Proof. We only sketch the proof. For more details, we refer the reader to [Reference Arzhantseva, Cashen and Tao1, §3] where similar arguments are given. We first claim that for every contracting element $g \in G$ , the group H is not contained in $E(g)$ . Assume on the contrary that $H \subset E(g)$ . Since G is not virtually cyclic, there is $u \in G \setminus E(g)$ . By malnormality, $uHu^{-1} \cap H$ is finite and cannot have finite index in H, which contradicts the fact that H is commensurated.

We now fix once and for all a contracting element $g \in G$ . We denote by A an orbit of ${\langle g\rangle }$ . It is $\alpha $ -contracting for some $\alpha \in {\mathbb {R}}^*_+$ . Moreover, there is $C \in {\mathbb {R}}_+$ such that for every $u \in G \setminus E(g)$ , we have $\operatorname {diam}(\pi _A(uA)) \leq C$ , see [Reference Yang50, Lemma 2.11]. We choose $n \in {\mathbb {N}}$ such that $d(o,{g^no})$ is very large compared to both $\alpha $ and C.

We claim that there is an element $h \in H\setminus E(g)$ such that both $g^nhg^{-n}$ and $g^{-n}hg^n$ belong to H. Since H is commensurated, both $g^{-n}Hg^n \cap H$ and $g^nHg^{-n} \cap H$ have finite index in H. Thus,

$$ \begin{align*} H_0 = (g^{-n}Hg^n) \cap (g^nHg^{-n}) \cap H \end{align*} $$

has finite index in H. It follows that $H_0$ is not contained in $E(g)$ . Indeed, otherwise, $E(g)$ should also contain H, contradicting our previous claim. Any element $h \in H_0 \setminus E(g)$ satisfies the conclusions of our second claim.

Consider now the element

$$ \begin{align*} f = (g^nhg^{-n})(g^{-n}h g^n)^{-1} = g^n (hg^{-2n} h^{-1}) g^n. \end{align*} $$

Note that f belongs to H. We claim that f is contracting. For simplicity, we let $g_0 = g^n$ and $g_1= hg^{-2n}h^{-1}$ so that $f = g_0g_1g_0$ . They respectively act ‘by translation’ on the $\alpha $ -contracting sets $A_0 = A$ and $A_1 = hA$ . Fix a point $x \in \pi _A(hA)$ . Since $\pi _A(hA)$ has diameter at most C, any geodesic $\gamma \colon {[0\, , T]} \to X$ from x to $fx$ fellow-travels for a long time with $A_0=A$ , $g_0A_1=g_0hA$ , and $g_0g_1A_0 = fA$ (see Figure 1).

Figure 1 The geodesic from x to $fx$ . The gray shapes are ‘axis’ of conjugates of g.

Note that the construction has been designed so that the end of $\gamma $ and the beginning of $f \gamma $ both fellow-travel with $fA$ in the ‘same direction’. Consequently, the concatenation of $\gamma $ and $f\gamma $ cannot backtrack much. We extend $\gamma $ to a bi-infinite ${\langle f\rangle }$ -invariant path, still denoted by $\gamma \colon {\mathbb {R}} \to X$ , which is characterized as follows: $\gamma (t + kT) = f^k \gamma (t)$ for every $t \in {\mathbb {R}}$ and $k \in {\mathbb {Z}}$ . One proves that $\gamma $ is a quasi-geodesic which fellow-travels for a long time with $f^kA$ for every $k \in {\mathbb {Z}}$ (see Figure 2). This suffices to show that f is contracting.

Figure 2 The ‘axis’ of f.

We are left to prove that H is not virtually cyclic. If H was virtually cyclic, it would be contained in $E(f)$ . This contradicts our first claim.

Proof. Let $(z_n)$ be a sequence of points of X converging to c. For every $n \in {\mathbb {N}}$ , we let $b_n = \iota (z_n)$ and denote by $\gamma _n \colon {[0\, , {\ell _n}]} \to X$ a geodesic from x to $z_n$ . Since X is proper, $(\gamma _n)$ converges, up to passing to a subsequence, to a geodesic ray $\gamma \colon {\mathbb {R}}_+ \to X$ starting at x. As $(b_n)$ converges uniformly on every compact subset to c, we check that $\gamma $ is a gradient ray for c.

Proof. It suffices to prove that $F^+$ is a Borel subset. Without loss of generality, we can assume that F is contained in $\partial X$ . Given $D \in {\mathbb {R}}_+$ and a compact subset $K \subset X$ , we write

$$ \begin{align*} F_{K,D} = {\{{c \in \partial X}\;:\ { \text{there exists } b \in F, \ {\|{c-b}\|_{K}} \leq D}\}}. \end{align*} $$

Since F is compact, $F_{K, D}$ is closed. Using again the fact that F is compact, we observe that

$$ \begin{align*} F^+ = \bigcup_{D \in {\mathbb{R}}_+} \bigcap_{K \subset X} F_{K,D}, \end{align*} $$

where K runs over all compact subsets of X. Hence the result.

