, \mu) \) be an essentially free p.m.p. action. Then for any $ \varepsilon > 0 $ there exists a measurable subset $ A \subset X $ such that $ \mu(T \cdot A) \ge 1 – \varepsilon $ and for almost all $ a \in A $ we have $ (T\cdot) a \cap (T\cdot b) = \emptyset $ for all $ b \in A\setminus\{a\} $.

To deduce hyperfiniteness from this result suppose that there exists a sequence $ T_1, \ldots, T_n, \ldots $ of tiles such that $ T_n $ is $ (T_{n-1} \cup T_{n-1}^{-1}, 1/n) $-invariant. Let $ T_n’ = T_n \setminus (T_n \Delta T_{n-1}T_n) $ so that $ |T_n’|/|T_n| \ge 1 – 1/n $.

Rokhlin’s lemma gives us subsets $ A_n \subset X $ such that $ \mu(T_nA_n) \ge 1-1/n $ and the $ T_na, a \in A_n $ are paiwise disjoint. Let $ F_n $ be the finite relation which is defined by
\[
xF_n y \Leftrightarrow \exists a \in A: x, y \in T_na
\]
on $ T_nA $ and which is the identity on $ X \setminus T_nA $. Clearly $ E = \bigcup_n F_n $ (because $ \Gamma = \bigcup_n T_n $); the problem with $ F_n $ is that we don’t have $ F_{n-1} \subset F_n $.

Let $ F_n’$ be defined as follows: a class $ [F_n]x $ contains exactly one « large » class for $ F_n’ $ consisting of all $ yF_nx $ such that $ [F_n’]y \subset [F_n]x $, and for all other points we take their $ F_{n-1}’ $-class. Clearly $ F_{n-1}’ \subset F_n’ $, and also $ T_n’a \subset [F_n’]a $ for all $ a \in A_n $. It follows from the latter that $ E = \bigcup_n F_n’ $.

Note that it is not known whether all amenable groups admit such a sequence of tiles; however, this is the case for all elementary amenable groups (Ornstein–Weiss) and for all residually finite amenable groups (Weiss). In general one has to use almost-tilings, for which the Rokhlin lemma also holds.

Proof of the Rokhlin lemma

The first step is to establish the following lemma.

For any $ \varepsilon > 0 $ and $ n \ge 1 $ there is a finite subset $ H \subset \Gamma $ which is $ T \cup T^{-1}, \varepsilon^2 $-invariant an a collection of measurable subsets $ U_1, \ldots, U_m \subset X $ such that :

for all $ i = 1, \ldots, n $ and $ u \not= u’ \in U_i $ the sets $ Hu, Hu’ $ are disjoint;

$ \mu\left( \bigcup_i U_i \right) \ge 1 – \varepsilon $;

$ \sum_i \mu(U_i) \le \varepsilon^{-1} $.

To prove this Ornstein and Weiss use the following two statements:

There is a cover of $ X $ by pairwise disjoint sets $ U_i, i \in \mathbb N $ satisfying 1) ;
Given $ \varepsilon > 0 $ and a cover $ X = \bigcup_{j \ge 1} V_j $ such that every $ x \in X $ belongs to exactly $ N $ of the $ V_j $ there exists a finite subsequence $ V_{j_i}, i=1, \ldots, N $ which satisfies 2) and 3).

The point a) is easily proven once it is known that every subset of positive measure in $ X $ contains a set satisfying 1) (choose a first set $ U_1 $ with positive measure, and then do the same in $ X \setminus U_1 $,…) which is more or less obvious.

