Proof. From the definition of Gromov product at infinity, it follows that

$$\begin{align*}({\alpha}\,|\,{\beta})_{p} \ge \liminf_{s,t\to\infty} ({\gamma_\alpha(s)}\,|\,{\gamma_\beta(t)})_{p}.\end{align*}$$

To simplify notation, consider p as a base point and write $|\cdot |$ for $d_M(p,\cdot )$ . Suppose that c is a point on $\gamma _\alpha $ and d a point on $\gamma _\beta $ so that $|c|>|a|$ and $|d|>|b|$ . We want to show that $({c}\,|\,{d})_{p} \ge ({a}\,|\,{b})_{p} - 2\nu $ .

Consider the triple of Gromov products $({c}\,|\,{d})_{p}$ , $({b}\,|\,{d})_{p} = |b|$ , and $({c}\,|\,{b})_{p}\le |b|$ . As is well known, it follows from $\nu $ –hyperbolicity that any one of such a triple is bounded below by $\nu $ less than the minimum of the other two, so we have

(2) $$ \begin{align} ({c}\,|\,{d})_{p} \ge ({c}\,|\,{b})_{p} - \nu.\end{align} $$

Considering next the triple $({c}\,|\,{b})_{p}$ , $({c}\,|\,{a})_{p} = |a|$ , and $({a}\,|\,{b})_{p} \le |a|$ , we conclude

(3) $$ \begin{align} ({c}\,|\,{b})_{p} \ge ({a}\,|\,{b})_{p} - \nu.\end{align} $$

Together (2) and (3) give $({c}\,|\,{d})_{p} \ge ({a}\,|\,{b})_{p} - 2\nu $ , as desired.

Proof. Recall $\Gamma $ is $\nu $ –hyperbolic, meaning that triangles are $\nu $ –thin. Let $(a,b,c) \in X$ be given. Fix biinfinite geodesics $[a,b], [b,c]$ , and $[a,c]$ in $\Gamma $ . Approximate this ideal triangle by a triangle in $\Gamma $ by choosing points $a'\ b'$ and $c' \in \Gamma $ on the geodesics $[a,b], [b,c]$ , and $[a,c]$ , respectively, satisfying $d_{\Gamma }(a', [a, c]) < \nu $ and $d_{\Gamma }(a', [b, c])> r + 2 \nu $ , and such that the same two inequalities also hold when the letters $a, b, c$ are cyclically permuted. Then $\pi _r(a,b,c)$ is a subset of

$$\begin{align*}S_r := N_{r + \nu}([a',b']) \cap N_{r + \nu}([b',c']) \cap N_{r + \nu}([c',a']) \end{align*}$$

so it suffices to show this set has diameter bounded by $Q(r)$ , for some suitable function Q.

Let p denote the map from the triangle with sides $[a',b'], [b', c']$ , and $[a', c']$ to a tripod witnessing that the triangle is $\nu $ –thin, and let Z be the preimage of the center, this is a set of three points with diameter at most $\nu $ . We now claim that $S_r$ lies in the $3r + 5 \nu $ -neighborhood of Z, which is enough to prove the lemma. To prove the claim, suppose $s \in S_r$ , so there exist points $x_1, x_2$ , and $x_3$ on $[a',b'], [b', c']$ , and $[a', c']$ , respectively, with $d_{\Gamma }(x_i, s) < r + \nu $ . Then for any $i = 1, 2, 3$ , there exists some j so that $p(x_i)$ and $p(x_j)$ lie on different prongs of the tripod, so there is a path between them passing through the midpoint m of the tripod. Thus, we have

$$\begin{align*}d_{\Gamma}(x_i, Z) = d_{\Gamma}(p(x_i), m) \leq d_{\Gamma}(p(x_i), p(x_j)) \leq d_{\Gamma}(x_i, x_j) + 2 \nu \leq 2r + 4 \nu ,\end{align*}$$

where the last inequality follows from the fact that $d_{\Gamma }(x_i, s) < r + \nu $ . This proves the claim.

Proof. Let $K\subset \Xi $ be compact. Increasing r makes $\pi _r(K)$ larger. Using Remark 2.4, we can therefore assume that for every $(a,b,c)$ , there is some $x\in \pi _r(a,b,c)$ .