Proof. Let $t\in I$ . Note that

$$ \begin{align*} d(q,{\gamma(t)}) \leq c(q, \gamma(t)) \leq c(q,y) + c(y,\gamma(t)) \leq c(q,y) + d(y,{\gamma(t)}). \end{align*} $$

The first inequality holds since $\gamma $ is a gradient line for c, while the last one follows from the fact that c is $1$ -Lipschitz. However, q being a projection of c on Y, we have $c(q,y) \leq 0$ . Consequently, $d(q,{\gamma (t)}) \leq d(y,{\gamma (t)})$ for every $y \in Y$ , which completes the proof.

Proof. The equivalences (iii) $\iff $ (iv) and (iv) $\iff $ (v) are standard properties of contracting sets, which only use the fact that $\gamma $ is a geodesic. Note that $c(x,y) = c(x, x') + c(x',y)$ for every $x,x' \in X$ and $y \in Y$ . The implication (i) $\Rightarrow $ (ii) follows from this observation. The proof of (ii) $\Rightarrow $ (iii) is by contraposition. Suppose that there exists $d \in {\mathbb {R}}_+$ such that $\gamma $ lies in $\mathcal N_d(Y)$ . Let $x = \gamma (0)$ . Let $t \in {\mathbb {R}}_+$ . We denote by $q_t$ a projection of $\gamma (t)$ onto Y. Using the fact that c is $1$ -Lipschitz, we get

$$ \begin{align*} c(x, q_t) \geq c(\gamma(0), \gamma(t)) - d({\gamma(t)},{q_t}) \geq t - d. \end{align*} $$

This inequality holds for every $t \in {\mathbb {R}}_+$ , hence the map $Y \to {\mathbb {R}}$ , sending y to $c(x,y)$ , is not bounded from above.

We now focus on implication (v) $\Rightarrow $ (vi). For every $n \in {\mathbb {N}}$ , we let $b_n = \iota (z_n)$ . Assume that $T < \infty $ . Let Q be the projection onto Y of $\gamma $ restricted to $[T, \infty )$ . Since Y is contracting, the diameter of Q is at most $\alpha $ . Let $q \in Q$ be a projection of $\gamma (T)$ on Y. We claim that there is $N \in {\mathbb {N}}$ , such that for every $n \geq N$ , the point $q_n$ stays at a distance at most $\alpha $ of Q. Assume on the contrary that it is not the case. Up to passing to a subsequence, $d(q_n, Q)> \alpha $ for every $n \in {\mathbb {N}}$ . Using Lemma 2.3, we observe that for every $t \geq T$ , for every $n \in {\mathbb {N}}$ ,

$$ \begin{align*} b_n(\gamma(t), q) \geq d(q,{\gamma(t)}) - 4\alpha. \end{align*} $$

After passing to the limit, we get $c(\gamma (t), q) \geq d(q,{\gamma (t)}) - 4\alpha $ for every $t \geq T$ . In particular, $t \mapsto c(\gamma (t), q)$ diverges to infinity as t tends to infinity, which contradicts the fact that $\gamma $ is a gradient line for c. This completes the proof of our claim and thus implies item (vi).

We finish the proof with implication (vi) $\Rightarrow $ (i). Assume now that $(q_n)$ is bounded. Let $q^*$ be an accumulation point of $(q_n)$ . As Y is closed, $q^*$ belongs to Y. Observe that for every $y \in Y$ , for every $n \in {\mathbb {N}}$ , we have

$$ \begin{align*} b_n(q^*,y) \leq b_n(q_n,y) + d({q^*},{q_n}) \leq d({q^*},{q_n}). \end{align*} $$

By construction, $b_n$ converges to c on every compact subset, and hence $c(q^*, y) \leq 0$ for every $y \in Y$ . Thus, $q^*$ is a projection of c onto Y. In particular, $c \notin \partial ^+Y$ .

Proof. Consider a gradient line $\gamma \colon {\mathbb {R}}_+ \to X$ from q to c. According to Lemma 3.9, q is a projection of any point $\gamma (t)$ on Y. Assume that, in contrast to our claim, $d(q,{q^*})> \alpha $ . Reasoning as in the proof of Proposition 3.11, we observe that $c(\gamma (t), q) \geq d(q,{\gamma (t)}) - 4\alpha $ for every $t \in {\mathbb {R}}_+$ , which contradicts the fact that $\gamma $ is a gradient line for c.

Proof. If c is the cocycle associated to a point $z \in X$ , then the statement is just a particular case of Lemma 2.3. Assume now that c belongs to $\partial X \setminus \partial ^+ Y$ . Fix a sequence $(z_n)$ of points in X converging to c. Denote by $q^*$ an accumulation point of $(q_n)$ , where $q_n$ stands for a projection of $z_n$ on Y. Denote by $\gamma (T)$ the exit point of $\gamma $ from $\mathcal N_\alpha (Y)$ , and Q the projection onto Y of $\gamma $ restricted to $[T, \infty )$ . According to Proposition 3.11, such an exit point exists. Moreover, Q has diameter at most $\alpha $ and $q^*$ is a projection of c on Y lying in the $\alpha $ -neighborhood of Q.