Point b) is a general fact in combinatorics, proven by Ornstein–Weiss using the following lemma : let $ X = \bigcup_i V_i $ be a cover by measurable such that every point belongs to exactly N of the $ V_i $s. Then for any $ V \subset X $ there exists $ V_i $ such that :
\[
(\ast\ast) \qquad \frac{\mu(V_i \cap V)}{\mu(V_i)} \le \mu(V).
\]
Now suppose that the $ U_i $ are ordered by measure (so that $ \mu(U_{i+1}) \le \mu(U_i) $) and use $ (\ast\ast) $ to construct a maximal sequence $ i_1 = 1, i_2, \ldots $ such that for all $ k \le n-1 $ ($ n \in \mathbb N \cup\{\infty\} $ the length of the sequence) $ \mu(\bigcup_{1\le l \le k} U_{i_l}) < 1 – \varepsilon $, $ \mu(U_{i_k} \cap \bigcup_{1\le l \le k-1} U_{i_l}) \le \mu(U_{i_k})\mu(\bigcup_{1 \le l \le k-1} U_{i_l}) $ and $ i_k $ is the smallest index with these properties. If $ n < +\infty $ this finishes the proof. Otherwise, if we had $ \mu(\bigcup_{l \ge 1} U_{i_l}) < 1 – \varepsilon $, by $ (\ast\ast) $ there would be an index $ j \in \mathbb N $ such that $ \mu(U_j \cap \bigcup_{l\ge 1} U_{i_l}) \le \mu(U_j)\mu(\bigcup_{1 \le l} U_{i_l}) $ and this contardicts the choice of the $ U_{i_k} $.

In addition, the fact that $ \mu(U_{i_k} \cap \bigcup_{1\le l \le k-1} U_{i_l}) \le (1 – \varepsilon) \mu(U_{i_k}) $ easily implies that $ \sum_{l \ge 1} \mu(U_{i_l}) \le \varepsilon^{-1} $.

For the second step let $ C \subset \Gamma $ such that $ | TC \Delta H | \le \varepsilon^2 | H | $ which exists because $ T $ tiles and H is $ TT^{-1}, \varepsilon^2 $-invariant. Let $ W_i = \bigcup_{j=1}^i HU_j $ and construct inductively sets $ W_i’ $ such that $ W_i’ $ contains all $ Tcu, u \in U_i, c \in C $ such that $ Tcu \cap U_{i+1} = \emptyset $. We get
\[
\mu(W_i \setminus W_i’) \le \varepsilon^2 \sum_{j=1}^i \mu(U_j)
\]
so that in the end $ \mu(W_m’) \ge 1 – \varepsilon – \varepsilon^2 \sum_{j=1}^i \mu(U_j) \ge 1 – 2\varepsilon $. The other conclusion of the Rokhlin lemma clearly holds for the set
\[
A = \bigcup_{i=1}^m \{ cu : u \in U_i, c \in C : (Tcu) \subset X \setminus \bigcup_{j \ge i+1} HU_j
\]
which satisfies $ TA = W_m’ $, and this finishes the proof.

Références

Gilbert Levitt, On the cost of generating an equivalence relation, Ergodic Theory and Dynamical systems.
Damien Gaboriau, Coût des relations d’équivalence et des groupes, Inventiones Mathematicae.
Donald Ornstein, Benjamin Weiss, Entropy and isomorphism theorems for actions of amenable groups, Journal d’analyse mathématique.
Alex Furman, A survey of measured group theory, in Geometry, Rigidity, and Group Actions, U. Chicago press.

Notes on the Abért–Nikolov theorem on rank gradient and cost (notes by Holger Kammeyer after his own lecture)

jraimbau — Mon, 28 Nov 2016 10:18:49 +0000

These are notes of a talk on a theorem of Abért–Nikolov: Let $\Gamma$ be a finitely generated group and let $(\Gamma_n)$ be a chain of finite index subgroups. Assume that the action of $\Gamma$ on the boundary $\partial T$ of the coset tree of $(\Gamma_n)$ is essentially free. Then the rank gradient of $\Gamma$ with respect to the chain $(\Gamma_n)$ equals the cost of the action of $\Gamma$ on $\partial T$.

The coset tree

Let $\Gamma$ be a countable, discrete group and let $\Gamma = \Gamma_0 \ge \Gamma_1 \ge \Gamma_2 \ge \cdots$ be a chain of finite index subgroups. For the moment we do not make any further assumptions on the chain $(\Gamma_n)$. So the subgroups $\Gamma_n$ must neither be normal, nor are they required to have trivial total intersection.

