Proof of Theorem 3.1

Assume that $G\curvearrowright (X,\mu _1)=\prod _{g\in G}(X_0,\mu _g)$ is non-singular. For every $t\in [0,1]$ we have that

$$ \begin{align*} \sum_{h\in G}H^2(\mu_h^t,\mu_{gh}^t)\leq t\sum_{h\in G}H^2(\mu_h, \mu_{gh}) \quad \text{for every } g\in G, \end{align*} $$

so that $G\curvearrowright (X,\mu _t)$ is non-singular for every $t\in [0,1]$ . The rest of the proof we divide into two steps.

Proof of Claim 1

Note that for every $g\in G$ we have that

$$ \begin{align*} (\mu_g^s)^r&=(1-r)\nu+r\mu_g^s=(1-r)\nu+r(1-s)\nu+rs\mu_g=\mu_g^{sr}, \end{align*} $$

so that $(\mu _s)_r=\mu _{sr}$ . Therefore, it suffices to prove that $G\curvearrowright (X,\mu _s)$ is weakly mixing for every $s<1$ , assuming that $G\curvearrowright (X,\mu _1)$ is conservative.

The claim is trivially true for $s=0$ . So assume that $G\curvearrowright (X,\mu _1)$ is conservative and fix $s\in (0,1)$ . Let $G\curvearrowright (Y,\eta )$ be an ergodic pmp action. Define $Y_0=X_0\times X_0\times \{0,1\}$ and define the probability measures $\unicode{x3bb} $ on $\{0,1\}$ by $\unicode{x3bb} (0)=s$ . Define the map $\theta \colon Y_0\rightarrow X_0$ by

(3.3) $$ \begin{align} \theta(x,x',j)=\begin{cases}x &\text{if } j=0,\\ x' &\text{if } j=1.\end{cases} \end{align} $$

Then for every $g\in G$ we have that $\theta _*(\mu _g\times \nu \times \unicode{x3bb} )=\mu _g^s$ . Write $Z=\{0,1\}^G$ and equip Z with the probability measure $\unicode{x3bb} ^{G}$ . We identify the Bernoulli action $G\curvearrowright Y_0^{G}$ with the diagonal action $G\curvearrowright X\times X\times Z$ . By applying $\theta $ in each coordinate we obtain a G-equivariant factor map

(3.4) $$ \begin{align} \Psi\colon X\times X\times Z\rightarrow X:\quad \Psi(x,x',z)_h=\theta(x_h,x^{\prime}_h,z_h). \end{align} $$

Then the map $\mathord {\textrm {id}}_Y\times \Psi \colon Y\times X\times X\times Z\rightarrow Y\times X$ is G-equivariant and we have that $(\mathord {\textrm { id}}_Y\times \Psi )_*(\eta \times \mu _1\times \mu _0\times \unicode{x3bb} ^G)=\eta \times \mu _s$ . The construction above is similar to [Reference Kosloff and SooKS20, §4].

Take $F\in L^{\infty }(Y\times X,\eta \times \mu _s)^{G}$ . Note that the diagonal action $G\curvearrowright (Y\times X,\eta \times \mu _1)$ is conservative, since $G\curvearrowright (Y,\eta )$ is pmp. The action $G\curvearrowright (X\times Z,\mu _0\times \unicode{x3bb} ^{G})$ can be identified with a pmp Bernoulli action with base space $(X_0\times \{0,1\},\nu \times \unicode{x3bb} )$ , so that it is mixing. By [Reference Schmidt and WaltersSW81, Theorem 2.3] we have that

$$ \begin{align*} L^{\infty}(Y\times X\times X\times Z,\eta\times \mu_1\times \mu_0\times \unicode{x3bb}^{G})^{G}=L^{\infty}(Y\times X,\eta\times \mu_1)^{G}\mathbin{\overline{\otimes}} 1\mathbin{\overline{\otimes}} 1, \end{align*} $$

which implies that the assignment $(y,x,x',z)\mapsto F(y, \Psi (x,x',z))$ is essentially independent of $x'$ and z. Choosing a finite set of coordinates $\mathcal {F}\subset G$ and changing, for $g\in \mathcal {F}$ , the value $z_g$ between $0$ and $1$ , we see that F is essentially independent of the $x_g$ -coordinates for $g\in \mathcal {F}$ . As this is true for any finite set $\mathcal {F}\subset G$ , we have that $F\in L^{\infty }(Y)^{G}\mathbin {\overline {\otimes }} 1$ . The action $G\curvearrowright (Y,\eta )$ is ergodic and therefore F is essentially constant. We conclude that $G\curvearrowright (X,\mu _s)$ is weakly mixing.

Proof of Claim 2

Again it suffices to assume that $G\curvearrowright (X,\mu _1)$ is not dissipative and to show that $G\curvearrowright (X,\mu _s)$ is conservative for every $s<1$ .

When $s=0$ , the statement is trivial, so assume that $G\curvearrowright (X,\mu _1)$ is not dissipative and fix $s\in (0,1)$ . Let $C\subset X$ denote the non-negligible conservative part of $G\curvearrowright (X,\mu _1)$ . As in the proof of Claim 1, write $Z=\{0,1\}^{G}$ and let $\unicode{x3bb} $ be the probability measure on $\{0,1\}$ given by $\unicode{x3bb} (0)=s$ . Writing $\Psi \colon X\times X\times Z\rightarrow X$ for the G-equivariant map (3.4). We claim that $\Psi _*((\mu _1\times \mu _0\times \unicode{x3bb} ^{G})|_{C\times X\times Z})\sim \mu _s$ , so that $G\curvearrowright (X,\mu _s)$ is a factor of a conservative non-singular action, and therefore must be conservative itself.

As $\Psi _*(\mu _1\times \mu _0\times \unicode{x3bb} ^{G})=\mu _s$ , we have that $\Psi _*((\mu _1\times \mu _0\times \unicode{x3bb} ^{G}) |_{C\times X\times Z})\prec \mu _s$ . Let $\mathcal {U}\subset X$ be the Borel set, uniquely determined up to a set of measure zero, such that $\Psi _*((\mu _1\times \mu _0\times \unicode{x3bb} ^{G}) |_{C\times X\times Z})\sim \mu _s |_{\mathcal {U}}$ . We have to show that $\mu _s(X\setminus \mathcal {U})=0$ . Fix a finite subset $\mathcal {F}\subset G$ . For every $t\in [0,1]$ define

$$ \begin{align*} (X_1,\gamma_1^{t})&=\prod_{g\in \mathcal{F}}(X_0,(1-t)\nu+t\mu_g),\\ (X_2,\gamma_2^{t})&=\prod_{g\in G \setminus \mathcal{F}}(X_0,(1-t)\nu+t\mu_g). \end{align*} $$

We shall write $\gamma _1=\gamma _1^1, \gamma _2=\gamma _2^1$ . Also define

$$ \begin{align*} (Y_1,\zeta_1)&=\prod_{g\in \mathcal{F}}(X_0\times X_0\times \{0,1\},\mu_g\times \nu\times \unicode{x3bb}),\\ (Y_2,\zeta_2)&=\prod_{g\in G\setminus \mathcal{F}}(X_0\times X_0\times \{0,1\},\mu_g\times \nu\times \unicode{x3bb}). \end{align*} $$

By applying the map (3.3) in every coordinate, we get factor maps $\Psi _j\colon Y_j\rightarrow X_j$ that satisfy $(\Psi _j)_*(\zeta _j)=\gamma _j^{s}$ for $j=1,2$ . Identify $X_1\times Y_2\cong X\times (X_0\times \{0,1\})^{G\setminus \mathcal {F}}$ and define the subset $C'\subset X_1\times Y_2$ by $C'=C\times (X_0\times \{0,1\})^{G\setminus \mathcal {F}}$ . Let $\mathcal {U}'\subset X$ be Borel such that

$$ \begin{align*} (\mathord{\textrm{id}}_{X_1}\times \Psi_2)_*((\gamma_1\times \zeta_2)|_{C'})\sim (\gamma_1\times \gamma_2^{s})|_{\mathcal{U}'}. \end{align*} $$

Identify $Y_1\times X_2\cong X\times (X_0\times \{0,1\})^{\mathcal {F}}$ and define $V\subset Y_1\times X_2$ by $V=\mathcal {U}'\times (X_0\times \{0,1\})^{\mathcal {F}}$ . Then we have that

$$ \begin{align*} (\Psi_1\times \mathord{\textrm{id}}_{X_2})_*((\zeta_1\times \gamma_2^s)|_{V})&\sim (\Psi_1\times \mathord{\textrm{ id}}_{X_2})_*(\mathord{\textrm{id}}_{Y_1}\times \Psi_2)_*((\gamma_1\times \zeta_1)|_{C'}\times \nu^{\mathcal{F}}\times \unicode{x3bb}^{\mathcal{F}})\\ &=\Psi_*((\zeta_1\times \zeta_2)|_{C\times X\times Z})\sim \mu_s|_{\mathcal{U}}. \end{align*} $$

