2.1. Translation between percolation and equivalence relations

Some of our results are stated in the language of percolation theory but all our proofs will be based on measured equivalence relations and graphings. An invariant (edge) percolation on a unimodular random graph $(\mathcal G,o)$ is a random triple $(\mathcal G,o,P)$ where P is a subset of edges of G and the distribution of the triple is invariant under the rerooting equivalence relation. The following proposition associates a probability-measure-preserving measured equivalence relation to a percolation is such a way that Theorem 1.1 can deduced from Theorem 1.2.

3.1. Proof of Theorem 3.2(i)

The essential idea behind Theorem 3.2(i) is that we have a natural vector in $L^{2}({\mathcal R}/{\mathcal S})$ which is given by the measurable field $\xi _{x}=\delta _{[x]_{{\mathcal S}}}.$ By a direct calculation, we have that

We then have to show that

Because $\nu $ is symmetric, the operator $\unicode{x3bb} _{{\mathcal S}}(\nu )$ is self-adjoint and the existence of the limit on the right-hand side follows from the spectral theorem [Reference Conway12, Theorems IX.2.2 and IX.2.3]. The limit on the right-hand side is also dominated by $\|\unicode{x3bb} _{{\mathcal S}}(\nu )\|$ . We prove a general Hilbert space theorem which, after a small amount of work, will prove the reverse inequality. First, let us alleviate any concerns about the measurable structure defined on $(\ell ^{2}([x]_{{\mathcal R}}/{\mathcal S}))_{x\in X}$ .

Proof. Fix a countable subgroup $\Gamma \leq [{\mathcal R}]$ so that $\Gamma x=[x]_{{\mathcal R}}$ for almost every x. We start by proving the following claim.

Claim. A vector $\xi =(\xi _{x})_{x}\in \prod _{x\in X}\ell ^{2}([x]_{{\mathcal R}}/{\mathcal S})$ is in ${\operatorname {Meas}}(\ell ^{2}([x]_{{\mathcal R}}/{\mathcal S}))$ if and only if the map $x\mapsto \xi _{x}([\phi (x)]_{{\mathcal S}})$ is measurable for every $\phi \in \Gamma $ . If $\xi $ is measurable then, by definition, $x\mapsto \xi _{x}([\phi (x)]_{{\mathcal S}})$ is measurable for every $\phi \in [{\mathcal R}]$ . In particular, this is true for $\phi \in \Gamma $ . Conversely, suppose that $x\mapsto \xi _{x}([\phi (x)]_{{\mathcal S}})$ is measurable for all $\phi \in \Gamma .$ Fix a $\psi \in [{\mathcal R}]$ . We then have to show that $x\mapsto \xi _{x}([\psi (x)]_{{\mathcal S}})$ is measurable. Since $\Gamma x=[x]_{{\mathcal R}}$ for almost every $x\in X$ , we may find a disjoint family of sets $(E_{\phi })_{\phi \in \ \Gamma }$ with $E_{\phi }\subseteq \{x\in X:\psi (x)=\phi (x)\}$ and so that $\bigsqcup E_{\phi }$ is a co-null subset of X. Then

$$ \begin{align*}\xi_{x}([\psi(x)]_{{\mathcal S}})=\sum_{\phi \in\Gamma}1_{E_{\phi}}(x)\xi_{x}([\phi(x)]_{{\mathcal S}})\end{align*} $$

for almost every x (the sum above converges since the $E_{\phi }$ are disjoint). Since $\Gamma $ is countable, this proves that $x\mapsto \xi _{x}([\psi (x)]_{{\mathcal S}})$ is measurable, and this proves the claim.

Having shown the claim, for $\phi \in \Gamma $ define $\zeta _{\phi }\in \prod _{x\in X}\ell ^{2}([x]_{{\mathcal R}}/{\mathcal S})$ by $\zeta _{\phi ,x}=\delta _{[\phi (x)]_{{\mathcal S}}}$ . Then

$$ \begin{align*}\ell^{2}([x]_{{\mathcal R}}/{\mathcal S})=\overline{\operatorname{span}\{\zeta_{\phi,x}:\phi \in\Gamma\}}^{\|\cdot\|_{2}}\end{align*} $$

for almost every $x\in X$ , and also $\xi _{x}([\phi (x)]_{{\mathcal S}})=\langle \xi _{x},\delta _{\phi (x)} \rangle $ for all $\phi \in [{\mathcal R}]$ . Moreover, for $\phi ,\psi \in \Gamma $ we have that

$$ \begin{align*}\langle \zeta_{\phi,x},\zeta_{\psi,x} \rangle=1_{S}(\phi(x),\psi(x)),\end{align*} $$

which is a measurable function of x. The lemma now follows from countability of $\Gamma $ and [Reference Takesaki37, Lemma IV.8.10].

We use the following well-known lemma (see, for example, [Reference Hutchcroft24, Equation 2.8] for a proof), which is the main way we will relate the operator norm of the Markov operator $\unicode{x3bb} _{{\mathcal S}}(\nu )$ to the growth of the matrix coefficients $\langle \unicode{x3bb} _{{\mathcal S}}(\nu )^{2n}\xi ,\xi \rangle $ .

Lemma 3.5. Let $\mathcal {H}$ be a Hilbert space and $T\in B(\mathcal {H})$ self-adjoint. Let K be an index set, and let $(\xi _{k})_{k\in K}$ be a K-tuple of vectors in $\mathcal {H}$ so that $\mathcal {H}=\overline {\operatorname {span}\{\xi _{k}:k\in K\}}.$ Then

$$ \begin{align*}\|T\|=\sup_{k\in K}\lim_{n\to\infty}\langle T^{2n}\xi_{k},\xi_{k} \rangle^{1/2n}.\end{align*} $$

In order to apply this in the context of a direct integral of Hilbert spaces, the following density criterion will be useful.

Lemma 3.6. Let $(X,\mu )$ be a standard probability space. Let $(\mathcal {H}_{x})_{x\in X}$ be a measurable family of Hilbert spaces over X and set $\mathcal {H}=\int _{X}^{\bigoplus }\mathcal {H}_{x}.$ Suppose we have a sequence $\xi _{n}=(\xi _{n,x})_{x\in X}\in {\operatorname {Meas}}(\mathcal {H}_{x})$ such that

$$ \begin{align*}\mathcal{H}_{x}=\overline{\operatorname{span}\{\xi_{n,x}:n\in {\mathbb N}\}}\end{align*} $$

for almost every $x\in X$ . Then

$$ \begin{align*}\mathcal{H}=\overline{\operatorname{span}\{1_{A}\xi_{n}:n\in {\mathbb N}, A\subseteq X\text{ is measurable}\}}.\end{align*} $$

Proof. Suppose that $\eta \in {\mathcal H}$ and that $\langle \eta ,1_{A}\xi _{n} \rangle =0$ for every $n\in {\mathbb N}$ and every measurable $A\subseteq X$ . Then for every measurable $A\subseteq X$ and every $n\in {\mathbb N}$ we have

$$ \begin{align*}\int_{A}\langle \eta_{x},\xi_{n,x} \rangle\,d\mu(x)=0.\end{align*} $$

Since this holds for every A, applying this with A being $\{x\in X:\operatorname {Re}(i^{j}\langle \eta _{x},\xi _{n,x} \rangle )>0\}$ for $j=0,1,2,3$ , and taking real and imaginary parts of the above integral shows that that for every $n\in {\mathbb N}$ we have that $\langle \eta _{x},\xi _{n,x} \rangle =0$ for almost every $x\in X$ . By countability, for almost every $x\in X$ we have $\langle \eta _{x},\xi _{n,x} \rangle =0$ for all $n\in {\mathbb N}$ . Since $\mathcal {H}_{x}=\overline {\operatorname {span}\{\xi _{n,x}:n\in {\mathbb N}\}}$ for almost every $x\in X$ , we deduce that $\eta _{x}=0$ for almost every $x\in X$ , that is $\eta =0$ as an element of ${\mathcal H}$ . Thus, we have shown that the only vector in ${\mathcal H}$ orthogonal to $\{1_{A}\xi _{n}: n\in {\mathbb N}, A\subseteq X\text { is measurable}\}$ is the zero vector, and this implies that

$$ \begin{align*}\mathcal{H}=\overline{\operatorname{span}\{1_{A}\xi_{n}:n\in {\mathbb N}, A\subseteq X\text{ is measurable}\}}.\\[-37pt]\end{align*} $$

Proof of Theorem 3.2(i)

Let $\xi \in L^{2}({\mathcal R}/{\mathcal S})$ be the measurable vector field given by ${\xi _{x}=\delta _{[x]_{{\mathcal S}}}.}$ By direct calculation,

By the spectral theorem, there is a probability measure $\eta $ on $[-\|\unicode{x3bb} _{S}(\nu )\|,\|\unicode{x3bb} _{S}(\nu )\|]$ so that

From this, we see that $\lim _{n\to \infty }\langle \unicode{x3bb} _{S}(\nu )^{2n}\xi ,\xi \rangle ^{{1}/{2n}}$ exists and is the $L^{\infty }$ -norm of t with respect to $\eta $ . Combining these results, we see that

$$ \begin{align*}\lim_{n\to\infty}\bigg(\int p_{2n,x,{\mathcal S}}\,d\mu(x)\bigg)^{1/2n}\end{align*} $$

exists. Call this limit $\rho ({\mathcal R}/{\mathcal S},\nu )$ as in the statement of the theorem.