Suppose that $d(x,Y) < \alpha $ or $d(p, q)> 4\alpha $ . According to Lemma 3.12, $d(p,{q^*})> 3 \alpha $ , and hence, $d(p, Q)> \alpha $ . It follows from the definition of contracting sets that ${d(\gamma , Y) < \alpha }$ . Let $p'$ (respectively $q'$ ) be a projection of the entry (respectively exit) point of $\gamma $ in $\mathcal N_\alpha (Y)$ —note that if $d(x, Y) < \alpha $ , then x is the entry point of $\gamma $ in $\mathcal N_\alpha (Y)$ , so we can choose $p' = p$ . Using again the contraction of Y, we see that $d(p,{p'}) \leq \alpha $ . Moreover, $q'$ belongs to Q, thus $d({q'},{q^*}) \leq 2\alpha $ . In addition,

$$ \begin{align*} d(x,{\gamma(T)}) \geq d(x,{p'}) + d({p'},{q'}) + d({q'},{\gamma(T)}) - 4\alpha. \end{align*} $$

The path $\gamma $ is a gradient line, thus $c(x, \gamma (T))= d(x,{\gamma (T)})$ , and hence

$$ \begin{align*} c(x, \gamma(T)) \geq d(x,{p'}) + d({p'},{q'}) + d({q'},{\gamma(T)}) - 4\alpha. \end{align*} $$

Combined with the fact that c is $1$ -Lipschitz, we get

(1) $$ \begin{align} c(x,q') \geq c(x, \gamma(T)) - d({q'},{\gamma(T)}) \geq d(x,{p'}) + d({p'},{q'}) - 4\alpha. \end{align} $$

We observed that $d(p,{p'}) \leq \alpha $ , while $d({q'},q) \leq d({q'},{q^*}) + d({q^*},q) \leq 4\alpha $ . The conclusion follows from equation (1) and the triangle inequality.

Proof. According to Proposition 3.11(iv), the set $\pi _Y(\gamma )$ is unbounded. Thus there exists $s \in {\mathbb {R}}_+$ and a projection q of $\gamma (s)$ onto Y such that $d(p,q)> \alpha $ . It follows that ${d(\gamma , Y) < \alpha }$ . Let $\gamma (t)$ be the entry point of $\gamma $ in $\mathcal N_\alpha (Y)$ . Using again the contraction of Y, we get $d(p,{\gamma (t)}) \leq 2\alpha $ . Since $\gamma $ is a gradient line, we have $c(x,\gamma (t)) = d(x,{\gamma (t)})$ . Combined with the triangle inequality and the fact that cocycles are $1$ -Lipschitz, we get $c(x,p) \geq d(x,p) - 4\alpha $ .

Proof. Consider an element $u \in G$ such that $\partial ^+A\cap \partial ^+(uA)$ is not empty. Let c be a cocycle in this intersection and $\gamma \colon {\mathbb {R}}_+ \to X$ a gradient ray for c. By Proposition 3.11(iii), there exist $d, T \in {\mathbb {R}}_+$ such that $\gamma $ restricted to $[T, \infty )$ is contained in $\mathcal N_d(A) \cap \mathcal N_d(uA)$ . In particular, the diameter of this intersection is infinite. It follows that $u \in E(g)$ , see [Reference Yang50, Lemma 2.12]. Recall that $E(g)$ , unlike G, is virtually cyclic. Thus there exists ${u \in G\setminus E(g)}$ . It follows from the above discussion that $\partial ^+A \cap \partial ^+(uA) = \emptyset $ .

Vocabulary. Let $\nu = (\nu _x)$ be a density on $(\bar X, \mathfrak A)$ . We make an abuse of vocabulary and say that a property on $(\bar X, \mathfrak A)$ holds $\nu $ -a.e. if it holds $\nu _x$ -a.e. for some (hence every) $x \in X$ .

Proof. Assume that $\nu $ is almost-fixed by G (if $\nu $ is fixed by G, the proof works in the exact same way). There exists $C \in {\mathbb {R}}_+^*$ such that for every $g \in G$ and $x \in X$ , we have

$$ \begin{align*} \frac 1C \nu_x\leq \nu_x^g \leq C \nu_x. \end{align*} $$

Let $g,g' \in G$ . Comparing the total masses of the above measures for $x = g'o$ , we get

$$ \begin{align*} \frac 1C{\|{\nu_{g'o}}\|} \leq \frac{{\|{\nu_{gg'o}}\|}}{{\|{\nu_{go}}\|}} \leq C {\|{\nu_{g'o}}\|}.\\[-45pt] \end{align*} $$