Let $\pi \colon X_1\times X_2\rightarrow X_2$ and $\pi '\colon Y_1\times X_2\rightarrow X_2$ denote the coordinate projections. Note that by construction we have that

(3.5) $$ \begin{align} \pi^{\prime}_*((\zeta_1\times \gamma_2^s)|_V) \sim \pi_*((\gamma_1\times \gamma_2^{s})|_{\mathcal{U}'})\sim \pi_*(\mu_s|_{\mathcal{U}}). \end{align} $$

Let $W\subset X_2$ be Borel such that $\pi _*(\mu _s |_{\mathcal {U}})\sim \gamma _2^s |_{W}$ . For every $y\in X_2$ define the Borel sets

$$ \begin{align*} \mathcal{U}_y=\{x\in X_1:(x,y)\in \mathcal{U}\}\quad \text{and} \quad\mathcal{U}^{\prime}_y=\{x\in X_1:(x,y)\in \mathcal{U}'\}. \end{align*} $$

As $\pi _*((\gamma _1\times \gamma _2^s) |_{\mathcal {U}'})\sim \gamma _2^s |_{W}$ , we have that

$$ \begin{align*} \gamma_1(\mathcal{U}^{\prime}_y)>0 \quad\text{for } \gamma_2^s\text{-a.e. } y\in W. \end{align*} $$

The disintegration of $(\gamma _1\times \gamma _2^s) |_{\mathcal {U}'}$ along $\pi $ is given by $(\gamma _1 |_{\mathcal {U}^{\prime }_y})_{y\in W}$ . Therefore, the disintegration of $(\zeta _1\times \gamma _2^s) |_{V}$ along $\pi '$ is given by $(\gamma _1 |_{\mathcal {U}^{\prime }_y}\times \nu ^{\mathcal {F}}\times \unicode{x3bb} ^{\mathcal {F}})_{y\in W}$ . We conclude that the disintegration of $(\Psi _1\times \mathord {\textrm {id}}_{X_2})_*((\zeta _1\times \gamma _2^s) |_V)$ along $\pi $ is given by $((\Psi _1)_*(\gamma _1 |_{\mathcal {U}^{\prime }_y}\times \nu ^{\mathcal {F}}\times \unicode{x3bb} ^{\mathcal {F}}))_{y\in W}$ . The disintegration of $\mu _s |_{\mathcal {U}}$ along $\pi $ is given by $(\gamma _2^s |_{\mathcal {U}_y})_{y\in W}$ . Since $\mu _s |_{\mathcal {U}}\sim (\Psi _1\times \mathord {\textrm { id}}_{X_2})_*((\zeta _1\times \gamma _2^s) |_V)$ , we conclude that

$$ \begin{align*} (\Psi_1)_*(\gamma_1|_{\mathcal{U}^{\prime}_y}\times \nu^{\mathcal{F}}\times \unicode{x3bb}^{\mathcal{F}})\sim \gamma_1^s|_{\mathcal{U}_y}\quad\text{for } \gamma_2^s\text{-a.e. } y\in W. \end{align*} $$

As $\gamma _1(\mathcal {U}^{\prime }_y)>0$ for $\gamma _2^s$ -a.e. $y\in W$ , and using that $\nu \sim \mu _e$ , we see that

$$ \begin{align*} \gamma_1^{s}\sim \nu^{\mathcal{F}}&\sim(\Psi_1)_*((\gamma_1\times \nu^{\mathcal{F}}\times \unicode{x3bb}^{\mathcal{F}})|_{\mathcal{U}^{\prime}_y\times X_0^{\mathcal{F}}\times \{1\}^{\mathcal{F}}})\\ &\prec (\Psi_1)_*(\gamma_1|_{\mathcal{U}^{\prime}_y}\times \nu^{\mathcal{F}}\times \unicode{x3bb}^{\mathcal{F}}). \end{align*} $$

for $\gamma _2^{s}$ -a.e. $y\in W$ . It is clear that also $(\Psi _1)_*(\gamma _1 |_{\mathcal {U}^{\prime }_y}\times \nu ^{\mathcal {F}}\times \unicode{x3bb} ^{\mathcal {F}})\prec \gamma _1^{s}$ , so that $\gamma _1^{s} |_{\mathcal {U}_y}\sim \gamma _1^{s}$ for $\gamma _2^s$ -a.e. $y\in W$ . Therefore, we have that $\gamma _1^s(X_1\setminus \mathcal {U}_y)=0$ for $\gamma _2^s$ -a.e. $y\in W$ , so that

$$ \begin{align*} \mu_s(\mathcal{U}\triangle (X_0^{{\mathcal{F}}}\times W))=0. \end{align*} $$

Since this is true for every finite subset $\mathcal {F}\subset G$ , we conclude that $\mu _s(X\setminus \mathcal {U})=0$ .

The conclusion of the proof now follows by combining both claims. Assume that ${G\curvearrowright (X,\mu _t)}$ is not dissipative and fix $s<t$ . Choose r such that $s<r<t$ .

$\nu \sim \mu _e$ . By Claim 2 we have that $G\curvearrowright (X,\mu _r)$ is conservative. Then by Claim 1 we see that $G\curvearrowright (X,\mu _s)$ is weakly mixing.

$\nu \prec \mu _e$ . As $\nu \prec \mu _e$ , the measures $\mu _e^{t}$ and $\mu _e$ are equivalent. We have that

$$ \begin{align*} \frac{d\mu_g^{t}}{d\mu_e^{t}}=\bigg((1-t)\frac{d\nu}{d\mu_e}+t\frac{d\mu_g}{d\mu_e}\bigg)\frac{d\mu_e}{d\mu_e^{t}}. \end{align*} $$

So if $\sup _{g\in G}|{\log}\ d\mu _g/d\mu _e(x)|<+\infty $ for a.e $x\in X_0$ , we also have that

$$ \begin{align*}\sup_{g\in G}|{\log}\ d\mu_g^{t}/d\mu_e^{t}(x)|<+\infty \quad\text{for a.e. }x\in X_0.\end{align*} $$

It follows from [Reference Björklund, Kosloff and VaesBV20, Proposition 4.3] that $G\curvearrowright (X,\mu _t)$ is conservative. Then by Claim 1 we have that $G\curvearrowright (X,\mu _s)$ is weakly mixing.

Proof. Let $E\colon L^0(X,[0,+\infty ))\rightarrow L^0(Y,[0,+\infty ))$ denote the conditional expectation map that is uniquely determined by

$$ \begin{align*} \int_Y E(F) H\, d\nu=\int_X F (H\circ \psi)\, d\mu \end{align*} $$

for all positive measurable functions $F\colon X\rightarrow [0,+\infty )$ and $H\colon Y\rightarrow [0,+\infty )$ . Since

$$ \begin{align*} \frac{dk^{-1}\nu}{d\nu}=\frac{d\psi_*(k^{-1}\mu)}{d\psi_*\mu}=E\bigg(\frac{dk^{-1}\mu}{d\mu}\bigg) \end{align*} $$

for every $k\in G$ , we have that

(3.6) $$ \begin{align} \sum_{k\in G}\eta_n(hk^{-1})\frac{dk^{-1}\nu}{d\nu}(y)=E\bigg(\sum_{k\in G}\eta_n(hk^{-1})\frac{dk^{-1}\mu}{d\mu}\bigg)(y) \quad\text{for a.e. } y\in Y. \end{align} $$

By Jensen’s inequality for conditional expectations, applied to the convex function ${t\mapsto 1/t}$ , we also have that

(3.7) $$ \begin{align} \frac{1}{E(\sum_{k\in G}\eta_n(hk^{-1}){dk^{-1}\mu}/{d\mu})(y)}\kern1pt{\leq}\kern1pt E\kern-1pt\bigg(\frac{1}{\sum_{k\in G}\eta_n(hk^{-1}){dk^{-1}\mu}/{d\mu}}\bigg)(y) \!\quad\text{for a.e. } y\kern1pt{\in}\kern1pt Y. \end{align} $$

Combining (3.6) and (3.7), we see that

$$ \begin{align*} &\sum_{h\in G}\eta_n^2(h)\int_{Y}\frac{d\nu(y)}{\sum_{k\in G}\eta_n(hk^{-1}){dk^{-1}\nu}/{d\nu}(y)}\\ &\quad\leq \sum_{h\in G}\eta_n^2(h)\int_Y E\bigg(\frac{1}{\sum_{k\in G}\eta_n(hk^{-1}){dk^{-1}\mu}/{d\mu}}\bigg)(y)\,d\nu(y)\\ &\quad=\sum_{h\in G}\eta_n^2(h)\int_X\frac{d\mu(x)}{\sum_{k\in G}\eta_n(hk^{-1}){dk^{-1}\mu}/{d\mu}(x)}, \end{align*} $$

which converges to $0$ as $\eta _n$ is strongly recurrent for $G\curvearrowright (X,\mu )$ .