Definition: The (right) coset tree $T$ of $(\Gamma, (\Gamma_i))$ has vertex set $\coprod_{n \ge 0} \Gamma_n \backslash \Gamma$ and an edge from $\Gamma_ng$ to $\Gamma_{m}h$ if and only if $m = n+1$ and $\Gamma_mh \subset \Gamma_ng$.

Not that $T$ has the canonical root $\Gamma_0$, so that the $n$-th level vertex set is just $\Gamma_n \backslash \Gamma$, and each node in $\Gamma_n \backslash \Gamma$ has precisely $[\Gamma_n \colon \Gamma_{n+1}]$ children.

A typical coset tree.

Definition: The boundary $\partial T$ of the coset tree $T$ consits of all infinite rays starting at the root $\Gamma_0$.

Thus in mathematical terms we have $\partial T = \underset{\leftarrow}{\lim} \Gamma_n \backslash \Gamma$ and this description makes sense not only in the category of sets but also in the category of topological spaces and of measure spaces. Each $\Gamma_n \backslash \Gamma$ carries the discrete topology and the uniform probability measure. Thus as a space, $\partial T$ is compact, totally disconneted and Hausdorff. It has a basis of the topology given by shadows where a shadow $\mathrm{sh}(\Gamma_n g)$ consists of all rays going through the vertex $\Gamma_n g$. The Borel probability measure $\mu$ on $\partial T$ is determined by the values $\mu (\mathrm{sh}(\Gamma_n g)) = \frac{1}{[\Gamma : \Gamma_n]}$. The group $\Gamma$ permutes the cosets in $\Gamma_n \backslash \Gamma$ by right multiplication preserving the child–parent relation. Thus $\Gamma$ acts on $T$ from the right by tree automorphisms and we obtain an induced probability measure preserving right action of $\Gamma$ on $\partial T$ by homeomorphisms.

Lemma: The action of $ \Gamma$ on $\partial T $ is ergodic.

Proof: If the chain stabilizes, the boundary is finite and the action is transitive so that the assertion is clear. Otherwise, Kuratowski’s theorem says that $\partial T$ is Borel-isomorphic to the unit interval with Lebesgue measure so that Lebesgue’s theorem applies: any measurable $A \subseteq \partial T$ is almost everywhere dense. In particular, if $\mu(A) > 0$, then for all $\varepsilon > 0$ there is some $\Gamma_n g_0$ with
\[ \mu (\mathrm{sh}(\Gamma_n g_0) \cap A) > (1 – \varepsilon) \mathrm{sh}(\Gamma_n g_0). \]
If $A$ is moreover $\Gamma$-invariant, then the same must hold for all $\Gamma_n g$. Adding up all these inequalities gives
\[ \mu(A) = \sum_{\Gamma_n g \,\in\, \Gamma_n \backslash \Gamma} \mu(\mathrm{sh}(\Gamma_n g) \cap A) > 1 – \varepsilon, \]
which implies $\mu(A) = 1$.

Lemma : Suppose each $\Gamma_n$ is normal in $\Gamma$ and that the total intersection $\bigcap_{n \ge 0} \Gamma_n$ is trivial. Then the action of $ \Gamma$ on $\partial T $ is free.

Proof: Given any nontrivial $g \in \Gamma$ there is $n > 0$ such that $g \notin \Gamma_n$. Since $\Gamma_n$ is a normal subgroup of $\Gamma$, the element $g$ permutes the set $\Gamma_n \backslash \Gamma$ without fixed points. Thus $g$ moves all rays in $\partial T$.

In view of this lemma the following concept is a natural generalization of the common assumptions on a chain $(\Gamma_n)$ given in the Lemma.

Definition: A chain $(\Gamma_n)$ of subgroups of $\Gamma$ is called Farber if the action of $ \Gamma$ on $\partial T $ is essentially free.

Here, as usual, « essentially free » means the set of points in $\partial T$ with nontrivial stabilizer has measure zero.