For each $(a,b,c)\in K$ , there is an open neighborhood U of $(a,b,c)$ in $\Xi $ , such that, for each point $(u, v, w) \in U$ , any geodesics joining points in $\{u,v,w\}$ come within $2 \nu + r$ of the point x. In particular, $x\in \pi _{r+2\nu }(u,v,w)$ . If $y\in \pi _r(u,v,w)$ , then $d_{\Gamma }(x,y)\le Q(r+2\nu )$ , so $\pi _r(U)$ has diameter at most $2 Q(r+2\nu )$ .

By compactness, we can cover K with finitely many neighborhoods U as in the last paragraph, so $\pi _r(K)$ is bounded.

Proof. Since $\Gamma $ is $\nu $ –hyperbolic, items (δ1) and (δ2) hold for any $\delta \ge 2\nu $ . For item (δ3), see [Reference Bridson and HaefligerBH99, III.H.3.17.(4)]. Item (δ4) follows from G-equivariance and the fact that $\pi _\delta (a,b,c)$ is nonempty when $\delta $ is large enough. For (δ5), suppose we are given a point z on a geodesic $\gamma $ joining a and b, take $c \in \partial G$ minimizing $\max \{({a}\,|\,{c})_{z}, ({b}\,|\,{c})_{z} \}$ . Cocompactness of the action of G allows one to bound this minimum from above, independently of $a, b$ , and c, and this can be used to give an upper bound on the distance from z to any geodesic joining a or b with c.

Notation 2.8. For the rest of the paper, we fix some $\delta>0$ so the conclusions of Lemma 2.7 hold, and denote the coarse projection $\pi _\delta $ by $\pi $ .

Proof. A standard argument, using only the fact that $\gamma $ is a geodesic in a $\delta $ -hyperbolic space, shows that it is enough to prove c is contained in the closed $(l + 3\delta )$ –neighborhood of $\gamma $ . We will not give the details, as this is classical. We write c as a concatenation $c_1\cdots c_k$ of geodesics, so each $c_i$ joins some $p_{i-1}$ to some $p_i$ . The endpoints of $\gamma $ are $p_0$ and $p_k$ . If $k\le 2$ , we are done by slimness of triangles, so we assume $k\ge 3$ .

Let x be the farthest point from $\gamma $ on c, and let $M=d(x,\gamma )$ . Without loss of generality, we suppose that $M>2\delta $ . It is then straightforward to show that x is within $2\delta $ of some breakpoint $p_i$ (consider the triangle made up of the segment $c_i$ containing x, together with geodesics joining the endpoints of $c_i$ to a point $x'\in \gamma $ closest to x).

Since $M>2\delta $ , the breakpoint $p_i$ cannot be either endpoint of the geodesic $\gamma $ ; in particular, $i\notin \{0,k\}$ . There are two cases, depending on whether or not $i\in \{1,k-1\}$ .

We suppose $i\notin \{1,k-1\}$ ; the case $i\in \{1,k-1\}$ is similar but easier. By the assumption that segments are long, $d(x,\{p_{i\pm 1}\})>2l+6\delta $ . Choose a geodesic $\sigma $ joining $p_{i-1}$ to $p_{i+1}$ . By the assumption on Gromov products in the corners, we have $({p_{i-1}}\,|\,{p_{i+1}})_{p_i}\le l$ . It follows that $d(x,\sigma )\leq l+\delta $ . Let y be a closest point to $p_{i-1}$ in $\gamma $ , and let z be a closest point to $p_{i+1}$ in $\gamma $ . Choose geodesics $[y,z]\subset \gamma $ , $[p_{i-1},y]$ , and $[p_{i+1},z]$ . The point x lies within $l+3\delta $ of some point w on the union of these three geodesics. We claim that $w\in [y,z]$ , so we have $M\le l+3\delta $ .