Proof. Since $\mu $ is G-quasi-invariant, we can define the following map:

$$ \begin{align*} \begin{array}{rccc} F \colon & G \times \bar X &\!\!\! \to\!\!\!\! & {\mathbb{R}}_+ \\ & (g,c) &\!\!\! \mapsto\!\!\!\! & \displaystyle \frac{d ({g^{-1}}_\ast \mu_{go})}{d\mu_o}(c) \end{array} \end{align*} $$

Claim 4.6. There exists $C \in {\mathbb {R}}_+^*$ such that for every $g_1, g_2 \in G$ , we have

$$ \begin{align*} \frac 1C\frac{F(g_1g_2,c)}{F(g_2,c) } \leq F(g_1,g_2c) \leq C \frac{F(g_1g_2,c)}{F(g_2,c) }, \quad \mu\text{-a.e.} \end{align*} $$

Let $g_1, g_2 \in G$ . The computation gives

(7) $$ \begin{align} F(g_1g_2,c) = \frac{d ({g_1^{-1}}_\ast \mu_{g_1g_2o})}{d\mu_{g_2o}}(g_2c) \ \frac{d ({g_2^{-1}}_\ast \mu_{g_2o})}{d\mu_o}(c), \quad \mu\text{-a.e.} \end{align} $$

Since $\mu $ is $\omega $ -quasi-conformal, there exists $C \in {\mathbb {R}}_+$ such that for every $x ,y\in X$ , for every $g \in G$ , we have

$$ \begin{align*} \frac 1C \frac {d\mu_x}{d\mu_y} \leq \frac{d\mu_{gx}}{d\mu_{gy}} \circ g \leq C \frac {d\mu_x}{d\mu_y}. \end{align*} $$

Hence,

$$ \begin{align*} \frac 1C \frac{d ({g^{-1}}_\ast \mu_{gy})}{d\mu_y} \leq \frac{d ({g^{-1}}_\ast \mu_{gx})}{d\mu_x} \leq C \frac{d ({g^{-1}}_\ast \mu_{gy})}{d\mu_y}. \end{align*} $$

It follows that the first factor in the right-hand side of (7) is $F(g_1,g_2c)$ —up to a multiplicative error that does not depend on $g_1$ or $g_2$ —while the second factor is exactly $F(g_2,c)$ . This completes the proof of our claim.

Let $g \in G$ and set $H_0 = H^g \cap H$ . Let $h \in H_0$ . According to our claim, we have

$$ \begin{align*} \frac 1C\frac{F(gh,c)}{F(h,c) } \leq F(g,hc) \leq C \frac{F(gh,c)}{F(h,c) }, \quad \mu\text{-a.e.} \end{align*} $$

Recall that $\mu $ is H-invariant, thus

$$ \begin{align*} F(h,c) = 1 \quad \text{and} \quad F(gh,c) = F(ghg^{-1}g,c) = F(g,c), \quad \mu\text{-a.e.} \end{align*} $$

Our previous inequalities becomes

$$ \begin{align*} \frac 1C F(g,c) \leq F(g,hc) \leq C F(g,c), \quad \mu\text{-a.e.} \end{align*} $$

We now define an auxiliary function $F_g \colon \bar X \to {\mathbb {R}}_+$ by

$$ \begin{align*} F_g(c) = \inf_{h \in H_0} F(g,hc). \end{align*} $$

By construction, $F_g$ is $H_0$ -invariant. Since the action of $H_0$ on $(\bar X, \mathfrak R, \mu _o)$ is ergodic, $F_g$ is constant. From now on, we denote by $F_g$ its essential value. It follows from our previous observation that

$$ \begin{align*} F_g \leq F(g,c) \leq C F_g, \quad \mu\text{-a.e.} \end{align*} $$

Coming back to the definition of F, this means that

$$ \begin{align*} F_g \leq \frac{d ({g^{-1}}_\ast \mu_{go})}{d\mu_o} \leq C F_g, \quad \mu\text{-a.e.} \end{align*} $$

Integrating these inequalities, we see that $F_g \leq {\|{\mu _{go}\|} } \leq C F_g$ . Hence,

$$ \begin{align*} \frac 1C {\|{\mu_{go}\|} } \leq \frac{d ({g^{-1}}_\ast \mu_{go})}{d\mu_o} \leq C {\|{\mu_{go}}\|}. \end{align*} $$

Recall that C does not depend on g. We have proved that there exists $C \in {\mathbb {R}}_+^*$ such that for every $g \in G$ ,

$$ \begin{align*} \frac 1C \mu_o \leq \mu^g_o \leq C \mu_o. \end{align*} $$

Using the quasi-conformality of $\mu $ , we conclude that $\mu $ is almost-fixed by G.

Proof. Assume that our claim fails. We can find a sequence $(r_n)$ diverging to infinity, a sequence $(z_n)$ of points in X, and a sequence $\nu ^n = (\nu ^n_x)$ of densities in $\mathcal D_0$ such that

$$ \begin{align*} \nu^n_o(\mathcal O_{z_n}(o,r_n)) \end{align*} $$

converges to zero. Since $\mathcal D_0$ is closed, up to passing to a subsequence, we may assume that $\nu ^n$ converges to a density $\nu $ in $\mathcal D_0$ . Note that for every $x,y \in X$ and $r \in {\mathbb {R}}_+$ , we have $\mathcal O_x(y,r) = \bar X$ , whenever $d(x,y) \leq r$ . Hence, $d(o,{z_n})$ necessarily diverges to infinity. Up to passing again to a subsequence, we can assume that $z_n$ converges to $b \in \partial X$ .