Proof. (1) We may assume that $t=1$ . So suppose that $G\curvearrowright (X,\mu _1)$ has an invariant mean and fix $s<1$ . Let $\unicode{x3bb} $ be the probability measure on $\{0,1\}$ that is given by ${\unicode{x3bb} (0)=s}$ . Then by [Reference Arano, Isono and MarrakchiAIM19, Proposition A.9] the diagonal action $G\curvearrowright (X\times X\times \{0,1\}^{G}, \mu _1\times \mu _0\times \unicode{x3bb} ^{G})$ has an invariant mean. Since $G\curvearrowright (X,\mu _s)$ is a factor of this diagonal action, it admits a G-invariant mean as well.

(2) It suffices to show that $G\curvearrowright (X,\mu _1)$ is amenable whenever there exists a ${t\in (0,1)}$ such that $G\curvearrowright (X,\mu _t)$ is amenable. Write $\unicode{x3bb} $ for the probability measure on $\{0,1\}$ given by $\unicode{x3bb} (0)=t$ . Then $G\curvearrowright (X,\mu _t)$ is a factor of the diagonal action $G\curvearrowright (X\times X\times \{0,1\}^{G},\mu _1\times \mu _0\times \unicode{x3bb} ^{G})$ , so by [Reference ZimmerZim78, Theorem 2.4] also the latter action is amenable. Since $G\curvearrowright (X\times \{0,1\}^{G},\mu _0\times \unicode{x3bb} ^{G})$ is pmp, we have that $G\curvearrowright (X,\mu _1)$ is amenable.

(3) We may again assume that $t=1$ . Suppose that $(\eta _n)$ is a strongly recurrent sequence of probability measures on G for the action $G\curvearrowright (X,\mu _1)$ . Fix $s<1$ and let $\unicode{x3bb} $ be the probability measure on $\{0,1\}$ defined by $\unicode{x3bb} (0)=s$ . As the diagonal action $G\curvearrowright (X\times \{0,1\}^{G},\mu _0\times \unicode{x3bb} ^{G})$ is pmp, the sequence $\eta _n$ is also strongly recurrent for the diagonal action $G\curvearrowright (X\times X\times \{0,1\},\mu _1\times \mu _0\times \unicode{x3bb} ^{G})$ . Since $G\curvearrowright (X,\mu _t)$ is a factor of $G\curvearrowright (X\times X\times \{0,1\}^{G},\mu _1\times \mu _0\times \unicode{x3bb} ^{G})$ , it follows from Lemma 3.5 that the sequence $\eta _n$ is strongly recurrent for $G\curvearrowright (X,\mu _t)$ .

Proof of Theorem 3.3

For every $t\in (0,1]$ write $\rho ^t$ for the Koopman representation

$$ \begin{align*} \rho^t\colon G\curvearrowright L^2(X,\mu_t):\;\;\;\; (\rho^t_g(\xi))(x)=\left(\frac{dg\mu_t}{d\mu_t}(x)\right)^{1/2}\xi(g^{-1}\cdot x). \end{align*} $$

Fix $s\in (0,1)$ and let $C>0$ be such that $\log (1-x)\geq -C x$ for every $x\in [0,s)$ . Then for every $t<s$ and every $g\in G$ we have that

$$ \begin{align*} \log(\langle \rho^t_g(1), 1\rangle)&=\sum_{h\in G}\log(1-H^2(\mu_{gh}^{t},\mu_h^{t}))\\ &\geq \sum_{h\in G}\log(1-tH^2(\mu_{gh},\mu_h))\\ &\geq -C t\sum_{h\in G}H^2(\mu_{gh},\mu_h). \end{align*} $$

Because $G\curvearrowright (X,\mu _1)$ is non-singular we get that

(3.8) $$ \begin{align} \langle \rho^t_g(1),1\rangle\rightarrow 1 \quad\text{as } t\rightarrow 0, \text{ for every } g\in G. \end{align} $$

We claim that there exists a $t'>0$ such that $G\curvearrowright (X,\mu _t)$ is non-amenable for every ${t<t'}$ . Suppose, to the contrary, that $t_n$ is a sequence that converges to zero such that ${G\curvearrowright (X,\mu _{t_n})}$ is amenable for every $n\in \mathbb {N}$ . Then it follows from [Reference NevoNev03, Theorem 3.7] that $\rho ^{t_n}$ is weakly contained in the left regular representation $\unicode{x3bb} _G$ for every $n\in \mathbb {N}$ . Write $1_G$ for the trivial representation of G. It follows from (3.8) that $\bigoplus _{n\in \mathbb {N}}\rho ^{t_n}$ has almost invariant vectors, so that

$$ \begin{align*} 1_G\prec \bigoplus_{n\in \mathbb{N}}\rho^{t_n}\prec \infty \unicode{x3bb}_G\prec \unicode{x3bb}_G, \end{align*} $$

which is in contradiction to the non-amenability of G. By Theorem 3.1 there exists a ${t_1\in [0,1]}$ such that $G\curvearrowright (X,\mu _t)$ is weakly mixing for every $t<t_1$ . Since every dissipative action is amenable (see, for example, [Reference Arano, Isono and MarrakchiAIM19, Theorem A.29]) it follows that $t_1\geq t'>0$ .

Write $Z_0=[0,1)$ and let $\unicode{x3bb} $ denote the Lebesgue probability measure on $Z_0$ . Let $\rho ^0$ denote the reduced Koopman representation

$$ \begin{align*} \rho^0\colon G\curvearrowright L^2(X\times Z_0^{G},\mu_0\times \unicode{x3bb}^G)\ominus \mathbb{C} 1: \quad (\rho^0_g(\xi))(x)=\xi(g^{-1}\cdot x). \end{align*} $$

As G is non-amenable, $\rho ^{0}$ has stable spectral gap. Suppose that for every $s>0$ we can find $0<s'<s$ such that $\rho ^{s'}$ is weakly contained in $\rho ^{s'}\otimes \rho ^{0}$ . Then there exists a sequence $s_n$ that converges to zero, such that $\rho ^{s_n}$ is weakly contained in $\rho ^{s_n}\otimes \rho ^{0}$ for every $n\in \mathbb {N}$ . This implies that $\bigoplus _{n\in \mathbb {N}}\rho ^{s_n}$ is weakly contained in $(\bigoplus _{n\in \mathbb {N}}\rho ^{s_n})\otimes \rho ^{0}$ . But by (3.8), the representation $\bigoplus _{n\in \mathbb {N}}\rho ^{s_n}$ has almost invariant vectors, so that $(\bigoplus _{n\in \mathbb {N}}\rho ^{s_n})\otimes \rho ^{0}$ weakly contains the trivial representation. This is in contradiction to $\rho ^{0}$ having stable spectral gap. We conclude that there exists an $s>0$ such that $\rho ^t$ is not weakly contained in $\rho ^t\otimes \rho ^0$ for every $t<s$ .

We prove that $G\curvearrowright (X,\mu _t)$ is strongly ergodic for every $t<\min \{t',s\}$ , in which case we can apply [Reference Marrakchi and VaesMV20, Lemma 5.2] to the non-singular action $G\curvearrowright (X,\mu _t)$ and the pmp action $G\curvearrowright (X\times Z_0^{G},\mu _0\times \unicode{x3bb} ^G)$ by our choice of $t'$ and s. After rescaling, we may assume that $G\curvearrowright (X,\mu _1)$ is ergodic and that $\rho ^{t}$ is not weakly contained in $\rho ^{t}\otimes \rho ^{0}$ for every $t\in (0,1)$ .