Cost and groupoid cost

Cost

Let $(X, \mu)$ be a standard Borel space and let $ \Gamma $ act on $ X $ preserving a probability measure, with finitely many ergodic components. « Lying in the same orbit » defines an equivalence relation $E$ on $X$. So $E$ is a measurable subset of $X \times X$ which we can picture as (the edge set of) a directed graph. Reflexivity, symmetry and transitivity say that the connected components of $E$ are complete as directed gaphs: each vertex carries a loop and any two distinct vertices in the same component are joined by precisely two edges, one in each direction. Since $\Gamma$ is countable, so are the complete directed graphs given by each component.

Apparently, we can reconstruct $E$ if we only know which pairs of points in $X$ are connected by some finite path along edges in $E$, regardless of their direction. This information, in turn, is captured by any measurable subset $S \subset E$ with the property that two points $x,y \in X$ are joined by an edge in $E$ if and only if there is a path between $x$ and $y$ along edges in $S$.

Definition:

A measurable subset $S \subseteq E$ is called a subgraph of $E$.

The $k$-th power $S^k$ of a subgraph $S \subseteq E$ is defined by requiring $(x,y) \in S^k$ if and only if there is an undirected path from $x$ to $y$ in $S$ of length $l$ with $0 \le l \le k$.

We say that a subgraph $S \subseteq E$ spans $E$ if $E = \bigcup_{k \ge 0} S^k$. In this case we write $E = \langle S \rangle$.

Note that $S^0$ is the set of loop edges in $X$, regardless of what $S$ is: each point $x \in X$ is joined to itself by the empty path of length zero. The rules of the game are to find spanning subgraphs $S$ of $E$ which are as « small » as possible. It makes sense to quantify « small » by the average number of edges starting in a vertex of $S$. The next definition makes this idea precise.

Definition: The edge measure $e(S)$ of a subgraph $S \subseteq E$ is given by
\[ e(S) = \int_X \deg_S(x) \,\mathrm{d} \mu(x) \quad \text{where} \quad \deg_S(x) = |\{y \in X \colon (x,y) \in S \}|. \]

Note that $e(S)$ can be infinite. The largest lower bound of the edge measures of spanning subgraphs of $E$ can be thought of measuring how much effort we must invest to generate the graph and thus the equivalence relation $E$ which the action of $ \Gamma $ on $ X $ defines.

Definition: The cost of the action of $ \Gamma $ on $ X $ is given by
\[ \mathrm{cost}(X, \Gamma) = \mathrm{cost} E = \inf_{\langle S \rangle = E} e(S). \]

As an example, say $\Gamma$ is finitely generated by $g_1, \ldots, g_n$. Then
\[ S = \bigcup_{i=1}^n \{ (x, xg_i) \colon x \in X \} \]
clearly defines a spanning subgraph of $E$. So we always have
\[ \mathrm{cost}(X, \Gamma) \le d(\Gamma) \]
where $d(\Gamma)$ is the minimal number of generators of $\Gamma$, also known as the rank of $\Gamma$. Note that if the action of $ \Gamma $ on $ X $ is essentially free, almost every connected component of this subgraph $S$ looks like the Cayley graph of $\Gamma$ with respect to the generators $g_1, \ldots, g_n$ but with no distinguished base point whatsoever.

The idea comes to mind that one can improve upon a spanning subgraph $S$ as above by picking a maximal tree in each component. The problem is that depending on $\Gamma$ there might be no measurable such choice which already hints at cost being quite a delicate invariant of the action. Let us mention that the cost of an action immediately yields an invariant for the group $\Gamma$.

Definition: We define $\mathrm{cost} \Gamma$ as the infimum of the numbers $\mathrm{cost}(X, \Gamma)$ running over all essentially free, ergodic, probability measure preserving actions of $\Gamma$ on a standard Borel space $(X, \mu)$.