We now turn to the proof that $\rho ({\mathcal R}/{\mathcal S},\nu )=\|\unicode{x3bb} _{{\mathcal S}}(\nu )\|$ . It follows from the logic in the preceding paragraph that $\rho ({\mathcal R}/{\mathcal S},\nu )\leq \|\unicode{x3bb} _{{\mathcal S}}(\nu )\|$ . Let $\Gamma $ be the subgroup of $[{\mathcal R}]$ generated by the support of $\nu $ . Since $\nu $ generates ${\mathcal R}$ , we have $[x]_{{\mathcal R}}=\Gamma x$ for almost every $x\in X$ . By Lemma 3.6,

has dense linear span in ${\mathcal H}$ . It thus suffices, by Lemma 3.5, to prove that

for every measurable $A\subseteq X$ and $\phi \in \Gamma $ . It is direct to see that

for every measurable $A\subseteq X$ . So it simply suffices to show that

(3.1)

To prove (3.1), fix $\phi \in \Gamma .$ Let $\phi ^{*}({\mathcal S})$ be the subrelation $\phi ^{*}({\mathcal S})=\{(\phi (x),\phi (y)):(x,y)\in ~{\mathcal S}\}$ . By direct computation,

Choose $k\in {\mathbb N}$ so that $c=\nu ^{*k}(\{\phi \})>0$ . Then for every $n\in {\mathbb N}$ we have that

$$ \begin{align*}p_{2(n+k),x,{\mathcal S}}&=\sum_{y\in [x]_{{\mathcal S}}}p_{2(n+k),x,y}\geq \sum_{y\in [x]_{{\mathcal S}}}p_{k,x,\phi(x)}p_{2n,\phi(x),\phi(y)}p_{k,\phi(y),y}\\ &\geq c^{2}\sum_{y\in [x]_{{\mathcal S}}}p_{2n,\phi(x),\phi(y)}=c^{2}p_{2n,\phi(x),\phi^{*}({\mathcal S})}. \end{align*} $$

Integrating both sides, we obtain

Thus,

This proves (3.1), and so completes the proof of Theorem 3.2(i).

As mentioned in the introduction, Theorem 3.2 and our later work in §4 can be used to recover a result due to Schramm used in [Reference Kozma31], [Reference Hutchcroft21, Lemma 6.4], and [Reference Hutchcroft22, §3]. Indeed, given a percolation of a connected, regular, transitive graph G by §2.1, one can build an inclusion ${\mathcal S}\leq {\mathcal R}$ of relations on a standard probability space as well as a symmetric probability measure $\nu $ on $[{\mathcal R}]$ so that, in the notation of [Reference Hutchcroft21, Lemma 6.4], we have

$$ \begin{align*}\int p_{n,x,{\mathcal S}}\,d\mu(x)={\mathbb E}[\tau_{p}(X_{0},X_{n})].\end{align*} $$

For $p\in (0,1)$ , consider Bernoulli(p) edge percolation, where each edge is kept with probability p; let ${\mathcal S}_{p}\leq {\mathcal R}_{p}$ be the corresponding inclusion of equivalence relations. Set

$$ \begin{align*}p_{c}=\inf\{p: \text{almost every connected component in Bernoulli }(p) \text{ percolation is finite}\}.\end{align*} $$

In the setup of the proof of Theorem 3.2(i), we have that

Since $\unicode{x3bb} _{{\mathcal S}}(\nu )$ is self-adjoint, the spectral theorem tells us that

is increasing in n and Theorem 3.2(i) characterizes its supremum as $\|\unicode{x3bb} _{{\mathcal S}}(\nu )\|$ . Additionally, for every $n\in {\mathbb N}$ we have by Cauchy–Schwarz that

where in the last step we use self-adjointness of $\unicode{x3bb} _{{\mathcal S}}(\nu )$ . The construction of the inclusion ${\mathcal S}\leq {\mathcal R}$ forces that the spectral radius of G is equal to the co-spectral radius of ${\mathcal T}\leq {\mathcal R}$ where ${\mathcal T}=\{(x,x):x\in X\}$ is the trivial relation. For Bernoulli(p) percolation with $p<p_{c}$ , finiteness of the clusters tells us that $[x]_{{\mathcal S}_{p}}$ is finite for almost every $x\in X$ . We will later show (see Proposition 4.1) that this implies that $\rho ({\mathcal R}/{\mathcal S}_{p},\nu )=\|\unicode{x3bb} _{{\mathcal S}_{p}}(\nu )\|\leq \|\unicode{x3bb} _{{\mathcal T}}(\nu )\|=\rho ({\mathcal R},\nu )$ . Using the standard monotone coupling of Bernoulli percolation [Reference Lyons and Peres33, §5.2], it is direct to see that $\|\unicode{x3bb} _{{\mathcal S}_{p}}(\nu )\|$ is semicontinuous and that $\|\unicode{x3bb} _{{\mathcal S}_{p_{c}}}(\nu )\|\leq \|\unicode{x3bb} _{{\mathcal T}}(\nu )\|$ . Thus, we obtain the estimate of Schramm for connected, transitive graphs. With minor modifications, our results work for random walks on finite cost graphings (see Example 3.9 of §3.2). In this manner, we can recover the same estimate of Schramm when G is a connected, locally finite, transitive graph. As mentioned in the introduction, in the context of percolation all our proofs can be rephrased without equivalence relations and can be written in the language of percolation theory.

3.2. The 2–3 method and proof of Theorem 3.2(ii)

We now explain how to deduce the existence of the pointwise limit defining the co-spectral radius. We first state the general Theorem behind this existence and then explain why it applies to our setting, as well as to more general situations. For notation, if $f,g\colon {\mathcal R}\to [0,\infty ]$ are measurable, then we define their convolution to be the function $f*g\colon {\mathcal R}\to [0,\infty ]$ given by

$$ \begin{align*}(f*g)(x,y)=\sum_{z\in [x]_{{\mathcal R}}}f(x,z)g(z,y).\end{align*} $$

Given a measurable $\pi \colon X\to {\mathbb C}$ , we say that $f\colon {\mathcal R}\to [0,\infty ]$ is $\pi $ -symmetric if $\pi (x)f(x,y)=\pi (y)f(y,x)$ for almost every $(x,y)\in {\mathcal R}$ . If $\pi =1$ , we just say that f is symmetric.

Theorem 3.7. Let ${\mathcal R}$ be a discrete, measure-preserving equivalence relation over a standard probability space $(X,\mu )$ , and fix $\pi \in L^{1}(X,\mu )$ with $\pi (x)\in (0,\infty )$ for almost every $x\in X$ . Let $f_{k}\colon {\mathcal R}\to [0,\infty ]$ for $k\geq 1$ be a sequence of measurable functions such that:

1. (a) $f_{k}$ is $\pi $ -symmetric for all $k\in {\mathbb N}$ ;

2. (b) for all $l,k\in {\mathbb N}$ we have $\sum _{y\in [x]_{{\mathcal R}}}(f_{l}*f_{k})(x,y)\leq \sum _{y\in [x]_{{\mathcal R}}}f_{l+k}(x,y)$ for almost every $x\in X$ ;

3. (c) for almost every $x\in X$ and every $k\in {\mathbb N}$ we have $0<\sum _{y\in [x]_{{\mathcal R}}}f_{k}(x,y)<\infty $ ;

4. (d) there is a measurable $D\colon X\to (0,\infty )$ so that

   $$ \begin{align*}\sum_{y\in [x]_{{\mathcal R}}}f_{l+k}(x,y)\geq D(x)^{l}\sum_{y\in [x]_{{\mathcal R}}} f_{k}(x,y)\end{align*} $$
   for almost every $x\in X$ and every $l,k\in {\mathbb N}$ .

Then

$$ \begin{align*}\widetilde{f}(x)=\lim_{k\to\infty}\bigg(\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\bigg)^{1/k}\end{align*} $$

exists and is positive almost surely. Further:

1. (i) for every $k\in {\mathbb N}$

   $$ \begin{align*}\int\pi(x)\frac{\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)}{\widetilde{f}(x)^{k}}\,d\mu(x)\leq \int \pi\,d\mu;\end{align*} $$

2. (ii) $\lim _{k\to \infty }(\int \pi (x)\sum _{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\,d\mu (x))^{1/k}=\|\widetilde {f}\|_{\infty }.$

The name ‘2–3’ refers to the way we are proving Theorem 3.7. In the proof we have two separate steps where we show that ‘typically’ $\sum _{y}f_{2k}(x,y)\sim (\sum _y f_k(x,y))^2$ and $(\sum _y f_{3k}(x,y))\sim (\sum _y f_k(x,y))^3$ . Since $2,3$ generate a multiplicative semigroup which is asymptotically dense on the logarithmic scale, we are able to deduce that the exponential growth of $(\sum _y f_k(x,y))$ has a definite rate.

Before jumping into the proof of Theorem 3.7, let us list several examples where it applies. For all of these examples, fix a discrete, measure-preserving equivalence relation ${\mathcal R}$ over a standard probability space $(X,\mu )$ and fix a ${\mathcal S}\leq {\mathcal R}$ .

Example 3.8. Fix a symmetric $\nu \in \operatorname {Prob}([{\mathcal R}])$ . Define

$$ \begin{align*}f_{k}(x,y)=p_{2k,x,y}1_{{\mathcal S}}(x,y).\end{align*} $$

It is direct to check that our hypotheses apply in this case with $\pi =1$ , $D(x)=p_{2,x,x}$ . In this case

$$ \begin{align*}\sum_{y\in [x]_{{\mathcal S}}}f_{k}(x,y)=p_{2k,x,{\mathcal S}},\end{align*} $$

and we recover the existence of the pointwise co-spectral radius $\rho ^{{\mathcal S}}_{\nu }$ . Moreover, item (ii) of Theorem 3.7 as well as Theorem 3.2(i) imply that

So we recover the operator norm of the Markov operator $\unicode{x3bb} _{{\mathcal S}}(\nu )$ as the $L^{\infty }$ -norm of $\rho ^{{\mathcal S}}_{\nu }$ .