By assumption G is not virtually cyclic and contains a contracting element. According to Proposition 3.15, there exists a contracting element $h \in G$ such that $b \notin \partial ^+A$ , where $A = {\langle h\rangle } o$ is $\alpha $ -contracting for some $\alpha \in {\mathbb {R}}_+^*$ . For every $n \in {\mathbb {N}}$ , we write $p_n$ for a projection of $z_n$ onto A. Up to passing to a subsequence, $(p_n)$ converges to a point $p \in A$ which is a projection of b onto A (Proposition 3.11). We introduce the following closed subset of $\bar X$ :

$$ \begin{align*} F = {\{{c \in \bar X}\;:\ { c(p,y) \leq 4\alpha \text{ for all } y \in Y}\}}. \end{align*} $$

We are going to prove that $\nu _o(F) = 1$ .

Let $c \in \bar X \setminus F$ . Suppose first that $c \notin \partial ^+A$ . Let q be a projection of c onto A. Since c is $1$ -Lipschitz, we have

$$ \begin{align*} c(p,y) \leq d(p,q) + c(q,y) \leq d(p,q)\quad \text{for all } y \in A. \end{align*} $$

As c does not belong to F, necessarily $d(p,q)> 4\alpha $ . In particular, there exists ${N_0 \in {\mathbb {N}}}$ , such that for every $n \geq N_0$ , we have $d({p_n},q)> 4\alpha $ . According to Corollary 3.13, ${\langle z_n,c\rangle _{p_n}} \leq 2\alpha $ . Note that the conclusion still holds if $c \in \partial ^+A$ . This is indeed a consequence of Corollary 3.14. Hence, in all cases, we have

$$ \begin{align*} {\langle z_n,c\rangle_{o}} \leq {\langle z_n,c\rangle_{p_n}} + d(o,{p_n}) \leq 2\alpha + d(o,{p_n}). \end{align*} $$

Recall that $(p_n)$ is bounded. Consequently, there is $N_1 \geq N_0$ such that for every $n \geq N_1$ , the set $\bar X \setminus F$ is contained in $\mathcal O_{z_n}(o,r_n)$ . In particular, $\nu ^n_o (\bar X \setminus F)$ converges to zero. Since $\bar X \setminus F$ is an open subset of $\bar X$ , we deduce that $\nu _o(\bar X \setminus F) = 0$ , that is, $\nu _o(F) = 1$ .

The group ${\langle h\rangle }$ has unbounded orbits. Thus, there is $g \in {\langle h\rangle }$ such that $d({gp},p)> 36\alpha $ . We claim that $F \cap gF$ is empty. Let $c \in F$ . Remember that c does not belong to $\partial ^+A$ by Proposition 3.11(ii). Choose a projection q of c onto A. It follows from Corollary 3.13 that

$$ \begin{align*} 4\alpha \geq c(p,q) \geq d(p,q) - 14\alpha. \end{align*} $$

Hence, $d(p,q) \leq 18\alpha $ . Suppose now that, in contrast to our claim, c also belongs to $gF$ . In particular, $g^{-1}q$ is a projection of $g^{-1}c$ onto A. Following the same argument as above, we get $d(p,{g^{-1}q}) \leq 18\alpha $ . Consequently, $d(p,{gp}) \leq 36\alpha $ , which is a contradiction. Observe that

$$ \begin{align*} \nu_o(gF) = {g^{-1}}_*\nu_o(F) = {\|{\nu_{go}}\|} \nu^g_{g^{-1}o}(F). \end{align*} $$

As an element of $\mathcal D_0$ , the density $\nu $ is G-quasi-invariant, and hence $\nu _o \ll \nu ^g_{g^{-1}o}$ . We proved that $\nu _o(F)> 0$ , thus $\nu _o(gF)> 0$ . Consequently,

$$ \begin{align*} \nu_o( F \cup gF) = \nu_o( F) + \nu_o(g F)> 1. \end{align*} $$

This contradicts the fact that $\nu _o$ is a probability measure.