Let $t\in (0,1)$ be arbitrary and define the map

$$ \begin{align*} \Psi\colon X\times X\times Z_0^{G}\rightarrow X:\quad \Psi(x,y,z)_h=\begin{cases}x_h &\text{if } z_h\leq t,\\ y_h &\text{if } z_h>t.\end{cases} \end{align*} $$

Then $\Psi $ is G-equivariant and we have that $\Psi (\mu _1\times \mu _0\times \unicode{x3bb} ^{G})=\mu _t$ . Suppose that ${G\curvearrowright (X,\mu _t)}$ is not strongly ergodic. Then we can find a bounded almost invariant sequence $f_n\in L^{\infty }(X,\mu _t)$ such that $\|f_n\|_2=1$ and $\mu _t(f_n)=0$ for every $n\in \mathbb {N}$ . Therefore, $\Psi _*(f_n)$ is a bounded almost invariant sequence for $G\curvearrowright (X\times X\times Z_0^{G},\mu _1\times \mu _0\times \unicode{x3bb} ^{G})$ . Let $E\colon L^{\infty }(X\times X\times Z_0^{G})\rightarrow L^{\infty }(X)$ be the conditional expectation that is uniquely determined by $\mu _1\circ E=\mu _1\times \mu _0\times \unicode{x3bb} ^{G}$ . By [Reference Marrakchi and VaesMV20, Lemma 5.2] we have that $\lim _{n\rightarrow \infty }\|(E\circ \Psi _*)(f_n)-\Psi _*(f_n)\|_2=0$ . As $\Psi $ is measure-preserving we get, in particular, that

(3.9) $$ \begin{align} \lim_{n\rightarrow \infty}\|(E\circ \Psi_*)(f_n)\|_2=1. \end{align} $$

Note that if $\mu _t(f)=0$ for some $f\in L^{2}(X,\mu _t)$ , we have that $\mu _1((E\circ \Psi _*)(f))=0$ . So we can view $E\circ \Psi _*$ as a bounded operator

$$ \begin{align*} E\circ\Psi_*\colon L^2(X,\mu_t)\ominus \mathbb{C} 1\rightarrow L^2(X,\mu_1)\ominus \mathbb{C} 1. \end{align*} $$

The claim is in direct contradiction to (3.9), so we conclude that $G\curvearrowright (X,\mu _t)$ is strongly ergodic.

Proof of claim

For every $g\in G$ , let $\varphi _g$ be the map

$$ \begin{align*} \varphi_g\colon L^2(X_0,\mu_g^t)\rightarrow L^2(X_0,\mu_g): \quad \varphi_g(F)=tF+(1-t)\nu(F)\cdot 1. \end{align*} $$

Then $E\circ \Psi _*\colon L^2(X_0,\mu _t)\rightarrow L^2(X,\mu _1)$ is given by the infinite product $\bigotimes _{g\in G}\varphi _g$ . For every $g\in G$ we have that

$$ \begin{align*} \|F\|_{2,\mu_g}=\|(d\mu_g^t/d\mu_g)^{-1/2}F\|_{2,\mu_g^t}\leq t^{-1/2}\|F\|_{2,\mu_g^t}, \end{align*} $$

so that the inclusion map $\iota _g \colon L^2(X_0,\mu _g^t)\hookrightarrow L^2(X_0,\mu _g)$ satisfies $\|\iota _g\|\leq t^{-1/2}$ for every $g\in G$ . We have that

$$ \begin{align*} \varphi_g(F)=t(F-\mu_g(F)\cdot 1)+\mu_t(F)\cdot 1 \quad\text{for every } F\in L^2(X_0,\mu_g^t). \end{align*} $$

So if we write $P_g^t$ for the projection map onto $L^2(X_0,\mu _g^t)\ominus \mathbb {C} 1$ , and $P_g$ for the projection map onto $L^2(X_0,\mu _g)\ominus \mathbb {C} 1$ , we have that

(3.10) $$ \begin{align} \varphi_g\circ P^t_g=t(P_g\circ \iota_g)\quad\text{for every } g\in G. \end{align} $$

For a non-empty finite subset $\mathcal {F}\subset G$ let $V(\mathcal {F})$ be the linear subspace of $L^2(X,\mu _t)\ominus \mathbb {C} 1$ spanned by

$$ \begin{align*}\bigg(\bigotimes_{g\in \mathcal{F}}L^2(X_0,\mu_g^t)\ominus \mathbb{C} 1\bigg)\otimes \bigotimes_{g\in G\setminus \mathcal{F}}1. \end{align*} $$

Then, using (3.10), we see that

$$ \begin{align*} \|(E\circ \Psi_*)(f)\|_2\leq t^{|\mathcal{F}|/2}\|f\|_2 \quad\text{for every } f\in V(\mathcal{F}). \end{align*} $$

Since $\bigoplus _{\mathcal {F}\neq \emptyset }V(\mathcal {F})$ is dense inside $L^2(X,\mu _t)\ominus \mathbb {C} 1$ , we have that

$$ \begin{align*} \|(E\circ\Psi_*)|_{L^2(X,\mu_t)\ominus\mathbb{C} 1} \|\leq t^{1/2}<1.\\[-3.1pc] \end{align*} $$

This also concludes the proof of Theorem 3.3.

Proof. Take $g\in G\setminus \{e\}$ . It suffices to show that $\mu (\{x\in X:g\cdot x=x\})=0$ . If g is elliptic, there exist disjoint infinite subtrees $T_1,T_2\subset T$ such that $g\cdot T_1=T_2$ . Note that

$$ \begin{align*} (X_1,\mu_1)=\prod_{e\in E(T_1)}(X_0,\mu_e) \quad\text{and}\quad (X_2,\mu_2)=\prod_{e\in E(T_2)}(X_0,\mu_e) \end{align*} $$

are non-atomic and that g induces a non-singular isomorphism $\varphi \colon (X_1,\mu _1)\rightarrow (X_2,\mu _2): \varphi (x)_e=x_{g^{-1}\cdot e}$ . We get that

$$ \begin{align*} \mu_1\times \mu_2(\{(x, \varphi(x)): x\in X_1\})=0. \end{align*} $$

A fortiori $\mu (\{x\in X:g\cdot x=x\})=0$ . If g is hyperbolic, let $L_g\subset T$ denote its axis on which it acts by non-trivial translation. Then $\prod _{e\in E(L_g)}(X_0,\mu _e)$ is non-atomic and by [Reference Berendschot and VaesBKV19, Lemma 2.2] the action $g^{\mathbb {Z}}\curvearrowright \prod _{e\in E(L_g)}(X_0,\mu _e)$ is essentially free. This implies that also $\mu (\{x\in X:g\cdot x=x\})=0$ .

Proof of Theorem 4.2

Define a family $(X_e)_{e\in E}$ of independent random variables on $(X,\mu )=\prod _{e\in E}(X_0,\mu _e)$ by

(4.3) $$ \begin{align} X_e(x)=\begin{cases}\log (d\mu_1/d\mu_0) (x_e)&\text{if } e \text{ is oriented towards } \rho,\\ \log (d\mu_0/d\mu_1)(x_e) &\text{if } e \text{ is oriented away from } \rho.\end{cases} \end{align} $$

For $v\in T$ we write

$$ \begin{align*} S_v=\sum_{e\in E([\rho, v])} X_e. \end{align*} $$

Then we have that

$$ \begin{align*} \frac{dg\mu}{d\mu}=\exp(S_{g\cdot \rho})\quad\text{for every }g\in G. \end{align*} $$

Since $G\subset \operatorname {Aut}(T)$ is a closed subgroup, for each $v\in T$ the stabilizer subgroup $G_v=\{g\in G:g\cdot v= v\}$ is a compact open subgroup of G.

Suppose that $1-H^2(\mu _0,\mu _1)<\exp (-\delta /2)$ . Then we have that

$$ \begin{align*} \int_X\sum_{v\in G\cdot \rho}\exp(S_v(x)/2)\,d\mu(x)=\sum_{v\in G\cdot \rho}(1-H^2(\mu_0,\mu_1))^{2d(\rho,v)}<+\infty, \end{align*} $$

by definition of the Poincaré exponent. Therefore, we have that $\sum _{v\in G\cdot \rho }\exp (S_v(x)/2)<+\infty $ for a.e. $x\in X$ . Let $\unicode{x3bb} $ denote the left invariant Haar measure on G and define ${L=\unicode{x3bb} (G_\rho )}$ , where $G_\rho =\{g\in G:g\cdot \rho =\rho \}$ . Then we have that

$$ \begin{align*} \int_{G}\frac{dg\mu}{d\mu}(x)\,d\unicode{x3bb}(g)=L\sum_{v\in G\cdot \rho }\exp(S_v(x))<+\infty\quad \text{for a.e. } x\in X. \end{align*} $$

We conclude that $G\curvearrowright (X,\mu )$ is dissipative up to compact stabilizers.