Example 3.9. Suppose that $\nu \colon {\mathcal R}\to [0,1]$ is measurable and that ${\sum _{y\in [x]_{{\mathcal R}}}\nu (x,y)=1}$ for almost every $x\in X$ . Moreover, assume that there is a $\pi \colon X\to (0,\infty )$ with ${\pi \in L^{1}(X,\mu )}$ so that $\nu $ is $\pi $ -symmetric. Consider the Markov chain on $[x]_{{\mathcal R}}$ with transition probabilities $\nu (x,y)$ , and for $k\in {\mathbb N}$ and $(x,y)\in {\mathcal R}$ let $p_{k,x,y}$ be the probability that the random walk corresponding to this Markov chain starting at x is at y after k steps. Set

$$ \begin{align*}f_{k}(x,y)=p_{2k,x,y}1_{S}(x,y),\end{align*} $$

and $p_{2k,x,{\mathcal S}}=\sum _{y\in [x]_{{\mathcal S}}}p_{2k,x,y}$ . Note that if $f,g\colon {\mathcal R}\to [0,\infty ]$ are $\pi $ -symmetric, then so is $f*g$ . Since $f_{k}=\nu ^{*2k}1_{S},$ we see that $f_{k}$ is $\pi $ -symmetric. Finally, observe that for every $x\in X,k\in {\mathbb N}$ ,

$$ \begin{align*}0<p_{2,x,x}^{k}\leq p_{2k,x,x}\leq \sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\leq 1.\end{align*} $$

So we deduce the existence of

$$ \begin{align*}\rho({\mathcal R}/{\mathcal S},\nu)=\lim_{k\to\infty}p_{2k,x,{\mathcal S}}^{{1}/{2k}}.\end{align*} $$

A good example to keep in mind is the following. Recall that the full pseudogroup, denoted $[[{\mathcal R}]]$ , of ${\mathcal R}$ is by definition the set of bimeasurable bijections $\phi \colon \operatorname {dom}(\phi )\to \operatorname {ran}(\phi )$ satisfying:

• • $\operatorname {dom}(\phi ),\operatorname {ran}(\phi )$ are measurable subsets of X;

• • $\phi (x)\in [x]_{{\mathcal R}}$ for almost every $x\in X$ .

We identify $\phi ,\psi \in [[{\mathcal R}]]$ if $\mu (\operatorname {dom}(\phi )\Delta \operatorname {dom}(\psi ))=0$ and if $\phi (x)=\psi (x)$ for almost every $x\in \operatorname {dom}(\phi )\cap \operatorname {dom}(\psi )$ . For $\phi ,\psi \in [[{\mathcal R}]]$ we use $\phi \circ \psi $ for the element of $[[{\mathcal R}]]$ whose domain is $\operatorname {dom}(\psi )\cap \psi ^{-1}(\operatorname {dom}(\phi ))$ and which satisfies $\phi \circ \psi (x)=\phi (\psi (x))$ for $x\in \operatorname {dom}(\psi )\cap \psi ^{-1}(\operatorname {dom}(\phi ))$ . For $\phi \in [[{\mathcal R}]]$ , we let $\phi ^{-1}$ be the element of $[[{\mathcal R}]]$ with $\operatorname {dom}(\phi ^{-1})=\operatorname {ran}(\phi )$ , $\operatorname {ran}(\phi ^{-1})=\operatorname {dom}(\phi )$ , and so that $\phi ^{-1}(\phi (x))=x$ for all $x\in \operatorname {dom}(\phi )$ . Given a countable $\Phi \subseteq [[{\mathcal R}]]$ and $x\in X$ we define a graph $G_{\Phi ,x}$ whose vertex set is $[x]_{{\mathcal R}}$ and whose edge set is $\bigcup _{\phi \in \Phi }\{\{y,\phi ^{\alpha }(y)\}:\alpha \in \{\pm 1\},y\in [x]_{{\mathcal R}}\cap \operatorname {dom}(\phi ^{\alpha })\}$ . We say that $\Phi $ is a graphing if $G_{\Phi ,x}$ is connected for almost every $x\in X$ . We define the cost of a graphing to be

$$ \begin{align*}c(\Phi)=\sum_{\phi\in \Phi}\mu(\operatorname{dom}(\phi)).\end{align*} $$

This definition is due to Levitt in [Reference Levitt32], and the cost of a relation (which is by definition the infimum of the cost of its graphings) was further systematically studied in [Reference Gaboriau18, Reference Gaboriau19]. If $\Phi $ is a finite cost graphing, then for almost every $x\in X$ define $\nu (x,y)={1}/{\deg _{G_{\Phi }}(x)}$ where $\deg _{G_{\Phi }}(x)$ is the degree of x in the graph $G_{\Phi ,x}$ . Observe that $\nu $ is $\deg _{G,\Phi }$ -symmetric. Since $\int \deg _{G_{\Phi }}\,d\mu \leq 2c(\Phi )<\infty $ , we have a well-defined co-spectral radius for the simple random walk associated to finite cost graphings.

Another good example is as follows. Consider a symmetric $\nu \in \operatorname {Prob}([{\mathcal R}])$ , and ${E\subseteq X}$ a measurable set with $\mu (E)>0$ . Assume that for almost every $x\in E$ , we have that $\sum _{y\in [x]_{{\mathcal R}}\cap E}p_{x,y}>0$ (e.g., this holds if the random walk is lazy).

For $x\in X$ , define $\nu |_{E}\colon {\mathcal R}\cap (E\times E)\to [0,1]$ by

$$ \begin{align*}\nu|_{E}(x,y)=\frac{p_{x,y}}{\sum_{y\in [x]_{{\mathcal R}}\cap E}p_{x,y}}.\end{align*} $$

As above this defines a Markov chain on $[x]_{{\mathcal R}}\cap E$ with transition probabilities given by $\nu |_{E}$ . We have that $\nu |_{E}$ is $\pi $ -symmetric, with $\pi (x,y)=\sum _{y\in [x]_{{\mathcal R}}\cap E}p_{x,y}$ . For $k\in {\mathbb N}$ , we let $p^{\nu |_{E}}_{k,x,y}$ be the probability that the random walk starting at x with these transition probabilities is at y after k steps. Setting $p_{2k,x,{\mathcal S}|_{E}}^{\nu |_{E}}=\sum _{y\in [x]_{{\mathcal S}}\cap E}p^{\nu |_{E}}_{2k, x,y}$ , we deduce the almost sure existence of the ‘conditional’ co-spectral radius

$$ \begin{align*}\rho^{{\mathcal S}|_{E}}(x)=\lim_{k\to\infty}(p_{2k,x,{\mathcal S}|_{E}}^{\nu|_{E}})^{{1}/{2k}}\end{align*} $$

for $x\in E$ .

Example 3.10. Fix a symmetric $\nu \in \operatorname {Prob}([{\mathcal R}])$ and a measurable $E\subseteq X$ with $\mu (E)>0.$ Define

$$ \begin{align*}f_{k}(x,y)=p_{2k,x,y}1_{{\mathcal S}}(x,y)1_{E}(x)1_{E}(y).\end{align*} $$

In this case for $x\in E$ we have

$$ \begin{align*}\sum_{y\in [x]_{{\mathcal S}}}f_{k}(x,y)=p_{2k,x,{\mathcal S},E}\end{align*} $$

where $p_{2k,x,{\mathcal S},E}$ is the probability that the random walk corresponding to $\nu $ starting at x is in $[x]_{{\mathcal S}}\cap E$ after $2k$ steps. All of our hypotheses apply in this case with X replaced with E, ${\mathcal R}$ replaced with ${\mathcal R}|_{E}$ , and $\pi =1$ . So we recover the existence of the pointwise local co-spectral radius

$$ \begin{align*}\rho_{E}^{{\mathcal S}}(x)=\lim_{k\to\infty}p_{2k,x,{\mathcal S},E}^{1/2k},\end{align*} $$

at least for $x\in E$ . See §3.3 for more details. In that section we will show more generally that

$$ \begin{align*}\lim_{k\to\infty}p_{2k,x,{\mathcal S},E}^{1/2k}\end{align*} $$

exists for almost every $x\in X$ ; see Corollary 3.17. This specific example will be important for us when we show that the spectral radius is almost surely constant in the case that ${\mathcal S}$ is either normal or ergodic.

We say that a $\nu \in \operatorname {Prob}([{\mathcal R}])$ is lazy if $\nu (\{\operatorname {id}\})>0$ .

Example 3.11. Fix a lazy, symmetric $\nu \in \operatorname {Prob}([{\mathcal R}])$ , and a measurable $E\subseteq X$ with $\mu (E)>0.$ Define for $(x,y)\in {\mathcal R}$ :

$$ \begin{align*}p_{2k,x,y,E}=\sum_{x_{1},\ldots,x_{2k-1}\in E}p_{x,x_{1}}p_{x_{1},x_{2}}\cdots p_{x_{2k-1},y},\end{align*} $$

$$ \begin{align*}p_{2k,x,{\mathcal S}|_{E}}=\sum_{\substack{x_{1},\ldots,x_{2k}\in E,\\ (x,x_{2k})\in {\mathcal S}}}p_{x,x_{1}}p_{x_{1},x_{2}}\cdots p_{x_{2k-1},x_{2k}}.\end{align*} $$

Define

$$ \begin{align*}f_{k}(x,y)=p_{2k,x,y,E}1_{E}(x)1_{E}(y)1_{S}(x,y).\end{align*} $$

Then

$$ \begin{align*}\sum_{y\in [x]_{{\mathcal S}}}f_{k}(x,y)=p_{2k,x,{\mathcal S}|_{E}}.\end{align*} $$

Again, it is direct to check that the hypotheses of Theorem 3.7 apply with $\pi =1$ . Note that laziness implies that

$$ \begin{align*}p_{2k,x,{\mathcal S}|_{E}}\geq p(x,x)^{2k}>0\end{align*} $$

for every $x\in E$ , and we always have

$$ \begin{align*}p_{2k,x,{\mathcal S}|_{E}}\leq 1.\end{align*} $$

Thus, (c) of Theorem 3.7 holds. So we deduce the existence of the ‘restricted’ co-spectral radius

$$ \begin{align*}\rho_{{\mathcal S}|_{E}}(x)=\lim_{k\to\infty}p_{2k,x,{\mathcal S}|_{E}}^{1/2k}\end{align*} $$

for $x\in E$ . It can be shown by the same method of proof as in Theorem 3.2(i) that

so

So we again recover the norm of a corner of the Markov operator as the essential supremum of the restricted co-spectral radius.