Proof. Let $\nu = (\nu _x)$ be an $\omega $ -conformal density. A classical computation shows that for every $g \in G$ and $r \in {\mathbb {R}}_+$ , we have

Shadows have been designed so that for every $c \in \mathcal O_{g^{-1}o}(o, r)$ , we have

$$ \begin{align*} d(o,{go}) -2 r \leq c(g^{-1}o,o) \leq d(o,{go}). \end{align*} $$

Consequently,

$$ \begin{align*} \nu^g_o (\mathcal O_{g^{-1}o}(o,r)) \leq \frac { e^{\omega d({go},o)}}{{\|{\nu_{go}}\|} } \nu_o (\mathcal O_o(go,r)) \leq e^{2\omega r}\nu^g_o (\mathcal O_{g^{-1}o}(o,r)). \end{align*} $$

The upper bound in equation (11) follows from the fact that $\nu ^g_o$ is a probability measure. Let us focus on the lower bound. To that end, we assume now that $\nu $ is N-invariant. As N is a normal subgroup of G, the density $\nu ^g$ is N-invariant as well. Since G contains a contracting element and is not virtually cyclic, the same holds for N (Lemma 2.8). The result now follows from Proposition 4.9 applied with the group N and the closed set ${\mathcal D_0 = \mathcal D(N,\omega )}$ .

Proof. Let $g \in G$ and $r \in {\mathbb {R}}_+$ . Reasoning as in the proof of Corollary 4.10, we see that

$$ \begin{align*} \nu_o (\mathcal O_o(go,r)) \geq {\|{\nu_{go}}\|} e^{-\omega d(o,{go})} \nu^g_o(\mathcal O_{g^{-1}o}(o,r)). \end{align*} $$

Since $\nu $ is almost-fixed by G, there is $\epsilon \in {\mathbb {R}}_+^*$ , which does not depend on g or r, such that $\nu ^g_o \geq \epsilon \nu _o$ . Consequently,

$$ \begin{align*} \nu_o(\mathcal O_o(go,r)) \geq \epsilon{\|{\nu_{go}}\|} e^{-\omega d(o,{go})} \nu_o (\mathcal O_{g^{-1}o}(o,r)). \end{align*} $$

The result now follows from Proposition 4.9 applied with $\mathcal D_0 = \{ \nu \}$ .

Notation 4.1. Given $\alpha \in {\mathbb {R}}^*_+$ and $L \in {\mathbb {R}}_+$ , we denote by $\mathcal T(\alpha , L)$ the set of all elements $g \in G$ such that the pair $(o,go)$ has an $(\alpha , L)$ -contracting tail.

Proof. Since $\tau $ is a contracting tail, there is a projection $p'$ of x on $\tau $ such that $d({p'},y) \geq ~L$ . The computation shows that

$$ \begin{align*} d(y,q) + {\langle x,c\rangle_{p'}} + {\langle p',c\rangle_{q}} = {\langle x,c\rangle_{y}} + {\langle x,y\rangle_{p'}} + {\langle y,c\rangle_{q}} + {\langle p',y\rangle_{q}}, \end{align*} $$

(it suffices to expand the definition of the Gromov products). On the one hand, we observed that ${\langle x,c\rangle _{p'}} \geq 0$ and ${\langle p',c\rangle _{q}}\geq 0$ . On the other hand, since $\tau $ is contracting, we have ${\langle x,y\rangle _{p'}} \leq 2\alpha $ (Lemma 2.3) and ${\langle y,c\rangle _{ q}} \leq 5\alpha $ (Corollary 3.13). It follows that

$$ \begin{align*} d(y,q) \leq r + 7 \alpha + {\langle p',y\rangle_{q}}. \end{align*} $$

We now discuss the relative positions of $p'$ , q, and y on $\tau $ . If $p'$ lies between q and y, then ${\langle p',y\rangle _{q }}= d({p'},q)$ while $d(y,q) = d(y,{p'}) + d({p'},q)$ . It forces $L\leq d(y,{p'})\leq r + 7 \alpha $ , which contradicts our assumption. Thus, q lies between $p'$ and y so that ${{\langle p',y\rangle _{q}}= 0}$ . Consequently, $d(y,q) \leq r + 7 \alpha $ , which completes the proof of item (i).

By assumption, p is another projection of x on $\tau $ , which is $\alpha $ -contracting, thus $d(p,{p'}) \leq 2 \alpha $ . Consequently,

$$ \begin{align*} d(p,q) \geq d({p'},y) - d(p,{p'}) - d(y,q) \geq L - (r+9\alpha)> 4\alpha. \end{align*} $$

According to Corollary 3.13, $d(\gamma ,\tau ) < \alpha $ . Moreover, the entry (respectively exit) point of $\gamma $ in $\mathcal N_\alpha (\tau )$ is $2\alpha $ -close to p (respectively $5\alpha $ -closed to q). Hence the result.

Proof. Let c be a cocycle in the intersection $\mathcal O_x(y_1, r) \cap \mathcal O_x(y_2,r)$ . Let $\gamma $ be a gradient arc from x to c. According to Lemma 4.14, the points $y_1$ and $y_2$ lie in the $(r+12\alpha )$ - neighborhood of $\gamma $ , while $p_1$ and $p_2$ are $2\alpha $ -closed to $\gamma $ . The result follows.

Proof. Let $h \in G$ be a contracting element. Denote by A the ${\langle h\rangle }$ -orbit of o. It is $\beta $ -contracting for some $\beta \in {\mathbb {R}}^*_+$ . Let $\alpha _0 \in {\mathbb {R}}^*_+$ be the parameter given by Lemma 2.7 applied with the element h, the point $z = o$ , and $d = 2\beta $ . Up to increasing the value of $\alpha _0$ , we can assume that $\alpha _0 \geq 2 \beta + d(o,{ho})$ .