Groupoid cost

In addition to examining (sub-)graphs, meaning measurable subsets of $X \times X$, it is useful to also study graphings by which we mean any measurable subset $M \subseteq X \times \Gamma$. We find it convenient to picture an element $(x,g) \in X \times \Gamma$ as an « arrow » in $X$ pointing from $x$ to $xg$. Note that (almost all) these arrows are determined by their initial and final point if and only if the action of $\Gamma$ on $ X $ is (essentially) free. We can sort the arrows in the subset $M$ either by initial point or by direction: either by the $X$- or by the $\Gamma$-coordinate. So interchangeably we think of $M$ as a family of subsets
\[ M_g = \{ x \in X \colon (x, g) \in M \} \subseteq X\]
parametrized by group elements $g \in \Gamma$, or as a family of subsets
\[ M_x = \{ g \in \Gamma \colon (x, g) \in M \} \subseteq \Gamma\]
parametrized by points $x \in X$. Guided by what we did above, we define the $k$-th power $M^k$ of $M$ by all the arrows we obtain by composing up to $k$ arrows from $M$ regardless of their direction. In more mathematical terms this means $(x,g) \in M^k$ if and only if there is $0 \le l \le k$ and a decomposition $g = g_1 \cdots g_l$ in $\Gamma$ such that for all $0 \le i \le l-1 $ either
\[ (x g_1 \cdots g_i, g_{i+1}) \in M \quad \text{or} \quad (x g_1 \cdots g_{i+1}, g_{i+1}^{-1}) \in M. \]
Note that $M^0 = X \times \{ 1 \}$, regardless of what $M$ is.

Definition: We say that a graphing $M \subset X \times \Gamma$ spans $X \times \Gamma$ if we have $X \times \Gamma = \bigcup_{k \ge 0} M^k$. In this case we write $\langle M \rangle = X \times \Gamma$.

Let $e$ be the measure on $X \times \Gamma$ given by the product of $\mu$ and the counting measure on $\Gamma$.

Definition: The groupoid cost of the action of $\Gamma$ on $ X $ is given by
\[ \mathrm{gcost}(X, \Gamma) = \inf_{\langle M \rangle = X \times \Gamma} e(M). \]

To explain the terminology, we observe that the set $X \times \Gamma$ has a groupoid structure: two arrows $(x_1, g_1), (x_2, g_2) \in X \times \Gamma$ can be composed if and only if $x_1 g_1 = x_2$, meaning the first arrow points to the initial point of the second, and in that case their composition is $(x_1, g_1 g_2)$. So a graphing spans $X \times \Gamma$ if and only if it generates $X \times \Gamma$ as groupoid.

Proposition: We have
\[ \mathrm{gcost}(X, \Gamma) \ge \mathrm{cost}(X, \Gamma) \]
with equality if the action is essentially free.

Proof: A graphing $M \subseteq X \times \Gamma$ defines a subgraph $\Phi(M) \subseteq E$ of the orbit relation $E \subseteq X \times X$ by setting
\[ \Phi(M) = \{ (x, xg) \colon (x,g) \in M \}. \]
Clearly $\Phi(M^k) = \Phi(M)^k$ so that $\Phi$ preserves the spanning property. We have $\deg_{\Phi(M)} (x) \le |M^x|$ with equality almost everywhere if the action of $\Gamma$ on $ X $ is essentially free. Integrating over $X$ gives the inequality. To obtain equality for essentially free actions one still has to show that each spanning subgraph can be obtained from a spanning graphing via $\Phi$; we skip this argument which needs some technical care but no unusual ideas.

Rank gradient and cost of the coset tree boundary

Let us now assume that $\Gamma$ is finitely generated so that there is an epimorphism $\varphi \colon F_k \rightarrow \Gamma$ from the free group on a finite number $k$ of letters. If $\Lambda \subseteq \Gamma$ is a subgroup of finite index, then by the Schreier theorem $\varphi^{-1}(\Lambda)$ is likewise free. It has index $[\Gamma \colon \Lambda]$ in $F_k$ so that an Euler characteristic argument reveals that the rank $l$ of $\varphi^{-1}(\Lambda)$ is given by $l-1 = [\Gamma \colon \Lambda](k-1)$. For the ranks $d(\Gamma_n)$ of the groups in our chain $(\Gamma_n)$ this means that the numbers
\[ \frac{d(\Gamma_n) -1}{[\Gamma \colon \Gamma_n]} \]
form a monotone non-increasing sequence of positive numbers.