Example 3.12. Let $\varepsilon>0$ and let $\eta \colon {\mathcal R}\to [\varepsilon ,+\infty )$ be a measurable bounded symmetric function. Define for $(x,y)\in {\mathcal R}$ :

$$ \begin{align*}f_k(x,y):=\sum_{x_{1},\ldots,x_{2k-1}\in [x]_{\mathcal R}}\eta_{x,x_{1}}\eta_{x_{1},x_{2}}\cdots\eta_{x_{2k-1},y}.\end{align*} $$

Theorem 3.7 yields almost sure existence of the growth exponent $\rho _\eta :=\lim _{k\to \infty } (\sum _{y\in [x]_{\mathcal R}} f_k(x,y))^{1/k}.$ Let $\Phi \subset {\mathcal R}$ be a symmetric graphing generating ${\mathcal R}$ . If we put $\eta =1_{\Phi }$ , we get the existence of the growth exponent for the number of trajectories of length $2k$ starting at x in the graph induced by $\Phi $ on the equivalence class of $[x]_{\mathcal R}$ . In particular, for every unimodular random graph $({\mathcal G},o)$ , the number of walks of length $2k$ starting from o has an exponential rate of growth almost surely. Note that such a result is definitely not true for any rooted bounded degree graph. This last example is discussed in more detail in §5.2.

Having explained why Theorem 3.7 implies the existence of the pointwise co-spectral radius, we now turn to the proof of Theorem 3.7. The following is the main technical lemma behind the proof.

Lemma 3.13. Let $(f_{k})_{k},\pi $ be as in Theorem 3.7. For $k\in {\mathbb N}$ , define $\widetilde {f}_{k}\colon X\to [0,\infty ]$ by

$$ \begin{align*}\widetilde{f}_{k}(x)=\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y),\end{align*} $$

and define

$$ \begin{align*}\phi_{k}(x)=\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\widetilde{f}_{k}(y)^{-1},\end{align*} $$

$$ \begin{align*}\psi_{k}(x)=\widetilde{f}_{k}(x)\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\bigg(\sum_{z\in [y]_{{\mathcal R}}}f_{k}(y,z)\widetilde{f}_{k}(z)\bigg)^{-1}\!.\end{align*} $$

Then:

1. (i) for every $k\in {\mathbb N}$ and almost every $x\in X$ ,

   $$ \begin{align*}\widetilde{f}_{k}(x)^{2}\leq \phi_{k}(x)\widetilde{f}_{2k}(x)\quad\text{and}\quad \widetilde{f}_{k}(x)^{3}\leq \psi_{k}(x)\widetilde{f}_{3k}(x);\end{align*} $$

2. (ii) for every $k\in {\mathbb N}$ ,

   $$ \begin{align*}\int \pi\phi_{k}\,d\mu=\int \pi\,d\mu=\int \pi\psi_{k}\,d\mu.\end{align*} $$

Note that in order to make sense of $\phi _{k},\psi _{k}$ we are using hypothesis (c).

Proof. (i) By Cauchy–Schwarz,

$$ \begin{align*} \widetilde{f}_{k}(x)^{2}&=\bigg(\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)^{1/2}\widetilde{f}_{k}(y)^{-1/2}f_{k}(x,y)^{1/2}\widetilde{f}_{k}(y)^{1/2}\bigg)^{2}\\&\leq \phi_{k}(x)\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\widetilde{f}_{k}(y)\\ &\leq \phi_{k}(x)\widetilde{f}_{2k}(x) \end{align*} $$

for almost every $x\in X$ , where in the last step we use hypothesis (b). Using Cauchy–Schwarz again,

$$ \begin{align*}\widetilde{f}_{k}(x)^{2}\leq \widetilde{f}_{k}(x)^{-1}\psi_{k}(x)\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\bigg(\sum_{z\in [y]_{{\mathcal R}}}f_{k}(y,z)\widetilde{f}_{k}(z)\bigg) \quad\text{for almost every } x\in X.\end{align*} $$

We can estimate the second sum using our hypothesis on $f_{k}$ :

$$ \begin{align*} \sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\bigg(\sum_{z\in [y]_{{\mathcal R}}}f_{k}(y,z)\widetilde{f}_{k}(z)\bigg)&=\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\bigg(\sum_{w\in [y]_{{\mathcal R}}}(f_{k}*f_{k})(y,w)\bigg)\\ &\leq \sum_{w,y\in [x]_{{\mathcal R}}}f_{k}(x,y)f_{2k}(y,w)\\ &=\sum_{w\in [x]_{{\mathcal R}}}(f_{k}*f_{2k})(x,w)\\ &\leq \sum_{w\in [x]_{{\mathcal R}}}f_{3k}(x,w)\\ &=\widetilde{f}_{3k}(x), \end{align*} $$

with all of the above inequalities and equalities holding for almost every $x\in X$ . So we have shown that

$$ \begin{align*}\widetilde{f}_{k}(x)^{2}\leq \widetilde{f}_{k}(x)^{-1}\psi_{k}(x)\widetilde{f}_{3k}(x) \quad\text{for almost every } x\in X,\end{align*} $$

and rearranging proves the desired inequality.

(ii) By the mass transport principle and $\pi $ -symmetry,

$$ \begin{align*}\int \pi(x)\phi_{k}(x)\,d\mu(x)=\int \sum_{y\in [x]_{{\mathcal R}}}\pi(x)f_{k}(x,y)\widetilde{f}_{k}(x)^{-1}\,d\mu(x)=\int \pi\,d\mu,\end{align*} $$

as

$$ \begin{align*}\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\widetilde{f}_{k}(x)^{-1}=1.\end{align*} $$

Similarly,

$$ \begin{align*}&\int \pi(x)\psi_{k}(x)\,d\mu(x)\\&\quad=\int \sum_{y\in [x]_{{\mathcal R}}}\pi(x)f_{k}(x,y)\widetilde{f}_{k}(y)\bigg(\sum_{z\in [x]_{{\mathcal R}}}f_{k}(x,z)\widetilde{f}_{k}(z)\bigg)^{-1}\,d\mu(x)=\int \pi\,d\mu,\end{align*} $$

as

$$ \begin{align*}\sum_{y\in [x]_{{\mathcal R}}}f_{k}(x,y)\widetilde{f}_{k}(y)\bigg(\sum_{z\in [x]_{{\mathcal R}}}f_{k}(x,z)\widetilde{f}_{k}(z)\bigg)^{-1}=1.\\[-46pt]\end{align*} $$

We now prove Theorem 3.7. It will be helpful to pass to limits along subsets of ${\mathbb N}$ which are ‘not too sparse’ in a multiplicative sense. We say that $A\subseteq {\mathbb N}$ is asymptotically dense on the logarithmic scale if

$$ \begin{align*}\lim_{\substack{x\to \infty\\ x\in {\mathbb R}}}\inf_{n\in A}|x-\log(n)|=0.\end{align*} $$

We leave it as an exercise for the reader to show that if $A\subseteq {\mathbb N}$ is asymptotically dense on the logarithmic scale, and if $(a_{k})_{k=1}^{\infty }$ is a sequence of non-negative real numbers for which there is a uniform $C>0$ with

$$ \begin{align*}a_{l+k}\geq C^{l}a_{k}\quad\text{for all } l,k\in {\mathbb N},\end{align*} $$

then

$$ \begin{align*}\limsup_{k\in A}a_{k}^{1/k}=\limsup_{k\to\infty}a_{k}^{1/k}\quad\text{and}\quad \liminf_{k\in A}a_{k}^{1/k}=\liminf_{k\to\infty}a_{k}^{1/k}.\end{align*} $$

Proof of Theorem 3.7

Adopt notation as in Lemma 3.13. Set

$$ \begin{align*}\overline{f}(x)=\limsup_{k\to\infty}\widetilde{f}_{k}(x)^{1/2k}\quad\text{and}\quad \underline{f}(x)=\liminf_{k\to\infty}\widetilde{f}_{k}(x)^{1/2k}.\end{align*} $$

For $p,q\in {\mathbb N}\cup \{0\},k\in {\mathbb N}$ we have, by Lemma 3.13 and induction, that

(3.2) $$ \begin{align} \widetilde{f}_{k}(x)^{2^{p}3^{q}}\leq C_{p,q,k}(x)\widetilde{f}_{2^{p}3^{q}k}(x)\quad\text{for almost every } x\in X, \end{align} $$

where

$$ \begin{align*}C_{p,q,k}(x)=\prod_{i=0}^{p-1}\phi_{2^{i}k}(x)^{2^{p-1-i}3^{q}}\prod_{j=0}^{q-1}\psi_{2^{p}3^{j}k}(x)^{3^{q-1-j}}.\end{align*} $$

By Lemma 3.13 and the fact that $\pi \in L^{1}(X,\mu )$ , both $\sum _{k}k^{-2}\pi \phi _{k}$ and $\sum _{k}k^{-2}\pi \psi _{k}$ converge almost everywhere. Since $\pi (x)>0$ almost surely, we see that for almost every $x\in X$ , there is a $k_{0}$ (depending upon x) so that for $k\geq k_{0}$ we have $\phi _{k}(x)\leq k^{2}, \psi _{k}(x)\leq k^{2}$ . So for almost every $x\in X$ , and for all $k\geq k_{0}$ , and all $p,q\in {\mathbb N}\cup \{0\}$ ,

$$ \begin{align*}C_{p,q,k}(x)\leq \prod_{i=0}^{p-1}(2^{i}k)^{2^{p-i}3^{q}}\prod_{j=0}^{q-1}(2^{p}3^{j}k)^{3^{q-1-j}2},\end{align*} $$

which implies

$$ \begin{align*}\frac{1}{2^{p}3^{q}}\log C_{p,q,k}(x)&\leq \sum_{i=0}^{p-1}2^{-i}\log(2^{i}k)+\sum_{j=0}^{q-1}2^{1-p}3^{-1-j}\log(2^{p}3^{j}k)\\ &\leq B+2\log(k), \end{align*} $$

where

$$ \begin{align*}B=\bigg(\max_{p\in {\mathbb N}\cup\{0\}}\frac{p}{2^{p}}\bigg)\log(2)+\log(2)\sum_{i=0}^{\infty}\frac{i}{2^{i}}+\frac{2\log(3)}{3}\sum_{j=0}^{\infty}\frac{j}{3^{j}}<\infty.\end{align*} $$