Let $L \in {\mathbb {R}}_+$ . Since h is contracting, the orbit map ${\mathbb {Z}} \to X$ sending n to $h^no$ is a quasi-isometric embedding. Thus, there is $N \in {\mathbb {N}}$ such that for every $n \geq N$ , we have $d(o,{h^no})\geq L+ 2\beta $ . We choose N minimal with the above property. In particular,

$$ \begin{align*} L + 2\beta \leq d(o,{h^No})\leq d(o,{h^{N-1}o}) + d(o,{ho}) < L + 2\beta + d(o,{ho}). \end{align*} $$

Hence, $h^N$ and $h^{-N}$ belong to $S(L, \alpha _0)$ .

Let $g \in G$ . There is $k \in {\mathbb {Z}}$ such that $h^ko$ is a projection of $g^{-1}o$ onto A. If $k \leq 0$ (respectively $k \geq 0$ ), we choose $u = h^N$ (respectively $u = h^{-N}$ ). We now prove that u satisfies the announced properties. We suppose that $k \geq 0$ . The other case works in the exact same way. Let $\gamma \colon {[0\, , T]} \to X$ be a geodesic from $g^{-1}o$ to $h^{-N}o$ . According to Lemma 2.3, $h^ko$ is $2\beta $ -close to the entry point $\gamma (t)$ of $\gamma $ in $\mathcal N_\beta (A)$ . By our choice of N, we have

$$ \begin{align*} T - t \geq d({\gamma(t)},{\gamma(T)}) \geq d({h^ko},{h^{-N}o}) - 2 \beta \geq d(o,{h^{N+k}o}) - 2\beta \geq L. \end{align*} $$

In particular, $t \leq T-L\leq T$ . It follows from Lemma 2.7 and our choice of $\alpha _0$ that:

• • $\gamma $ restricted to ${[{T-L}\, , T]}$ is $\alpha _0$ -contracting;

• • there is $s \in {[{T-L}\, , T]}$ such that $d({\gamma (s)},o) \leq \alpha _0$ .

Using the triangle inequality, we observe that

$$ \begin{align*} {| {d({\gamma(s)},{h^{-N}o}) - d(o,{h^{-N}o})}|} \leq d({\gamma(s)},o) \leq \alpha_0. \end{align*} $$

On the one hand, $d({\gamma (s)},{h^{-N}o}) = T - s$ . On the other hand, $h^{-N} \in S(L, \alpha _0)$ . Thus, ${| {(T-L) - s}|} \leq 2\alpha _0$ . Consequently, the triangle inequality yields

$$ \begin{align*} d(o,{\gamma(T-L)}) \leq d(o,{\gamma(s)}) + d({\gamma(s)},{\gamma(T-L)}) \leq 3\alpha_0. \end{align*} $$

Observe now that the path $\tau = g\gamma $ satisfies the conclusion of the lemma with the parameter $\alpha = 3\alpha _0$ .

Proof. We choose for $\alpha $ the parameter given by Lemma 4.16. Let $\omega , a \in {\mathbb {R}}_+$ and $(\epsilon , r_0) \in {\mathbb {R}}_+^* \times {\mathbb {R}}_+$ . The action of G on X is proper, so there is $M \in {\mathbb {N}}$ such that

$$ \begin{align*} {| {{\{{g \in G}\;:\ {d(o,{go}) \leq 2a + 12\alpha}\}}}|} \leq M. \end{align*} $$

In addition, we set

$$ \begin{align*} r_1 = \max\{r_0, a + 3\alpha\}. \end{align*} $$

Let $\nu = (\nu _x)$ be an $\omega $ -conformal density such that $(G, \nu )$ satisfies the shadow principle with parameters $(\epsilon , r_0)$ . Let $r \geq r_1$ , $L> r + 13\alpha $ and $\ell \in {\mathbb {R}}_+$ . According to Lemma 4.16, for every $g \in S(\ell , a)$ , there is $u_g \in S(L,\alpha )$ such that $gu_g$ belongs to $\mathcal T(\alpha , L)$ . Moreover, there is a geodesic $\tau _g \colon {[0\, , {T_g}]} \to X$ joining o to $gu_go$ , whose restriction to ${[{T_g-L}\, , {T_g}]}$ is $\alpha $ -contracting, and such that $go$ is $\alpha $ -close to $z_g = \tau _g(T_g-L)$ . In particular,

$$ \begin{align*} {| {d(o,{go}) + d(o,{u_go}) - d(o,{gu_go})}|} \leq 2\alpha. \end{align*} $$

Since $g \in S(\ell , a)$ and $u_g \in S(L,\alpha )$ , we get $gu_g \in S(\ell + L, a + 3\alpha )$ . Hence, $\mathcal O_o(gu_go,r)$ is contained in $A_{\ell +L}(\alpha , r, L)$ .