Now assume that $1-H^2(\mu _0,\mu _1)>\exp (-\delta /2)$ . We start by proving that $G\curvearrowright (X,\mu )$ is infinitely recurrent. By [Reference Arano, Isono and MarrakchiAIM19, Theorem 8.17] we can find a non-elementary closed compactly generated subgroup $G'\subset G$ such that $1-H^2(\mu _0,\mu _1)>\exp (-\delta (G')/2)$ . Let $T'\subset T$ be the unique minimal $G'$ -invariant subtree. Then $G'$ acts cocompactly on $T'$ and we have that $\delta (G')=\dim _{H}\partial T'$ . Let X and Y be independent random variables with distributions $(\log d\mu _1/d\mu _0)_*\mu _0$ and $(\log d\mu _0/d\mu _1)_*\mu _1$ , respectively. Set $Z=X+Y$ and write

$$ \begin{align*} \varphi(t)=\mathbb{E}(\exp(tZ)). \end{align*} $$

The assignment $t\mapsto \varphi (t)$ is convex, $\varphi (t)=\varphi (1-t)$ for every t and $\varphi (1/2)= (1-H^2(\mu _0,\mu _1))^2$ . We conclude that

$$ \begin{align*} \inf_{t\geq 0}\varphi(t)=(1-H^2(\mu_0,\mu_1))^{2}. \end{align*} $$

Write $R_k$ for the sum of k independent copies of Z. By the Chernoff–Cramér theorem, as stated in [Reference Lyons and PemantleLP92], there exists an $M\in \mathbb {N}$ such that

(4.4) $$ \begin{align} \mathbb{P}(R_M\geq 0)>\exp(-M\delta(G')). \end{align} $$

Below we define a new unoriented tree S. This means that the edge set of S consists of subsets $\{v,w\}\subset V(S)$ . Fix a vertex $\rho '\in T'$ and define the unoriented tree S as follows.

• • S has vertices $v\in T'$ so that $d_{T'}(\rho ', v)$ is divisible by M.

• • There is an edge $\{v,w\}\in E(S)$ between two vertices $v,w\in S$ if $d_{T'}(v,w)=M$ and $[\rho ',v]_{T'}\subset [\rho ',w]_{T'}$ .

Here the notation $[\rho ',v]_{T'}$ means that we consider the line segment $[\rho ',v]$ as a subtree of $T'$ . We have that $\dim _H\partial S=M\dim _H \partial T'= M\delta (G')$ . Form a random subgraph $S(x)$ of S by deleting those edges $\{v,w\}\in E(S)$ where

$$ \begin{align*} \sum_{e\in E([v,w]_{T'})}X_e(x_e)<0. \end{align*} $$

This is an edge percolation on S, where each edge remains with probability ${p=\mathbb {P}(R_M\geq 0)}$ . So by (4.4) we have that $p\exp (\dim _H S)>1$ . Furthermore, if $\{v,w\}$ and $\{v',w'\}$ are edges of S so that $E([v,w]_{T'})\cap E([v',w']_{T'})=\emptyset $ , their presence in $S(x)$ constitutes independent events. So the percolation process is a quasi-Bernoulli percolation as introduced in [Reference LyonsLyo89]. Taking $w\in (1,p\exp (\dim _H S))$ and setting $w_n=w^{-n}$ , it follows from [Reference LyonsLyo89, Theorem 3.1] that percolation occurs almost surely, that is, $S(x)$ contains an infinite connected component for a.e. $x\in X$ . Writing

$$ \begin{align*} S^{\prime}_v(x)=\sum_{e\in E([\rho',v]_{T'})}X_e(x_e), \end{align*} $$

this means that for a.e. $x\in (X,\mu )$ we can find a constant $a_x>-\infty $ such that $S^{\prime }_v(x)>a_x$ for infinitely many $v\in T'$ . As $T'/G'$ is finite, there exists a vertex $w\in T'$ such that

(4.5) $$ \begin{align} \sum_{v\in G'\cdot w}\exp(S^{\prime}_v(x))=+\infty \quad\text{with positive probability.} \end{align} $$

Therefore, by Kolmogorov’s zero–one law, we have that $\sum _{v\in G'\cdot w}\exp (S^{\prime }_v(x))=+\infty $ almost surely. Since a change of root results in a conjugate action, we may assume that $\rho =w$ . Then (4.5) implies that $\sum _{v\in G\cdot \rho }\exp (S_v(x))=+\infty $ for a.e. $x\in X$ . Writing again L for the Haar measure of the stabilizer subgroup $G_\rho =\{g\in G:g\cdot \rho = \rho \}$ , we see that

$$ \begin{align*} \int_{G}\frac{dg\mu}{d\mu}\,d\unicode{x3bb}(g)=L\sum_{v\in G\cdot \rho}\exp(S_v)=+\infty\quad\text{almost surely.} \end{align*} $$

We conclude that $G\curvearrowright (X,\mu )$ is infinitely recurrent. We prove that $G\curvearrowright (X,\mu )$ is weakly mixing using a phase transition result from the previous section. Define the measurable map

$$ \begin{align*} \psi\colon X_0\rightarrow (0,1]:\quad \psi(x)=\min\{d\mu_1/d\mu_0(x),1\}. \end{align*} $$

Let $\nu $ be the probability measure on $X_0$ determined by

$$ \begin{align*} \frac{d\nu}{d\mu_0}(x)=\rho^{-1}\psi(x)\quad \text{where } \rho=\int_{X_0}\psi(x)\,d\mu_0(x). \end{align*} $$

Then we have that $\nu \sim \mu _0$ and for every $s>1-\rho $ the probability measures

$$ \begin{align*} \eta_0^s&=s^{-1}(\mu_0-(1-s)\nu),\\ \eta_1^s&=s^{-1}(\mu_1-(1-s)\nu) \end{align*} $$

are well defined. We consider the non-singular actions $G\curvearrowright (X,\eta _s)=\prod _{e\in E}(X_0,\eta _e^s)$ , where

$$ \begin{align*} \eta_e^s=\begin{cases}\eta_0^s &\text{if } e \text{ is oriented towards } \rho,\\ \eta_1^s &\text{if } e \text{ is oriented away from } \rho.\end{cases} \end{align*} $$

By the dominated convergence theorem we have that $H^2(\eta _0^s,\eta _1^s)\rightarrow H^2(\mu _0,\mu _1)$ as $s\rightarrow 1$ . So we can choose s close enough to $1$ , but not equal to $1$ , such that $1-H^2(\eta _0^s,\eta _1^s)>\exp (-\delta /2)$ . By the first part of the proof we have that $G\curvearrowright (X,\eta _s)$ is infinitely recurrent. Note that

$$ \begin{align*} \mu_j=(1-s)\nu+s\eta_j^s\quad\text{for } j=0,1. \end{align*} $$

Since we assumed that $G\subset \operatorname {Aut}(T)$ is closed, all the stabilizer subgroups $G_{v}=\{g\in G:g\cdot v=v\}$ are compact. By Remark 3.4 we conclude that $G\curvearrowright (X,\mu )$ is weakly mixing.

Let $G\curvearrowright (Y,\nu )$ be an ergodic pmp action. To determine the Krieger flow and the flow of weights of $\beta \colon G\curvearrowright X\times Y$ we use a similar approach to [Reference Arano, Isono and MarrakchiAIM19, Theorem 10.4] and [Reference Vaes and WahlVW17, Proposition 7.3]. First we determine the Krieger flow and then we deal with the flow of weights.

As before, let $G'\subset G$ be a non-elementary compactly generated subgroup such that ${1-H^2(\mu _0,\mu _1)>\exp (-\delta (G')/2)}$ . By [Reference Arano, Isono and MarrakchiAIM19, Theorem 8.7] we may assume that $G/G'$ is not compact. Let $T'\subset T$ be the minimal $G'$ -invariant subtree. Let $v\in T'$ be as in Lemma 4.5 below so that

(4.6) $$ \begin{align} \bigcap_{g\in G} \left(E(gT')\cup E([v,g^{-1}\cdot v])\right)=\emptyset. \end{align} $$

Since changing the root yields a conjugate action, we may assume that $\rho =v$ . Let $(Z_0,\zeta _0)$ be a standard probability space such that there exist measurable maps $\theta _0,\theta _1\colon Z_0\rightarrow X_0$ that satisfy $(\theta _0)_*\zeta _0=\mu _0$ and $(\theta _1)_*\zeta _0=\mu _1$ . Write

$$ \begin{align*}(Z,\zeta)&=\prod_{e\in E(T)\setminus E(T')}(Z_0,\zeta_0),\\(X_1,\rho_1)&=\prod_{e\in E(T)\setminus E(T')}(X_0,\mu_e), \\(X_2,\rho_2)&=\prod_{e\in E(T')}(X_0,\mu_e).\end{align*} $$