Fix $k_{1}\geq k_{0}$ . Since $A=\{2^{p}3^{q}k_{1}:p,q\in {\mathbb N}\}$ is asymptotically dense on the logarithmic scale, we have by (3.2) and hypothesis (d) that

(3.3) $$ \begin{align} \underline{f}(x)=\liminf_{p+q\to \infty}\widetilde{f}_{2^{p}3^{q}k_{1}}(x)^{{1}/{2^{p}3^{q}k_{1}}}&\geq \liminf_{p+q\to\infty}C_{p,q,k_{1}}(x)^{-{1}/{2^{p}3^{q}k_{1}}}\widetilde{f}_{k_{1}}(x)^{1/k_{1}}\nonumber\\ &\geq e^{-B/k_{1}}k_{1}^{-2/k_{1}}\widetilde{f}_{k_{1}}(x)^{1/k_{1}}. \end{align} $$

Letting $k_{1}\to \infty $ , we see that $\underline {f}\geq \overline {f}$ almost everywhere, and this proves that $\widetilde {f}$ exists almost everywhere. Note that (3.3) applied with $k_{1}=k_{0}$ shows that ${\underline {f}(x)\geq e^{-B/k_{0}}k_{0}^{-2/k_{0}}\widetilde {f}_{k_{0}}(x)^{1/k_{0}}.}$ Thus, $\widetilde {f}(x)>0$ for almost every x.

(i) By (3.2), we have for $p,k\in {\mathbb N}$ and almost every $x\in X$ that

$$ \begin{align*} \frac{\widetilde{f}_{k}(x)}{\widetilde{f}_{2^{p}k}(x)^{{1}/{2^{p}}}}\leq C_{p,0,k}(x)^{{1}/{2^{p}}}&=\bigg(\prod_{i=0}^{p-1}\phi_{2^{i+1}k}(x)^{2^{p-1-i}}\bigg)^{1/2^{p}}\\ &\leq \frac{1}{2^{p}}\bigg(1+\sum_{i=0}^{p-1}2^{p-1-i}\phi_{2^{i+1}k}(x)\bigg), \end{align*} $$

where in the last step we use the arithmetic-geometric mean inequality. Lemma 3.13 implies that

$$ \begin{align*}\int \pi(x)\frac{\widetilde{f}_{k}(x)}{\widetilde{f}_{2^{p}k}(x)^{{1}/{2^{p}}}}\,d\mu(x)\leq \int \pi\,d\mu.\end{align*} $$

Letting $p\to \infty $ and applying Fatou’s lemma proves (i).

(ii) Part (i) shows that

$$ \begin{align*}\int \pi(x) \widetilde{f}_{k}(x)\,d\mu(x)\leq \|\widetilde{f}\|_{\infty}^{k}\int \pi\,d\mu,\end{align*} $$

for every $k\in {\mathbb N}$ . Since $\pi \in L^{1}(X,\mu ),$

$$ \begin{align*}\limsup_{k\to\infty}\bigg(\int \pi(x)\widetilde{f}_{k}(x)\,d\mu(x)\bigg)^{{1}/{k}}\leq \|\widetilde{f}\|_{\infty}.\end{align*} $$

We now prove the reverse inequality. Fatou’s lemma implies that for all $r\in {\mathbb N}$ ,

$$ \begin{align*}\int \pi(x)\widetilde{f}(x)^{r}\,d\mu(x)&\leq \liminf_{k\to \infty}\int \pi(x)\widetilde{f}_{k}(x)^{{r}/{k}}\,d\mu(x)\\&\leq \liminf_{k\to\infty}\bigg(\int \pi(x)\widetilde{f}_{k}(x)\,d\mu(x)\bigg)^{{r}/{k}}\bigg(\int \pi(x)\,d\mu(x)\bigg)^{1-{r}/{k}}\!,\end{align*} $$

where in the last step we use Holder’s inequality for $k>r$ . Since $\pi \in L^{1}(X,\mu )$ ,

$$ \begin{align*}\bigg(\int \pi(x)\widetilde{f}(x)^{r}\,d\mu(x)\bigg)^{{1}/{r}}\leq \liminf_{k\to\infty}\bigg(\int \pi(x) \widetilde{f}_{k}(x)\,d\mu(x)\bigg)^{{1}/{k}}.\end{align*} $$

Letting $r\to \infty $ and using that $0<\pi (x)$ for almost every x and that $\pi \in L^{1}(X,\mu )$ shows that

$$ \begin{align*}\|\widetilde{f}\|_{\infty}\leq \liminf_{k\to\infty}\bigg(\int \pi(x)\widetilde{f}_{k}(x)\,d\mu(x)\bigg)^{{1}/{k}}\!.\\[-42pt]\end{align*} $$

3.3. Normal subrelations and the proof of Theorem 3.2(iii)

In this subsection we prove that the co-spectral radius $\rho ^{{\mathcal S}}(x)$ is almost surely constant when ${\mathcal S}$ is either ergodic normal (the case when ${\mathcal S}$ is ergodic is fairly direct; see Corollary 3.17(a)). We prove a common generalization of the cases where ${\mathcal S}$ is normal or ergodic, for which we need partial one-sided normalizers.

One example to keep in mind for intuition is as follows. Suppose that ${\mathcal R}$ is the orbit equivalence relation of a measure-preserving action of G on $(X,\mu )$ . Suppose that $H\leq G$ , and let ${\mathcal S}=\{(x,ax):a\in N,x\in X\}$ . For $g\in G$ , we let $\phi _{g}\in [{\mathcal R}]$ be given by ${\phi _{g}(x)=gx}$ . If g is in the normalizer of H inside G, then by direct calculation ${\phi _{g}([x]_{{\mathcal S}})=[x]_{{\mathcal S}}}$ for almost every $x\in X$ . So $\phi _{g}\in PN_{{\mathcal R}}^{(1)}({\mathcal S})$ . More generally, ${\phi _{g}|_{E}\in PN_{{\mathcal R}}^{(1)}({\mathcal S})}$ for every $g\in G$ in the normalizer of H. We can thus think of the partial one-sided normalizers as a generalization of normalizers to the setting of the full pseudogroup.

Another source of elements in the partial normalizer is as follows. Recall that the set of endomorphisms of ${\mathcal R}$ over ${\mathcal S}$ , denoted by $\operatorname {End}_{{\mathcal R}}({\mathcal S})$ , is constituted by the measurable functions $\phi \colon X\to X$ such that for almost every $x\in X$ the following two conditions hold:

• • $\phi (x)\in [x]_{{\mathcal R}}$ ;

• • $\phi ([x]_{{\mathcal S}})\subseteq [\phi (x)]_{{\mathcal S}}$ .

Such functions need not be injective modulo null sets, and when they fail to be injective modulo null sets are not measure-preserving. Indeed, one can use the mass-transport principle to show that for $\phi \in \operatorname {End}_{{\mathcal R}}({\mathcal S})$ we have that

$$ \begin{align*}\frac{d\phi_{*}\mu}{d\mu}(x)=|\phi^{-1}(\{x\})| \quad\text{for almost every } x\in X\end{align*} $$

(we will not use this fact so will not give a full proof of it). However, as we will see shortly (see Lemma 3.16), for each $\phi \in \operatorname {End}_{{\mathcal R}}({\mathcal S})$ we may find a partition modulo null sets $(E_{i})_{i\in I}$ of X into countably many sets so that $\phi |_{E_{i}}$ is injective for each $i\in I$ .

This last example in fact provides the motivation for our consideration of the one-sided partial normalizers, as this is one way to define normal subrelations.

From the above definition, one can imagine attempting to prove that the co-spectral radius is almost surely constant for a normal subrelation by trying to understand how it varies after applying endomorphisms of ${\mathcal S}\leq {\mathcal R}$ . However, as alluded to above, endomorphisms can have undesirable properties (namely, lack of injectivity and failure to be measure-preserving) and the possibility of these undesirability properties makes them ill-suited for our proofs. A good tradeoff for our purposes is to drop being everywhere defined and gain being measure-preserving and injective. This is precisely the purpose of the one-sided partial normalizers.

We have a natural way for elements of $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ to act on $L^{p}(X,\mu )$ for $1\leq p\leq \infty $ . Namely, if $\phi \in PN^{(1)}_{{\mathcal R}}({\mathcal S})$ we define $\alpha _{\phi }\in B(L^{p}(X,\mu ))$ via

$$ \begin{align*}(\alpha_{\phi}f)(x)=1_{\operatorname{ran}(\phi)}(x)f(\phi^{-1}(x)).\end{align*} $$

This allows us to say what it means for $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ to act ergodically. It simply means that if $f\in L^{p}(X,\mu )$ and $\alpha _{\phi }(f)=1_{\operatorname {ran}(\phi )}f$ for every $\phi \in PN^{(1)}_{{\mathcal R}}({\mathcal S})$ , then f is essentially constant. We leave it as an exercise for the reader to follow the usual arguments to verify that this is equivalent to saying that there exists a countable $C\subseteq PN^{(1)}_{{\mathcal R}}({\mathcal S})$ so that if a measurable subset of X is invariant under C, then its measure is either $0$ or $1$ . Note that $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ automatically acts ergodically if ${\mathcal S}$ is ergodic, because $[{\mathcal S}]\subseteq PN^{(1)}_{{\mathcal R}}({\mathcal S})$ . The following lemma shows that $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ also acts ergodically if ${\mathcal S}$ is normal and ${\mathcal R}$ is ergodic.