Definition: The rank gradient of $\Gamma$ with respect to $(\Gamma_n)$ is given by
\[ \mathrm{rg}(\Gamma, (\Gamma_n)) = \lim_{n \rightarrow \infty} \frac{d(\Gamma_n) – 1}{[\Gamma \colon \Gamma_n]}. \]

This notion is due to Lackenby and here is its dynamic interpretation.

Theorem (Abért–Nikolov, 2012): For any chain $(\Gamma_n)$ we have
\[ \mathrm{gcost}(\partial T, \Gamma) = \mathrm{rg}(\Gamma, (\Gamma_n)) + 1. \]

Before we come to the proof let us combine this theorem with the proposition.

Corollary: For any Farber chain $(\Gamma_n)$ we have
\[ \mathrm{cost}(\partial T, \Gamma) = \mathrm{rg}(\Gamma, (\Gamma_n)) + 1. \]

Proof of Theorem: We first show $\mathrm{gcost}(\partial T, \Gamma) \le \mathrm{rg}(\Gamma, (\Gamma_n)) + 1$. For all $\varepsilon > 0$ we find some $n$ with $\frac{d(\Gamma_n) -1}{[\Gamma \colon \Gamma_n]} \le \mathrm{rg}(\Gamma, (\Gamma_n)) + \varepsilon$. Thus the integer
\[ d = \lfloor(\mathrm{rg}(\Gamma, (\Gamma_n)) + \varepsilon)[\Gamma \colon \Gamma_n]\rfloor + 1 \]
gives an upper bound for $d(\Gamma_n)$. Say $\Gamma_n$ is generated by $g_1, \ldots, g_d$ and let $1 = \gamma_1, \ldots, \gamma_{[\Gamma \colon \Gamma_n]}$ be a system of representatives for $\Gamma_n \backslash \Gamma$. We define a graphing $M \subseteq \partial T \times \Gamma$ by setting
\[
M_g = \mathrm{sh}(\Gamma_n) \text{if } g = g_i \text{ for some } i \ge 1 \text{ or } g = \gamma_i \text{ for some } i > 1
\]
and $ M_g = \emptyset $ otherwise. We claim that $\langle M \rangle = \partial T \times \Gamma$. Indeed, let $(x, g) \in \partial T \times \Gamma$ and let $\gamma_a$ and $\gamma_b$ be the representatives from the list for which $x \in \mathrm{sh}(\Gamma_n \gamma_a)$ and $\gamma_a g \gamma_b^{-1} \in \Gamma_n$. Hence we can write $g$ as a word of the form $\gamma_a^{-1} g^{\pm 1}_{i_1} \cdots g^{\pm 1}_{i_k} \gamma_b$. With respect to this factorization of $g$ one easily verifies the criterion to conclude $(x,g) \in M^{k+2}$ proving the claim. By definition $M_g$ equals $\mathrm{sh}(\Gamma_n)$ for precisely $d + [\Gamma \colon \Gamma_n] – 1$ elements in $\Gamma$ and is empty otherwise. Hence the graphing $M$ has measure
\[ e(M) = \frac{d + [\Gamma \colon \Gamma_n] – 1}{[\Gamma \colon \Gamma_n]}. \]
It follows that
\[ \mathrm{gcost}(\partial T, \Gamma) \le \frac{d-1}{[\Gamma \colon \Gamma_n]} + 1 \le \mathrm{rg}(\Gamma, (\Gamma_n)) + 1 + \varepsilon. \]

The reverse inequality $\mathrm{gcost}(\partial T, \Gamma) \ge \mathrm{rg}(\Gamma, (\Gamma_n)) + 1$ is somewhat harder. Given $\varepsilon 0$, there exists a graphing $M$ which spans $\partial T \times \Gamma$ and has measure $e(M) \le \mathrm{gcost}(\partial T, \Gamma) + \frac{\varepsilon}{2}$. The first thing to do now is to construct yet another graphing $N \subseteq \partial T \times \Gamma$ which is close to $M$ in the sense that $e(N \triangle M) \le \frac{\varepsilon}{2}$ and has the convenient property that each $N_g$ is a finite union of shadows which is nonempty only for finitely many $g \in \Gamma$. Since the shadows form a countable basis of the topology of $\partial T$, it is conceivable that such a « finite approximation » to $M$ exists. So we shall allow ourselves to skip the precise technical construction. Since $N$ is made up from only finitely many shadows altogether, there exists a large enough $n$ such that each $N_g$ is in fact a finite union of level-$n$ shadows of the form $\mathrm{sh}(\Gamma_n h)$.