By the first part of the proof we have that $G'\curvearrowright (X_2,\rho _2)$ is infinitely recurrent. Define the pmp map

$$ \begin{align*} \Psi\colon (Z,\zeta)\rightarrow (X_1,\rho_1):\;\;\;\; (\Psi(z))_e=\begin{cases}\theta_0(z_e) &\text{ if } e \text{ is oriented towards } \rho,\\ \theta_1(z_e) &\text{ if } e \text{ is oriented away from } \rho.\end{cases} \end{align*} $$

Consider

$$ \begin{align*} U=\{e\in E(T): e \text{ is oriented towards } \rho\}. \end{align*} $$

Since $gU\triangle U=E(T)([\rho ,g\cdot \rho ])\subset E(T')$ for any $g\in G'$ , the set $(E(T)\setminus E(T'))\cap U$ is $G'$ -invariant. Therefore, $\Psi $ is a $G'$ -equivariant factor map. Consider the Maharam extensions

$$ \begin{align*} G'\curvearrowright Z\times X_2\times Y\times \mathbb{R}\quad\text{and}\quad G\curvearrowright X\times Y\times \mathbb{R} \end{align*} $$

of the diagonal actions $G'\curvearrowright Z\times X_2\times Y$ and $G'\curvearrowright X\times Y\times \mathbb {R}$ , respectively. Identifying $(X,\mu )=(X_1,\rho _1)\times (X_2,\rho _2)$ , we obtain a $G'$ -equivariant factor map

$$ \begin{align*} \Phi\colon Z\times X_2\times Y\times \mathbb{R}\rightarrow X_1\times X_2\times Y\times \mathbb{R}:\quad \Phi(z,x,y,t)=(\Psi(z),x,y,t). \end{align*} $$

Take $F\in L^{\infty }(X\times Y\times \mathbb {R})^{G}$ . By [Reference Arano, Isono and MarrakchiAIM19, Proposition A.33] the Maharam extension $G'\curvearrowright X_2\times Y\times \mathbb {R} $ is infinitely recurrent. Since $G'\curvearrowright Z$ is a mixing pmp generalized Bernoulli action we have that $F\circ \Phi \in L^{\infty }(Z\times X_2\times Y\times \mathbb {R})^{G}\subset 1\mathbin {\overline {\otimes }} L^{\infty }(X_2\times Y\times \mathbb {R})^{G}$ by [Reference Schmidt and WaltersSW81, Theorem 2.3]. Therefore, F is essentially independent of the $E(T)\setminus E(T')$ -coordinates. Thus, for any $g\in G$ the assignment

$$ \begin{align*} (x,y,t)\mapsto F(g\cdot x,y,t)=F(x,y,t-\log(dg^{-1}\mu/d\mu)(x)) \end{align*} $$

is essentially independent of the $E(T)\setminus E(gT')$ -coordinates. Since $\log (dg^{-1}\mu /d\mu )$ only depends on the $E([\rho ,g^{-1}\cdot \rho ])$ -coordinates, we deduce that F is essentially independent of the $E(T)\setminus (E(gT')\cup E([\rho ,g^{-1}\cdot \rho ]))$ -coordinates, for every $g\in G$ . Therefore, by (4.6), we have that $F\in 1\mathbin {\overline {\otimes }} L^{\infty }(Y\times \mathbb {R})$ .

So we have proven that any G-invariant function $F\in L^{\infty }(X\times Y\times \mathbb {R})$ is of the form $F(x,y,t)=H(y,t)$ , for some $H\in L^{\infty }(Y\times \mathbb {R})$ that satisfies

$$ \begin{align*} H(y,t)=H(g\cdot y, t+\log(dg^{-1}\mu/d\mu)(x)) \quad\text{for a.e. } (x,y,t)\in X\times Y\times \mathbb{R}. \end{align*} $$

Since $0$ is in the essential range of the maps $\log (dg\mu /d\mu )$ , for every $g\in G$ , we see that $H(g\cdot y,t)=H(y,t)$ for a.e. $(y,t)\in Y\times \mathbb {R}$ . By ergodicity of $G\curvearrowright Y$ , we conclude that H is of the form $H(y,t)=P(t)$ , for some $P\in L^{\infty }(\mathbb {R})$ that satisfies

(4.7) $$ \begin{align} P(t)=P(t+\log(dg^{-1}\mu/d\mu)(x))\quad\text{for a.e. } (x,t)\in X\times \mathbb{R}, \text{ for every } g\in G. \end{align} $$

Let $\Gamma \subset \mathbb {R}$ be the subgroup generated by the essential ranges of the maps $\log (dg\mu /d\mu )$ , for $g\in G$ . If $\Gamma =\{0\}$ we can identify $L^{\infty }(X\times Y\times \mathbb {R})^{G}\cong L^{\infty }(\mathbb {R})$ . If $\Gamma \subset \mathbb {R}$ is dense, then it follows that P is essentially constant so that the Maharam extension $G\curvearrowright X\times Y\times \mathbb {R}$ is ergodic, that is, the Krieger flow of $G\curvearrowright X\times Y$ is trivial. If $\Gamma =a\mathbb {Z}$ , with $a>0$ , we conclude by (4.7) that we can identify $L^{\infty }(X\times Y\times \mathbb {R})^{G}\cong L^{\infty }(\mathbb {R}/a\mathbb {Z})$ , so that the Krieger flow of $G\curvearrowright X\times Y$ is given by $\mathbb {R}\curvearrowright \mathbb {R}/a\mathbb {Z}$ . Finally, note that the closure of $\Gamma $ equals the closure of the subgroup generated by the essential range of the map

$$ \begin{align*} X_0\times X_0\rightarrow \mathbb{R}\colon \quad (x,x')\mapsto \log(d\mu_0/d\mu_1)(x)-\log(d\mu_0/d\mu_1)(x'). \end{align*} $$

So we have calculated the Krieger flow in every case, concluding the proof of the theorem in the case where G is unimodular.

When G is not unimodular, let $G_0=\ker \Delta $ be the kernel of the modular function. Let $G\curvearrowright X\times Y\times \mathbb {R}$ be the modular Maharam extension and let $\alpha \colon G_0\curvearrowright X\times Y\times \mathbb {R}$ be its restriction to the subgroup $G_0$ . Then we have that

$$ \begin{align*} L^{\infty}(X\times Y\times \mathbb{R})^{G}\subset L^{\infty}(X\times Y \times \mathbb{R})^{\alpha}. \end{align*} $$

By [Reference Arano, Isono and MarrakchiAIM19, Theorem 8.16] we have that $\delta (G_0)=\delta $ , and we can apply the argument above to conclude that $L^{\infty }(X\times Y\times \mathbb {R})^{\alpha }\subset 1\mathbin {\overline {\otimes }} 1\mathbin {\overline {\otimes }} L^{\infty }(\mathbb {R})$ . So for every $F\in L^{\infty }(X\times Y\times \mathbb {R})^{G}$ there exists a $P\in L^{\infty }(\mathbb {R})$ such that

(4.8) $$ \begin{align} P(t)\kern1.3pt{=}\kern1.3pt P(t\kern1.3pt{+}\log(dg^{-1}\mu/d\mu)(x)\kern1.3pt{+}\kern1.3pt\log(\Delta(g))) \!\quad\text{for a.e. } (x,t)\kern1.3pt{\in}\kern1.3pt X\kern1.3pt{\times}\kern1.3pt \mathbb{R}, \text{ for every } g\kern1.3pt{\in}\kern1.3pt G. \end{align} $$

Let $\Pi $ be the subgroup of $\mathbb {R}$ generated by the essential range of the maps

$$ \begin{align*} x\mapsto \log(dg^{-1}\mu/d\mu)(x)+\log(\Delta(g)) \quad\text{with } g\in G. \end{align*} $$

As $0$ is contained in the essential range of $\log (dg^{-1}\mu /d\mu )$ , for every $g\in G$ , we get that $\log (\Delta (G))\subset \Pi $ . Therefore, $\Pi $ also contains the subgroup $\Gamma \subset \mathbb {R}$ defined above. Thus, the closure of $\Pi $ equals the closure of $\Sigma $ , where $\Sigma \subset \mathbb {R}$ is the subgroup as in the statement of the theorem. From (4.8) we conclude that we may identify $L^{\infty }(X\times Y\times \mathbb {R})^{G}\cong L^{\infty }(\mathbb {R})^{\Sigma }$ , so that the flow of weights of $G\curvearrowright X\times Y$ is as stated in the theorem.