Lemma 3.16. Suppose that ${\mathcal S}$ is normal in ${\mathcal R}$ . Then there is a countable set $C\subseteq PN^{(1)}_{{\mathcal R}}({\mathcal S})$ so that

$$ \begin{align*}[x]_{{\mathcal R}}=\bigcup_{\phi\in C}\phi(x)\end{align*} $$

for almost every $x\in X$ .

Proof. By [Reference Feldman, Sutherland and Zimmer15, Theorem 2.2], we may find a countable set $D\subseteq \operatorname {End}_{{\mathcal R}}({\mathcal S})$ so that

$$ \begin{align*}[x]_{{\mathcal R}}=\bigcup_{\phi\in D}[\phi(x)]_{{\mathcal S}}\end{align*} $$

for almost every $x\in X$ . Fix a countable subgroup $H\subseteq [{\mathcal S}]$ so that $[x]_{{\mathcal S}}=Hx$ for almost every $x\in X$ . Then for almost every $x\in X$ we have that

$$ \begin{align*}[x]_{{\mathcal R}}=\bigcup_{\phi\in D,\psi\in H}(\psi\circ \phi)(x).\end{align*} $$

Fix a countable subgroup $G\subseteq [{\mathcal R}]$ so that $[x]_{{\mathcal R}}=Gx$ for almost every $x\in X$ . Then for each $\phi \in D$ we may write, up to sets of measure zero,

$$ \begin{align*}\operatorname{dom}(\phi)=\bigcup_{g\in G}E_{g,\phi}\end{align*} $$

where $E_{g,\phi }\subseteq \{x\in \operatorname {dom}(\phi ):\phi (x)\in gx\}$ . For each $g\in G$ , $\phi \in D$ , define $\phi _{g}\in [[{\mathcal R}]]$ by declaring that $\operatorname {dom}(\phi _{g})=E_{g,\phi }$ and that $\phi _{g}(x)=gx$ for $x\in E_{g,\phi }$ . Observe that ${\phi _{g}\in PN^{(1)}_{{\mathcal R}}({\mathcal S})}$ . Then

$$ \begin{align*}[x]_{{\mathcal R}}=\bigcup_{\substack{\psi\in H,\\ \phi\in D,g\in G \text{ with } x\in E_{g,\phi}}}\psi\circ \phi_{g}(x).\end{align*} $$

Thus,

$$ \begin{align*}C=\{\psi\circ \phi_{g}:g\in G,\phi\in D,\psi\in H\}\end{align*} $$

is the desired countable set.

By the preceding lemma and the inclusion $[{\mathcal S}]\subseteq PN^{(1)}_{{\mathcal R}}({\mathcal S})$ , the condition that $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ acts ergodically on $(X,\mu )$ is satisfied if either ${\mathcal S}$ is ergodic or ${\mathcal S}$ is normal and ${\mathcal R}$ is ergodic. We will show that $\rho ^{{\mathcal S}}$ is almost surely invariant under the partial one-sided normalizers, and is thus essentially constant when $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ acts ergodically.

Since the elements of $PN^{(1)}_{{\mathcal R}}({\mathcal S})$ are only partially defined, we will need to relate the co-spectral radius to the modification given by Example 3.10 by restricting to a subset. It will not be hard to show from Theorem 3.7 that $\rho _{E}^{{\mathcal S}}(x)$ as defined in Example 3.10 exists for almost every $x\in X$ . To deduce the existence of $\rho ^{{\mathcal S}}_{E}(x)$ for almost every $x\in X$ , as well as some of its basic properties, we need to recall the ${\mathcal S}$ -saturation of a measurable set. Given a discrete, measure-preserving equivalence relation ${\mathcal S}$ on a standard probability space and a measurable $E\subseteq X$ , the ${\mathcal S}$ -saturation of E is the (unique up to measure zero sets) measurable subset $\widetilde {E}\subseteq X$ satisfying the following properties:

• • $E\subseteq \widetilde {E}$ ;

• • for almost every $x\in \widetilde {E}$ we have $[x]_{{\mathcal S}}\subseteq \widetilde {E}$ ;

• • for almost every $x\in \widetilde {E}$ , there is a $y\in E$ with $x\in [y]_{{\mathcal S}}$ .

If $\Gamma \leq [{\mathcal S}]$ is a countable subgroup which generates ${\mathcal S}$ , then a model for $\widetilde {E}$ can be given by

$$ \begin{align*}\widetilde{E}=\bigcup_{\phi\in \Gamma}\phi(E).\end{align*} $$

It will also be helpful to relate $\rho ^{{\mathcal S}}_{E}$ to a quantity more operator-theoretic in nature. For a measurable set $E\subseteq X$ of positive measure, and $x\in X$ , we let

$$ \begin{align*}p_{2k,x,{\mathcal S},E}=\sum_{y\in [x]_{{\mathcal S}}\cap E}p_{2k,x,y}.\end{align*} $$

We also let

$$ \begin{align*}\rho_{E}({\mathcal R}/{\mathcal S},\nu)=\lim_{k\to\infty}\bigg(\frac{1}{\mu(E)}\int_{E}p_{2k,x,{\mathcal S},E}\,d\mu(x)\bigg)^{1/2k}.\end{align*} $$

Define $\zeta _{E}\in L^{2}({\mathcal R}/{\mathcal S})$ by $(\zeta _{E})_{x}={1_{E}(x)}/{\sqrt {\mu (E)}}\delta _{[x]_{{\mathcal S}}}$ . Then

so it follows from the spectral theorem that the limit defining $\rho _{E}({\mathcal R}/{\mathcal S},\nu )$ exists. We now proceed to show that $\lim _{k\to \infty }p_{2k,x,{\mathcal S},E}^{{1}/{2k}}$ exists.

Proof. Note that the sequence of functions $f_{k}\colon E\to [0,\infty )$ given by

$$ \begin{align*}f_{k}(x,y)=p_{2k,x,y}1_{{\mathcal S}}(x,y)1_{E}(x)1_{E}(y)\end{align*} $$

satisfies the hypothesis of Theorem 3.7 with ${\mathcal R}$ replaced by ${\mathcal R}|_{E}$ . This proves that $\rho ^{{\mathcal S}}_{E}(x)$ exists for almost every $x\in E$ . Since

Theorem 3.7 also proves (b).

To prove that $\rho ^{{\mathcal S}}_{E}(x)$ exists for almost every x, let $\widetilde {E}$ be the ${\mathcal S}$ -saturation of E. Note that $\rho ^{{\mathcal S}}_{E}(x)=0$ for almost every $x\in \widetilde {E}^{c}$ . For almost every $x\in \widetilde {E}$ , we have that $\rho ^{{\mathcal S}}_{E}(y)$ exists for all $y\in [x]_{{\mathcal S}}\cap E$ . Fix such an $x\in \widetilde {E}$ , and let $y\in [x]_{{\mathcal S}}\cap E.$ Then there is an $\ell \in {\mathbb N}$ with $p_{\ell ,x,y}>0$ . So for all $k\geq 2\ell $ ,

$$ \begin{align*}p_{2k,x,{\mathcal S},E}=\sum_{z\in [x]_{{\mathcal S}}\cap E}p_{2k,x,z}\geq p_{\ell,x,y}^{2}\sum_{z\in [x]_{{\mathcal S}}\cap E}p_{2(k-\ell),y,z}=p_{\ell,x,y}^{2}p_{2(k-\ell),y,{\mathcal S},E}. \end{align*} $$

This shows that for almost every $x\in \widetilde {E}$ we have

$$ \begin{align*}\liminf_{k\to \infty}p_{2k,x,{\mathcal S},E}^{{1}/{2k}}\geq \rho^{{\mathcal S}}_{E}(y).\end{align*} $$

The proof that

$$ \begin{align*}\limsup_{k\to\infty}p_{2k,x,{\mathcal S},E}^{{1}/{2k}}\leq \rho^{{\mathcal S}}_{E}(y)\end{align*} $$

is similar.

It will be helpful to study how $\rho _{E}({\mathcal R}/{\mathcal S},\nu )$ varies as a function of E. We can embed measurable subsets (modulo null sets) of $(X,\mu )$ into $L^{1}(X,\mu )$ via identifying each set with its indicator function. We thus have a natural distance on measurable sets modulo null sets defined by

$$ \begin{align*}d(E,F)=\mu(E\Delta F).\end{align*} $$

We show that $\rho _{E}({\mathcal R}/{\mathcal S},\nu )$ is semicontinuous as function of E. For later use, it will also be useful to know that $\rho _{E}({\mathcal R}/{\mathcal S},\nu )$ as a function of ${\mathcal S}$ . We now define a topology on subrelations of ${\mathcal R}$ making this precise.

For ${\mathcal R}$ a discrete, probability-measure-preserving equivalence relation on $(X,\mu )$ , $\phi \in [[{\mathcal R}]]$ , and ${\mathcal S}\leq {\mathcal R}$ , set

$$ \begin{align*}F_{{\mathcal S},\phi}=\{x\in \operatorname{dom}(\phi):\phi(x)\in [x]_{{\mathcal S}}\}.\end{align*} $$

For a given subrelation ${\mathcal S}\leq {\mathcal R}$ , a finite set $\Omega \subseteq [[{\mathcal R}]]$ , and $\varepsilon>0$ , let ${\mathcal O}_{\Omega ,\varepsilon }({\mathcal S})$ consist of all subrelations ${\mathcal S}'\leq {\mathcal R}$ such that

$$ \begin{align*}\mu(F_{{\mathcal S},\phi}\Delta F_{{\mathcal S}',\phi})<\varepsilon \quad\text{for all } \phi \in \Omega.\end{align*} $$

We define a topology on subrelations of ${\mathcal R}$ (modulo null sets) by declaring that the family ${\mathcal O}_{\Omega ,\varepsilon }({\mathcal S})$ form a neighborhood basis of ${\mathcal S}$ . Suppose that I is countable, and that $\Phi =(\phi _{i})_{i\in I}$ in $[[{\mathcal R}]]$ is such that $[x]_{{\mathcal R}}=\{\phi _{i}(x):i\in I,x\in \operatorname {dom}(\phi _{i})\}$ for almost every $x\in X$ . In this case, given $(\alpha _{i})_{i\in I}\in (0,1]^{I}$ with $\sum _{i}\alpha _{i}=1$ , we can define a metric on subrelations of ${\mathcal R}$ by

$$ \begin{align*}d({\mathcal S},{\mathcal S}')=\sum_{i}\alpha_{i}\mu(F_{{\mathcal S},\phi_{i}}\Delta F_{{\mathcal S}',\phi_{i}}).\end{align*} $$

It can be check that this metric induces the topology defined above (moreover, this metric is complete, though we will not need this). We now prove our promised semicontinuity.