We define a finite, directed, labeled graph $G$ as follows. The vertex set is $V = \Gamma_n \backslash \Gamma$ and for each $g \in \Gamma$ we connect $w \in V$ with $wg \in V$ by an edge of label $g$ if and only if $\mathrm{sh}(w) \subseteq N_g$. The graph $G$ has the canonical base point $v = \Gamma_n \in V$. This data clearly defines a homomorphism of groups
\[
\begin{array}{cc}
\varphi \colon \pi_1(G,v) & \longrightarrow \Gamma \\
l = (e_1, \ldots, e_k) & \longmapsto \mathrm{label}(e_1)^{\pm 1} \cdots \mathrm{label}(e_k)^{\pm 1}
\end{array}
\]
which multiplies the labels along a loop of edges, inverting the label whenever we travel through an edge in reverse direction. We claim that the image of the homomorphism $\varphi$ is precisely $\Gamma_n$. Indeed, for each $l \in \pi_1(G,v)$ we have $v \varphi(l) = v$ by the construction of the graph $G$. Thus $\varphi(l) \in \mathrm{Stab}(v) = \Gamma_n$. Let $h \in \Gamma_n$ be any element and pick some ray $x \in \mathrm{sh}(v)$. Since $N$ spans $\partial T \times \Gamma$, there is a factorization $h = g_1 \cdots g_k$ with $g_i \in \Gamma$ such that for all $0 \le i \le k-1$ either
\[ (x g_1 \cdots g_i, g_{i+1}) \in N \quad \text{or} \quad (x g_1 \cdots g_{i+1}, g_{i+1}^{-1}) \in N. \]
For $0 \le i \le k$ let $w_i \in V = \Gamma_n \backslash \Gamma$ be the level-$n$ vertex in the coset tree $T$ through which the ray $x \gamma_1 \cdots \gamma_i$ passes. Then we have $w_0 = w_k = v$ and for each $0 \le i \le k-1$ either $w_i$ is connected to $w_{i+1}$ by an edge in $G$ with label $g_{i+1}$ or $w_{i+1}$ is connected to $w_i$ by an edge of label $g_{i+1}^{-1}$. Hence these edges form a loop $l$ with $\varphi(l) = h$ proving the claim.

Thus $\Gamma_n$ is a quotient group of $\pi_1(G,v)$. The latter group is free of rank $1 – \chi(G)$ where $\chi(G)$ is the Euler characterisitc of the graph $G$. Note that
\[ \textstyle e(N) = \sum_{g \in \Gamma} \ \sum_{\mathrm{sh}(\Gamma_n h) \subseteq N_g} \frac{1}{[\Gamma \colon \Gamma_n]} \]
so that $e(N) [\Gamma \colon \Gamma_n]$ is the number of edges in $G$ while the number of vertices in $G$ is of course $[\Gamma \colon \Gamma_n]$. Putting pieces together we obtain
\[ d(\Gamma_n) \le d(\pi_1(G,v)) = 1 + e(N)[\Gamma \colon \Gamma_n] -[\Gamma \colon \Gamma_n]. \]
By subadditivity of the measure $e$ applied to $N \subseteq M \cup (N \triangle M)$ we conclude
\[ \textstyle \mathrm{rg}(\Gamma, (\Gamma_i)) \le \frac{d(\Gamma_n) – 1}{[\Gamma \colon \Gamma_n]} = e(N) – 1 < \mathrm{gcost}(\partial T, \Gamma) + \varepsilon – 1. \]

References

Miklós Abért, Nikolay Nikolov, Rank gradient, cost of groups and the rank versus Heegaard genus, J. Eur. Math. Soc.