Proof. Let $k\in H$ be a hyperbolic element and let $L\subset T$ be its axis, on which k acts by a non-trivial translation. Then $L\subset S$ , as one can show for instance as in the proof of [Reference Caprace and de MedtsCM11, Proposition 3.8]. Pick any vertex $v\in L$ . We claim that this vertex will satisfy (4.9). Take any $w\in V(T)\setminus \{v\}$ . As $G/H$ is not compact, one can show as in [Reference Arano, Isono and MarrakchiAIM19, Theorem 9.7] that there exists a $g\in G$ such that $g\cdot w\notin S$ . Since k acts by translation on L, there exists an $n\in \mathbb {N}$ large enough such that

$$ \begin{align*} [v,k\cdot v]\subset [v,k^ng\cdot v]\quad\text{and}\quad [v,k^{-1}\cdot v]\subset [v,k^{-n}g\cdot v], \end{align*} $$

so that in particular we have that $w\notin [v,k^ng\cdot v]\cap [v,k^{-n}g\cdot v]=\{v\}$ . Since S is H-invariant, we also have that $k^ng\cdot w\notin S$ and $k^{-n}g\cdot w\notin S$ and we conclude that

$$ \begin{align*} w\notin ((k^ng)^{-1}S\cup[v,k^ng\cdot v])\cap((k^{-n}g)^{-1}S\cup [v,k^{-n}g\cdot v]).\\[-3.1pc] \end{align*} $$

Proof of Proposition 4.3

Define the family $(X_e)_{e\in E}$ of independent random variables on $(X,\mu )$ by (4.3) and write

$$ \begin{align*} S_v=\sum_{e\in E([\rho,v])}X_e. \end{align*} $$

Claim. There exists a $\delta>0$ such that

$$ \begin{align*} \mu(\{x\in X:S_v(x)\leq -\delta\quad\text{for every } v\in T\setminus \{\rho\}\})>0. \end{align*} $$

Proof of claim

Note that $\mathbb {E}(\exp (X_e/2))=1-H^2(\mu _0,\mu _1)$ for every $e\in E$ . Define a family of random variables $(W_n)_{n\geq 0}$ on $(X,\mu )$ by

$$ \begin{align*} W_n=\sum_{\substack{v\in T\\d(v,\rho)=n}}\exp(S_v/2). \end{align*} $$

Using that $1-H^2(\mu _0,\mu _1)=(q-1)^{-1/2}$ , one computes that

$$ \begin{align*} \mathbb{E}(W_{n+1}|\;S_v,\; d(v,\rho)\leq n)=W_n\quad\text{for every } n\geq 1. \end{align*} $$

So the sequence $(W_n)_{n\geq 0}$ is a martingale, and since it is positive it converges almost surely to a finite limit when $n\rightarrow +\infty $ . Write $\Sigma _n=\{v\in T:d(v,\rho )=n\}$ . As ${W_n\geq \max _{v\in \Sigma _n}\exp (S_v/2)}$ we conclude that there exists a positive constant $C<+\infty $ such that

$$ \begin{align*} \mathbb{P}(S_v\leq C\text{ for every }v\in T)>0. \end{align*} $$

For any vertex $w\in T$ , write $T_w=\{v\in T:[\rho ,w]\subset [\rho ,v]\}$ : the set of children of w, including w itself. Using the symmetry of the tree and changing the root from $\rho $ to $w\in T$ , we also have that

(4.10) $$ \begin{align} \mathbb{P}(S_v-S_w\leq C\text{ for every } v\in T_w)>0\quad\text{for every } w\in T. \end{align} $$

Set $\nu _0=(\log d\mu _1/d\mu _0)_*\mu _0$ and $\nu _1=(\log d\mu _0/d\mu _1)_*\mu _1$ . Because $1-H^2(\mu _0,\mu _1)\neq 0$ we have that $\mu _0\neq \mu _1$ , so that there exists a $\delta>0$ such that

$$ \begin{align*} \nu_0*\nu_1((-\infty,-\delta))>0. \end{align*} $$

Here $\nu _0*\nu _1$ denotes the convolution product of $\nu _0$ with $\nu _1$ . Therefore, there exists $N\in \mathbb {N}$ large enough such that

(4.11) $$ \begin{align} \mathbb{P}(S_w\leq -C-\delta \text{ for every } w\in \Sigma_N \text{ and } S_{w'}\leq-\delta \text{ for every } w'\in \Sigma_n \text{ with } n\leq N)>0. \end{align} $$

Since for any $w\in \Sigma _N$ and $w'\in \Sigma _n$ with $n\leq N$ , we have that $S_v-S_w$ is independent of $S_{w'}$ for every $v\in T_w$ , and since $\Sigma _N$ is a finite set, it follows from (4.10) and (4.11) that

$$ \begin{align*} \mathbb{P}(S_v\leq -\delta \quad\text{for every } v\in T\setminus\{\rho\})>0. \end{align*} $$

This concludes the proof of the claim.

Let $\delta>0$ be as in the claim and define

$$ \begin{align*} \mathcal{U}=\{x\in X:S_v(x)\leq -\delta\text{ for every } v\in T\setminus \{\rho\}\}, \end{align*} $$

so that $\mu (\mathcal {U})>0$ . Let $G_\rho $ be the stabilizer subgroup of $\rho $ . Note that for every $g,h\in G$ we have that $S_{hg\cdot \rho }(x)=S_{g\cdot \rho }(h^{-1}\cdot x)+S_{h\cdot \rho }(x)$ for a.e. $x\in X$ , so that for $h\in G$ we have that

$$ \begin{align*} h\cdot \mathcal{U}\subset \{x\in X:S_{hg\cdot \rho}(x)\leq -\delta+S_{h\cdot \rho}(x)\text{ for every } g\notin G_{\rho}\}. \end{align*} $$

It follows that if $h\notin G_{\rho }$ , we have that

$$ \begin{align*} \mathcal{U}\cap h\cdot \mathcal{U}\subset \{x\in X:S_{h\cdot \rho}(x)\leq -\delta\text{ and }S_{h\cdot \rho}(x)\geq \delta\}=\emptyset. \end{align*} $$

Since $G\subset \operatorname {Aut}(T)$ is closed, we have that $G_\rho $ is compact. So the action $G\curvearrowright (X,\mu )$ is not infinitely recurrent. Let $\unicode{x3bb} $ denote the left invariant Haar measure on G. By an adaptation of the proof of [Reference Björklund, Kosloff and VaesBV20, Proposition 4.3], the set

$$ \begin{align*} D=\bigg\{ x\in X: \int_{G}\frac{dg\mu}{d\mu}(x)\,d\unicode{x3bb}(g)<+\infty\bigg\}=\bigg\{ x\in X: \int_{G}\exp(S_{g\cdot \rho}(x))\,d\unicode{x3bb}(g)<+\infty\bigg\} \end{align*} $$

satisfies $\mu (D)\in \{0,1\}$ . Since $G\curvearrowright (X,\mu )$ is not infinitely recurrent, it follows from [Reference Arano, Isono and MarrakchiAIM19, Proposition A.28] that $\mu (D)>0$ , so that we must have that $\mu (D)=1$ . By [Reference Arano, Isono and MarrakchiAIM19, Theorem A.29] the action $G\curvearrowright (X,\mu )$ is dissipative up to compact stabilizers.

Proof of Proposition 4.4

It follows from Theorem 4.2 and Proposition 4.3 that the action $G\curvearrowright (X,\mu )$ , given by (4.2), is dissipative when $1-H^2(\mu _0,\mu _1)\leq (2d-1)^{-1/2}$ and weakly mixing when $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/2}$ . So it remains to show that $G\curvearrowright (X,\mu )$ is non-amenable when $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/2}$ and strongly ergodic when $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/4}$ .

Assume first that $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/2}$ . By taking the kernel of a surjective homomorphism $\mathbb {F}_d\rightarrow \mathbb {Z}$ we find a normal subgroup $H_1\subset \mathbb {F}_d$ that is free on infinitely many generators. By [Reference Roblin and TapieRT13, Théorème 0.1] we have that $\delta (H_1)=(2d-1)^{-1/2}$ . Then, using [Reference SullivanSul79, Corollary 6], we can find a finitely generated free subgroup $H_2\subset H_1$ such that $H_1=H_2*H_3$ for some free subgroup $H_3\subset H_1$ and such that $1-H^2(\mu _0,\mu _1)>\exp (-\delta (H_2)/2)$ . Let $\psi \colon H_1\rightarrow H_3$ be the surjective group homomorphism uniquely determined by

$$ \begin{align*} \psi(h)=\begin{cases}e &\text{if } h\in H_2,\\ h&\text{if } h\in H_3.\end{cases} \end{align*} $$

We set $N=\ker \psi $ , so that $H_2\subset N$ and we get that $1-H^2(\mu _0,\mu _1)>\exp (-\delta (N)/2)$ . Therefore, $N\curvearrowright (X,\mu )$ is ergodic by Theorem 4.2. Also we have that $H_1/N\cong H_3$ , which is a free group on infinitely many generators. Therefore, $H_1\curvearrowright (X,\mu )$ is non-amenable by [Reference Marrakchi and VaesMV20, Lemma 6.4]. A posteriori also $\mathbb {F}_d\curvearrowright (X,\mu )$ is non-amenable.