Proof. From

and the spectral theorem, we have

$$ \begin{align*}\rho_{E}({\mathcal R}/{\mathcal S},\nu)=\sup_{k}\bigg(\frac{1}{\mu(E)}\int_{E} p_{2k,x,{\mathcal S},E}\,d\mu(x)\bigg)^{1/2k}.\end{align*} $$

It thus suffices to prove that

$$ \begin{align*}(E, {\mathcal S})\mapsto \int_{E} p_{2k,x,{\mathcal S},E}\,d\mu(x) \end{align*} $$

is continuous. Let $\operatorname {id}_{E}\in [[{\mathcal R}]]$ be defined by $\operatorname {dom}(\operatorname {id}_{E})=E$ and $\operatorname {id}_{E}(x)=x$ for every $x\in E$ . Then $ \int _{E} p_{2k,x,{\mathcal S},E}\,d\mu (x)$ can be rewritten as

$$ \begin{align*}\sum_{\phi\in \operatorname{supp}(\nu^{*2k})}\nu^{*2k}(\phi)\mu(E\cap F_{{\mathcal S},\operatorname{id}_{E}\phi}).\end{align*} $$

Note that each term in this sum is a continuous as a function of $(E,{\mathcal S})$ . Since $\sum _{\phi \in \operatorname {supp}(\nu ^{*2k})}\nu ^{*2k}(\phi )=1,$ it follows that the sum converges uniformly. Thus,

$$ \begin{align*}(E,{\mathcal S})\mapsto \int_{E} p_{2k,x,{\mathcal S},E}\,d\mu(x) \end{align*} $$

is continuous.

For technical reasons, in order to show that $\rho ^{{\mathcal S}}$ is invariant under the partial normalizer of ${\mathcal S}$ inside of ${\mathcal R}$ , it will be helpful to reduce to the case that $\nu $ is lazy. We will briefly need to adopt notation for how the co-spectral radius depends upon the measure. So for ${\mathcal S}\leq {\mathcal R},X$ as in Theorem 3.20 and a countably supported $\nu \in \operatorname {Prob}([{\mathcal R}])$ , we use $\rho ^{{\mathcal S}}_{\nu }(x)$ for the co-spectral radius at x defined using $\nu $ . Similarly, for $n\in {\mathbb N}$ we use $p_{n,x,{\mathcal S}}^{\nu }$ for the probability that the random walk starting at x associated to $\nu $ is in $[x]_{{\mathcal S}}$ after n steps.

It will be helpful to use the following lemma which shows that the local co-spectral radius associated to $\nu $ is uniformly close to the local co-spectral radius associated to ${(1-t)\nu +t\delta _{\operatorname {id}}}$ as $t\to 0$ .

Lemma 3.19. Let $(X,\mu )$ be a standard probability space, and ${\mathcal S}\leq {\mathcal R}$ discrete, probability-measure-preserving equivalence relations defined over X. For a symmetric and countably supported $\nu \in \operatorname {Prob}([{\mathcal R}])$ and $t\in [0,1]$ , define $\nu _{t}=(1-t)\nu +t\delta _{\operatorname {id}}$ . Then

$$ \begin{align*}\|\rho_{\nu_{t}}^{{\mathcal S}}-\rho_{\nu}^{{\mathcal S}}\|_{\infty}\leq t(1+\rho({\mathcal R}/{\mathcal S},\nu)),\quad\text{for all } t\in [0,1].\end{align*} $$

Proof. To simplify notation, set $\rho =\rho ({\mathcal R}/{\mathcal S},\nu )$ . Let $\zeta _{E}$ be defined as in the discussion preceding Corollary 3.17. Let $\eta \in \operatorname {Prob}([-\rho ,\rho ])$ be the spectral measure of $\unicode{x3bb} (\nu )$ with respect to $\zeta _{E}$ . Then for every measurable set $E\subseteq X$ of positive measure we have that $\rho _{E}({\mathcal R}/{\mathcal S},\nu _{t}),\rho _{E}({\mathcal R}/{\mathcal S},\nu )$ are the $L^{\infty }$ -norms of $s\mapsto s$ , $s\mapsto (1-t)s+t$ with respect to $\eta $ . In particular, for all measurable $E\subseteq X$ of positive measure,

$$ \begin{align*}|\rho_{E}({\mathcal R}/{\mathcal S},\nu_{t})-\rho_{E}({\mathcal R}/{\mathcal S},\nu)|\leq t(1+\rho).\end{align*} $$

If E is almost surely ${\mathcal S}$ -invariant, then $\rho ^{{\mathcal S}}_{E}=\rho ^{{\mathcal S}}1_{E}$ . So Corollary 3.17(b) implies that

$$ \begin{align*}|\|\rho^{{\mathcal S}}_{\nu_{t}}1_{E}\|_{\infty}-\|\rho^{{\mathcal S}}_{\nu}1_{E}\|_{\infty}|\leq t(1+\rho)\quad\text{for all measurable } E\subseteq X \text{ which are } {\mathcal S}\text{-invariant}.\end{align*} $$

To prove the lemma, fix $c>0$ and let

$$ \begin{align*}E=\{x:\rho_{\nu}^{{\mathcal S}}(x)+ct(1+\rho)\leq \rho^{{\mathcal S}}_{\nu_{t}}(x)\}.\end{align*} $$

Note that E is ${\mathcal S}$ -invariant, by ${\mathcal S}$ -invariance of $\rho ^{{\mathcal S}}_{\nu }$ and $\rho ^{{\mathcal S}}_{\nu _{t}}$ . Assume, for the sake of contradiction, that $\mu (E)>0$ . Then by the preceding paragraph we have

$$ \begin{align*}\|\rho^{{\mathcal S}}_{\nu_{t}}1_{E}\|_{\infty}\leq t(1+\rho)+\|\rho^{{\mathcal S}}_{\nu}1_{E}\|_{\infty}.\end{align*} $$

On the other hand, the definition of E forces

$$ \begin{align*}\|\rho^{{\mathcal S}}_{\nu_{t}}1_{E}\|_{\infty}\geq ct(1+\rho)+\|\rho^{{\mathcal S}}_{\nu}1_{E}\|_{\infty}.\end{align*} $$

Since $c>1$ , we obtain a contradiction. Thus,

$$ \begin{align*}\rho^{{\mathcal S}}_{\nu_{t}}-\rho^{{\mathcal S}}_{\nu}\leq ct(1+\rho)\end{align*} $$

almost everywhere. A similar proof shows that $\rho ^{{\mathcal S}}_{\nu }-\rho ^{{\mathcal S}}_{\nu _{t}}\leq ct(1+\rho )$ almost everywhere. Thus,

$$ \begin{align*}\|\rho^{{\mathcal S}}_{\nu_{t}}-\rho^{{\mathcal S}}_{\nu}\|_{\infty}\leq ct(1+\rho)\end{align*} $$

for all $c>1$ , and the proof is completed by letting $c\to 1$ .

The above reduction to the lazy case and the semicontinuity in Lemma 3.18 allow us to give a general formula for the local co-spectral radius in terms of the co-spectral radius. We will ultimately use this to prove invariance under partial normalizers of ${\mathcal S}\leq {\mathcal R}$ by restrict to sets where we have uniform lower bound on transition probabilities $p_{k,x,\phi (x)}$ for $\phi \in PN^{(1)}_{{\mathcal R}}({\mathcal S})$ .

Theorem 3.20. Let $(X,\mu )$ be a standard probability space, and let ${\mathcal S}\leq {\mathcal R}$ be discrete, probability-measure-preserving equivalence relations defined over X, and fix a symmetric and countably supported $\nu \in \operatorname {Prob}([{\mathcal R}])$ . If $E\subseteq X$ has positive measure, then

$$ \begin{align*}\rho^{{\mathcal S}}_{E}=\rho^{{\mathcal S}}1_{\widetilde{E}},\end{align*} $$

where $\widetilde {E}$ is the ${\mathcal S}$ -saturation of E.

Proof. By Lemma 3.19, we may assume that $\nu $ is lazy. Observe that by ${\mathcal S}$ -invariance of $\widetilde {E}$ , we have that $\rho ^{{\mathcal S}}1_{\widetilde {E}}=\rho ^{{\mathcal S}}_{\widetilde {E}}$ almost everywhere. Hence, it is enough to show that $\rho ^{{\mathcal S}}_{E}=\rho ^{{\mathcal S}}_{\widetilde {E}}$ almost everywhere. Clearly $\rho ^{{\mathcal S}}_{E}\leq \rho ^{{\mathcal S}}_{\widetilde {E}}$ so it suffices to show that reverse inequality holds almost everywhere. For $k\in {\mathbb N}$ , let

$$ \begin{align*}E_{k}=\{x\in X:p_{2k,x,{\mathcal S},E}>0\}.\end{align*} $$

Then $E_{k}$ is an increasing sequence of sets with $E_{k}\supseteq E$ . By symmetry and laziness of $\nu $ we know that $\bigcup _{k}E_{k}=\widetilde {E}$ up to sets of measure zero. By Lemma 3.18, it suffices to show that for almost every $k\in {\mathbb N}$ we have that $\rho ^{{\mathcal S}}_{E_{k}}\leq \rho ^{{\mathcal S}}_{E}$ almost surely.