Consider now $g,g' \in S(\ell , a)$ such that $\mathcal O_o(gu_go,r) \cap \mathcal O_o(g'u_{g'}o,r)$ is non-empty. By construction, $go$ and $g'o$ are $\alpha $ -close to $z_g$ and $z_{g'}$ . Note that $z_g = \tau _g(T_g -L)$ is the (unique) projection of o on $\tau _g$ restricted to ${[{T_g -L}\, , {T_g}]}$ , which is $\alpha $ -contracting. A similar statement holds for $\tau _{g'}$ . Using Lemma 4.15 and the triangle inequality, we observe that

$$ \begin{align*} d({go},{g'o}) \leq d({z_g},{z_{g'}}) + 2\alpha & \leq {| { d(o,{z_g}) - d(o,{z_{g'}})}|} + 10\alpha \\ & \leq {| { d(o,{go}) - d(o,{g'o})}|} + 12\alpha \\ & \leq 2a + 12\alpha. \end{align*} $$

It follows from our choice of M that any cocycle $c \in \bar X$ belongs to at most M shadows of the form $\mathcal O_o(gu_go,r)$ , where $g \in S(\ell ,a)$ . Consequently,

(12) $$ \begin{align} \sum_{g \in S(\ell, a)} \nu_o (\mathcal O_o(gu_go,r)) \leq M\nu_o(A_{\ell+L}(\alpha, r, L)). \end{align} $$

Recall that $(G, \nu )$ satisfies the shadow principle with parameters $(\epsilon , r_0)$ . Hence, for every $g \in S(\ell , a)$ , we have

$$ \begin{align*} \nu_o (\mathcal O_o(gu_go,r)) \geq \epsilon {\|{\nu_{gu_go}}\|} e^{-\omega d(o,{gu_go})} \geq \epsilon e^{-2\omega (L+\alpha)}{\|{\nu_{go}}\|} e^{-\omega d(o,{go})}. \end{align*} $$

The second inequality follows from equation (2). Consequently, equation (12) becomes

$$ \begin{align*} \sum_{g \in S(\ell, a)} {\|{\nu_{go}}\|} e^{-\omega d(o,{go})} \leq \bigg(\frac{M e^{2\omega\alpha}}\epsilon\bigg) e^{2\omega L}\nu_o(A_{\ell+L}(\alpha, r, L)).\\[-52pt] \end{align*} $$

Proof. Denote by $\alpha , r_1, C$ the parameters given by Proposition 4.17 applied with $\omega $ , a, and $(\epsilon ,r_0)$ . We choose $r \geq r_1$ and $L> \max \{ 2r, r + 15\alpha \}$ . Let $\nu = (\nu _x)$ be an $\omega $ -conformal density such that $(G, \nu )$ satisfies the shadow principle with parameters $(\epsilon , r_0)$ . By Proposition 4.17, we have

$$ \begin{align*} \sum_{g \in S(\ell, a)} {\|{\nu_{go}}\|} e^{-\omega d(o,{go})} \leq Ce^{2\omega L}\nu_o ( A_{\ell+L}( \alpha, r, L)) \quad \text{for all } \ell \in {\mathbb{R}}_+. \end{align*} $$

Consider now $x,y,z \in X$ . By the very definition of shadows, if $z \in \mathcal O_x(y,r)$ , then

$$ \begin{align*} d(x,y) - d(x,z) \leq {\langle x,z\rangle_{y}} \leq r. \end{align*} $$

Hence, for every $\ell \in {\mathbb {R}}_+$ , the set $A_{\ell +L}(\alpha , r, L)$ lies in $\bar X \setminus B(o, \ell + L - 2r)$ . The result follows from the fact that $L \geq 2r$ .

Proof. According to Proposition 4.3, there exists a G-invariant, $\omega _G$ -conformal density $\nu = (\nu _x)$ . By Corollary 4.10, the pair $(G, \nu )$ satisfies the shadow principle for some parameters $(\epsilon , r_0) \in {\mathbb {R}}_+^* \times {\mathbb {R}}_+$ . Fix $a \in {\mathbb {R}}_+^*$ . Applying Corollary 4.18, there exists $C \in {\mathbb {R}}_+$ such that for every $\ell \in {\mathbb {R}}_+$ ,

$$ \begin{align*} \sum_{g \in S(\ell, a)} {\|{\nu_{go}}\|} e^{-\omega_G d(o,{go})} \leq C \nu_o ( \bar X \setminus B(o, \ell)) \leq C. \end{align*} $$

However, since $\nu $ is G-invariant, ${\|{\nu _{go}}\|} = {\|{\nu _o}\|} = 1$ for every $g \in G$ . Hence, we get

$$ \begin{align*} {| {S(\ell, a)}|} \leq Ce^{\omega_G a}e^{\omega_G \ell} \quad \text{for all } \ell \in {\mathbb{R}}_+. \end{align*} $$

The result follows by summing this inequality over $\ell \in a {\mathbb {N}}$ .