Let $\pi $ be the Koopman representation of the action $\mathbb {F}_d\curvearrowright (X,\mu )$ :

$$ \begin{align*} \pi\colon G\curvearrowright L^2(X,\mu):\quad (\pi_g(\xi))(x)=\left(\frac{dg\mu}{d\mu}(x)\right)^{1/2}\xi(g^{-1}\cdot x). \end{align*} $$

Proof of claim

Let $\eta $ denote the canonical symmetric measure on the generator set of $\mathbb {F}_d$ and define

$$ \begin{align*} P=\sum_{g\in \mathbb{F}_d}\eta(g)\pi_g. \end{align*} $$

The $\eta $ -spectral radius of $\alpha \colon \mathbb {F}_d\curvearrowright (X,\mu )$ , which we denote by $\rho _\eta (\alpha )$ , is by definition the norm of P, as a bounded operator on $L^2(X,\mu )$ . By [Reference Arano, Isono and MarrakchiAIM19, Proposition A.11] we have that

$$ \begin{align*} \rho_\eta(\alpha)&=\lim_{n\rightarrow \infty}\langle P^n(1), 1\rangle^{1/n}\\ &=\lim_{n\rightarrow \infty}\bigg(\sum_{g\in \mathbb{F}_d}\eta^{*n}(g)(1-H^2(\mu_0,\mu_1))^{2|g|}\bigg)^{1/n}, \end{align*} $$

where $|g|$ denotes the word length of a group element $g\in \mathbb {F}_d$ . By [Reference Arano, Isono and MarrakchiAIM19, Theorem 6.10] we then have that

$$ \begin{align*} \rho_\eta(\alpha)=\frac{(1-H^2(\mu_0,\mu_1))^2}{2d}\left((2d-1)+(1-H^2(\mu_0,\mu_1))^{-4}\right) \end{align*} $$

if $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/4}$ , and

$$ \begin{align*} \rho_\eta(\alpha)= \frac{\sqrt{2d-1}}{d} \end{align*} $$

if $1-H^2(\mu _0,\mu _1)\leq (2d-1)^{-1/4}$ . Therefore, if $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/4}$ , we have that $\rho _\eta (\alpha )>\rho _\eta (\mathbb {F}_d)$ , where $\rho _\eta (\mathbb {F}_d)$ denotes the $\eta $ -spectral radius of the left regular representation. This implies that $\alpha $ is not weakly contained in the left regular representation (see, for instance, [Reference Anantharaman-DelarocheAD03, §3.2]).

Now assume that $1-H^2(\mu _0,\mu _1)>(2d-1)^{-1/4}$ . As in the proof of Theorem 4.2 there exist probability measures $\nu , \eta _0$ and $\eta _1$ on $X_0$ that are equivalent to $\mu _0$ and a number $s\in (0,1)$ such that

$$ \begin{align*} \mu_j=(1-s)\nu+s\eta_j\quad\text{for } j=0,1, \end{align*} $$

and such that $1-H^2(\eta _0,\eta _1)>(2d-1)^{-1/4}$ . Consider the non-singular action

$$ \begin{align*} \mathbb{F}_d\curvearrowright (X,\eta)=\prod_{e\in E(T)}(X_0,\eta_e)\quad \text{where } \eta_e=\begin{cases}\eta_0&\text{if } e \text{ is oriented towards } \rho,\\ \eta_1 &\text{if } e \text{ is oriented away from } \rho.\end{cases} \end{align*} $$

By Theorem 4.2 the action $\mathbb {F}_d\curvearrowright (X,\eta )$ is ergodic. Write $\rho $ for the Koopman representation associated to $\mathbb {F}_d\curvearrowright (X,\eta )$ . By the claim, $\rho $ is not weakly contained in the left regular representation. Let $\unicode{x3bb} $ be the probability measure on $\{0,1\}$ given by ${\unicode{x3bb} (0)=s}$ . Let $\rho ^0$ be the reduced Koopman representation of the pmp generalized Bernoulli action ${\mathbb {F}_d\curvearrowright (X\times \{0,1\}^{E(T)},\nu ^{E(T)}\times \unicode{x3bb} ^{E(T)})}$ . Then $\rho ^0$ is contained in a multiple of the left regular representation. Therefore, as $\rho $ is not weakly contained in the left regular representation, $\rho $ is not weakly contained in $\rho \otimes \rho ^{0}$ .

Define the map

$$ \begin{align*} \Psi\colon X\times X\times \{0,1\}^{E(T)}\rightarrow X\colon \;\;\;\; \Psi(x,y,z)_e=\begin{cases}x_e &\text{if } z_e=0,\\ y_e &\text{if } z_e=1.\end{cases} \end{align*} $$

Then $\Psi $ is $\mathbb {F}_d$ -equivariant and we have that $\Psi _*(\eta \times \nu ^{E(T)}\times \unicode{x3bb} ^{E(T)})=\mu $ . Suppose that $\mathbb {F}_d\curvearrowright (X,\mu )$ is not strongly ergodic. Then there exists a bounded almost invariant sequence $f_n\in L^{\infty }(X,\mu )$ such that $\|f_n\|_2=1$ and $\mu (f_n)=0$ for every $n\in \mathbb {N}$ . Therefore, $\Psi _*(f_n)$ is a bounded almost invariant sequence for the diagonal action ${\mathbb {F}_d\curvearrowright (X\times X\times \{0,1\}^{E(T)},\eta \times \nu ^{E(T)}\times \unicode{x3bb} ^{E(T)})}$ . Let $E\colon L^{\infty }(X\times X\times \{0,1\}^{E(T)})\rightarrow L^{\infty }(X)$ be the conditional expectation that is uniquely determined by $\mu \circ E=\eta \times \nu ^{E(T)}\times \unicode{x3bb} ^{E(T)}$ . By [Reference Marrakchi and VaesMV20, Lemma 5.2] we have that $\lim _{n\rightarrow \infty }\|(E\circ \Psi _*)(f_n)-\Psi _{*}(f_n)\|_2=0$ , and in particular we get that

(4.12) $$ \begin{align} \lim_{n\rightarrow \infty}\|(E\circ \Psi_*)(f_n)\|_2=1. \end{align} $$

But just as in the proof of Theorem 3.3 we have that

$$ \begin{align*} \left\|(E\circ\Psi_*)\big|_{L^2(X,\mu)\ominus \mathbb{C}1}\right\|<1, \end{align*} $$

which is in contradiction with (4.12). We conclude that $\mathbb {F}_d\curvearrowright (X,\mu )$ is strongly ergodic.

Proof. Let $H\subset \operatorname {Aut}(T)$ be the closure of G. Then $\delta (H)=\delta $ and we can apply Theorem 4.2 to the non-singular action $H\curvearrowright (X,\mu )$ .

If $1-H^2(\mu _0,\mu _1)>\exp (-\delta /2)$ , then $H\curvearrowright X$ is ergodic. As $G\subset H$ is dense, we have that

$$ \begin{align*} L^{\infty}(X)^{G}=L^{\infty}(X)^{H}=\mathbb{C}1, \end{align*} $$

so that $G\curvearrowright X$ is ergodic. Let $H\curvearrowright X\times \mathbb {R}$ be the Maharam extension associated to  ${H\curvearrowright X}$ . Again, as $G\subset H$ is dense, we have that

$$ \begin{align*} L^{\infty}(X\times \mathbb{R})^{G}=L^{\infty}(X\times \mathbb{R})^{H}. \end{align*} $$

Note that the subgroup generated by the essential ranges of the maps $\log (dg^{-1}\mu /d\mu )$ , with $g\in G$ , is the same as the subgroup generated by the essential ranges of the maps $\log (dh^{-1}\mu /d\mu )$ , with $h\in H$ . Then one determines the Krieger flow of $G\curvearrowright X$ as in the proof of Theorem 4.2.

If $1-H^2(\mu _0,\mu _1)<\exp (-\delta /2)$ , the action $H\curvearrowright (X,\mu )$ is dissipative up to compact stabilizers. By [Reference Arano, Isono and MarrakchiAIM19, Theorem A.29] each ergodic component is of the form $H\curvearrowright H/K$ for a compact subgroup $K\subset H$ . Therefore, each ergodic component of $G\curvearrowright (X,\mu )$ is of the form $G\curvearrowright H/K$ , for some compact subgroup $K\subset H$ .