For $m\in {\mathbb N}$ , set $E_{k,m}=\{x\in X:p_{2k,x,{\mathcal S},E}>1/m\}$ . Then the $E_{k,m}$ are increasing and

$$ \begin{align*}E_{k}=\bigcup_{m=1}^{\infty}E_{k,m},\end{align*} $$

so again by Lemma 3.18 it suffices to show that $\rho ^{{\mathcal S}}_{E_{k,m}}\leq \rho ^{{\mathcal S}}_{E}$ almost surely. Fix an $x\in E_{k,m}$ so that both $\rho ^{{\mathcal S}}_{E}(x),\rho ^{{\mathcal S}}_{E_{k,m}}(x)$ exist. Then for every $l\in {\mathbb N}$ ,

$$ \begin{align*}p_{2(l+k),x,{\mathcal S},E}\geq \sum_{y\in [x]_{{\mathcal S}}\cap E_{k,m}}p_{2l,x,y}p_{2k,y,{\mathcal S},E}\geq \frac{1}{m}p_{2l,x,{\mathcal S},E_{k,m}}.\end{align*} $$

Taking $2l$ th roots of both sides and letting $l\to \infty $ shows that

$$ \begin{align*}\rho^{{\mathcal S}}_{E}\geq \rho^{{\mathcal S}}_{E_{k,m}}.\end{align*} $$

Since $\rho ^{{\mathcal S}}_{E},\rho ^{{\mathcal S}}_{E_{k,m}}$ exist almost everywhere, this completes the proof.

We now have amassed enough results to show that the co-spectral radius does not increase after applying partial normalizing elements.

Proposition 3.21. Let $(X,\mu )$ be a standard probability space, let ${\mathcal S}\leq {\mathcal R}$ be discrete, probability-measure-preserving equivalence relations defined over X, and fix a symmetric countably supported $\nu \in \operatorname {Prob}([{\mathcal R}])$ which generates ${\mathcal R}$ . Then for every $\phi \in PN^{(1)}_{{\mathcal R}}({\mathcal S})$ we have

$$ \begin{align*}\rho^{{\mathcal S}}(\phi(x))\geq \rho^{{\mathcal S}}(x)\end{align*} $$

for almost every $x\in \operatorname {dom}(\phi )$ .

Proof. First assume that $\nu $ is lazy. Fix $\phi \in PN^{(1)}_{{\mathcal R}}({\mathcal S})$ . Replacing X with a co-null set, we may assume that $\phi ([x]_{{\mathcal S}}\cap \operatorname {dom}(\phi ))\subseteq [\phi (x)]_{{\mathcal S}}$ for every $x\in X$ . Since $\nu $ is lazy and generating, for almost every $x\in \operatorname {dom}(\phi )$ we have that $p_{k,x,\phi (x)}>0$ for all large k. So, up to sets of measure zero,

$$ \begin{align*}\operatorname{dom}(\phi)=\bigcup_{k=1}^{\infty}\{x\in \operatorname{dom}(\phi):p_{k,x,\phi(x)}>0\text{ for all } k\geq n\}.\end{align*} $$

Since $\nu $ is lazy, this union is increasing and thus it follows that we may find a sequence $\delta _{n}\to 0$ of positive numbers and an increasing sequence of integers $r_{n}$ so that

$$ \begin{align*}\sum_{n=1}^{\infty}\mu(\{x\in \operatorname{dom}(\phi):p_{r_{n},x,\phi(x)}\leq \delta_{n}\})<\infty.\end{align*} $$

Set $F_{n}=\{x\in \operatorname {dom}(\phi ):p_{r_{n},x,\phi (x)}>\delta _{n}\}$ and $E_{n}=\phi (F_{n}).$ By the Borel–Cantelli lemma, for almost every $x\in \operatorname {dom}(\phi )$ we have that $x\in F_{k}$ for all sufficiently large $k.$ By Theorem 3.20, for almost every $x\in \operatorname {dom}(\phi )$ the following conditions are satisfied for all sufficiently large k:

• • $x\in F_{k}$ ;

• • $\rho ^{{\mathcal S}}(\phi (x))=\rho ^{{\mathcal S}}_{E_{k}}(\phi (x))$ ;

• • $\rho ^{{\mathcal S}}(x)=\rho ^{{\mathcal S}}_{F_{k}}(x)$ .

Fix an $x\in \operatorname {dom}(\phi )$ which satisfies the above three conditions, and fix k such that the above three bulleted items hold. Then for all $l\in {\mathbb N}$ ,

$$ \begin{align*}p_{2l,\phi(x),{\mathcal S},E_{k}}=\sum_{z\in [\phi(x)]_{S}\cap E_{k}}p_{2l,\phi(x),z}.\end{align*} $$

Since $\phi ([x]_{{\mathcal S}}\cap \operatorname {dom}(\phi ))\subseteq [\phi (x)]_{{\mathcal S}}$ , and $\phi (x)\in E_{k},x\in F_{k}$ , we have for all $l\geq r_{k}$ that

$$ \begin{align*}p_{2l,\phi(x),{\mathcal S},E_{k}}\geq \sum_{y\in [x]_{S}\cap F_{k}}p_{2l,\phi(x),\phi(y)}\geq \delta_{k}^{2}\sum_{y\in [x]_{S}\cap F_{k}}p_{2(l-r_{k}),x,y}=\delta_{k}^{2}p_{2(l-r_{k}),x,{\mathcal S},F_{k}},\end{align*} $$

where in the second to last step we use symmetry and the fact that $x\in F_{k}$ . Thus,

$$ \begin{align*}\rho^{{\mathcal S}}(\phi(x))=\rho^{{\mathcal S}}_{E_{k}}(\phi(x))\geq \rho^{{\mathcal S}}_{F_{k}}(x)=\rho^{{\mathcal S}}(x).\end{align*} $$

So

$$ \begin{align*}\rho^{{\mathcal S}}\circ \phi|_{\operatorname{dom}(\phi)}\geq \rho^{{\mathcal S}}|_{\operatorname{dom}(\phi)}\end{align*} $$

almost everywhere. The general case follows from the lazy case by using Lemma 3.19.

We are now ready to prove that the co-spectral radius does not change under the partial one-sided normalizers.

Proof. Let C be as in Lemma 3.16. By the preceding proposition and countability, for every $t\in [0,1]$ .

$$ \begin{align*}E_{t}=\{x\in X:\rho^{{\mathcal S}}(x)\geq t\}\end{align*} $$

is invariant under C, and so by ergodicity of ${\mathcal R}$ has measure $0$ or $1$ for every $t\in [0,1]$ . If $s=\sup \{t:\mu (E_{t})=1\}$ , then $\rho ^{{\mathcal S}}\geq s$ almost everywhere, and $\rho ^{{\mathcal S}}(x)<s+{1}/{n}$ for almost every x and every $n\in {\mathbb N}$ . Thus, $\rho ^{{\mathcal S}}(x)=s$ for almost every x.

We close this section with an example illustrating the fact that $\rho ^{{\mathcal S}}(x)$ may fail to be essentially constant if we only assume that ${\mathcal R}$ is ergodic. Thus, we need to assume something special about the inclusion ${\mathcal S}\leq {\mathcal R}$ .

Example 3.23. Let ${\mathcal R}$ be an ergodic, discrete, measure-preserving equivalence relation on $(X,\mu )$ . Let $E\subseteq X$ be a measurable set of positive measure. Let

$$ \begin{align*}{\mathcal S}={\mathcal R}\cap (E\times E)\cup\{(x,x):x\in E^{c}\}.\end{align*} $$

We claim that the following statements hold.

Claim

1. (i) For almost every $x\in E^{c}$ , we have $\rho ^{{\mathcal S}}(x)=\rho ({\mathcal R},\nu )$ .

2. (ii) For almost every $x\in E$ , we have $\rho ^{{\mathcal S}}(x)=1.$

In particular, if ${\mathcal R}$ is not hyperfinite, then by [Reference Bowen, Hoff and Ioana10, Lemma 2.2.] there is a $\nu \in \operatorname {Prob}([{\mathcal R}])$ so that $\rho ({\mathcal R},\nu )<1$ (this also follows from condition (GM) in [Reference Kaimanovich26] being equivalent to hyperfiniteness) and this gives an example where the co-spectral radius is not essentially constant.

Proof of claim

Let $\mathcal {T}=\{(x,x):x\in X\}$ . Let $\unicode{x3bb} _{x}(\nu )$ be the Markov operator associated to $\nu $ acting on $\ell ^{2}([x]_{{\mathcal R}})$ . Then for all $x\in X$ we have

and so the co-spectral radius of ${\mathcal R}$ with respect to ${\mathcal T}$ agrees with the spectral radius of $[x]_{{\mathcal R}}$ with respect to $\nu $ , that is, $\rho ^{{\mathcal T}}(x)=\|\unicode{x3bb} _{x}(\nu )\|$ . If $(x,y)\in {\mathcal R}$ , then $\unicode{x3bb} _{x}(\nu ),\unicode{x3bb} _{y}(\nu )$ are both operators on $\ell ^{2}([x]_{{\mathcal R}})$ and $\unicode{x3bb} _{x}(\nu )=\unicode{x3bb} _{y}(\nu )$ . So, by ergodicity, the operator norm $\|\unicode{x3bb} _{x}(\nu )\|$ is essentially constant. Since $\rho ({\mathcal R}/{\mathcal T},\nu )=\rho ({\mathcal R},\nu )$ , it follows from Theorem 3.2(ii) that $\|\unicode{x3bb} _{x}(\nu )\|$ is almost surely equal to $\rho ({\mathcal R},\nu )$ .

(i) For almost every $x\in E^{c}$ we have $\rho ^{{\mathcal S}}(x)=\rho ^{{\mathcal T}}(x)$ , so this follows by the preceding paragraph.

(ii) For almost every $x\in E$ we have $\rho ^{{\mathcal S}}(x)=\rho _{E}^{{\mathcal R}}(x)$ , with notation as in Corollary 3.17. So this follows from Theorem 3.20.