Indeed, suppose that $w\in [p_{i-1},y]$ (the case $w\in [p_{i+1},z]$ being identical). Now we have

$$ \begin{align*} 0 \leq d(x,y)-d(p_{i-1},y) & \leq d(x,w)+d(w,y) - (d(p_{i-1},w)+d(w,y))\\ & = d(x,w)-d(p_{i-1},w)\\ & \leq d(x,w)-(d(x,p_{i-1})-d(x,w))\\ & = 2d(x,w)-d(x,p_{i-1})\\ & \leq 2(l+3\delta) - d(x,p_{i-1}) < 0 \end{align*} $$

a contradiction. We have thus established that $M \le l+ 3\delta $ , and so c lies in the $l+3\delta $ –neighborhood of  $\gamma $ .

Proof. Choose any $s_0$ in S, and let $\gamma _0 = \gamma _{s_0}$ be a biinfinite geodesic as in the hypothesis, parameterized so that $\gamma _0(0)$ is within H of $s_0$ . Since the points $\gamma _0(\pm \frac {R}{2})$ lie in $B_R(s)$ , there are points $s_{\pm 1}\in S$ whose distances from $\gamma _0(\pm \frac {R}{2})$ are at most H. Since $d_{\Gamma }(s_0,s_{\pm 1})\le \frac {R}{2} + 2 H$ and $d_{\Gamma }(s_{-1},s_1)\ge R- 2 H$ , we deduce $({s_{-1}}\,|\,{s_1})_{s_0}\leq 3 H$ . In particular, we have the following estimates:

(4) $$ \begin{align} d_{\Gamma}(s_0,s_{\pm 1}) \ge \frac{R}{2}-2H\mbox{ and }({s_{-1}}\,|\,{s_1})_{s_0}\leq 3 H. \end{align} $$

Now, we inductively find $s_i$ and $\gamma _i$ for all integers i.

For clarity, we focus on $i>1$ . The construction for $i<0$ is entirely analogous. Suppose we have chosen points $s_{-1},s_0,\ldots s_{i-1}$ in S, and that for each positive $j\le i-1$ , we have chosen a biinfinite geodesic $\gamma _j = \gamma _{s_j}$ , and some $t_j$ within $4H$ of $-\frac {R}{2}$ so that

$$ \begin{align*} d_{\Gamma}(\gamma_{i-1}(0),s_{i-1})& \le H,\mbox{ and }\\ d_{\Gamma}(\gamma_{i-1}(t_{i-1}),s_{i-2})& \le H. \end{align*} $$

Since $R>8H$ , the number $t_{i-1}$ is negative. The point $\gamma _{i-1}(\frac {R}{2})$ lies in the R–ball around $s_{i-1}$ , so we may choose a point $s_i\in S$ so that $d_{\Gamma }(s_i,\gamma _{i-1}(\frac {R}{2}))\le H$ . Let $\gamma _i$ be the geodesic $\gamma _{s_i}$ provided by the hypothesis of the lemma. We can assume that $\gamma _i(0)$ is within H of $s_i$ . The distance $d_{\Gamma }(s_{i-1},s_i)$ differs from $\frac {R}{2}$ by at most $2H$ . Thus, for some $t_i$ of absolute value in $[\frac {R}{2}-4H,\frac {R}{2}+4H]$ , we have $d_{\Gamma }(\gamma _i(t_i),s_{i-1})\le H$ . We parameterize $\gamma _i$ so that $t_i<0$ . This completes the inductive construction.

From the construction, we have

(5) $$ \begin{align} d_{\Gamma}(s_i,s_{i+1})\le \frac{R}{2} + 2 H \end{align} $$

and

$$ \begin{align*} d_{\Gamma}(s_{i-1},s_{i+1}) \ge \frac{R}{2}+ |t_i| - 2 H \ge R - 6 H \end{align*} $$

(the lower bound when $i=0$ is slightly better). This implies a bound on Gromov products

(6) $$ \begin{align} ({s_{i-1}}\,|\,{s_{i+1}})_{s_i} \le 5 H. \end{align} $$

Let $c_k$ be a piecewise geodesic formed by concatenating geodesics

$$\begin{align*}[s_{-k},s_{-k+1}]\cdots [s_{k-1},s_k]. \end{align*}$$

We verify the hypotheses of Lemma 2.10 with $l = 5H$ . The inequality (6) gives the bound on Gromov products in the corners. The inequality (5) gives that the segments $[s_i,s_{i+1}]$ have length at least $\frac {R}{2} - 2H> 10 H + 8 \delta = 2 l + 8\delta $ as required. Thus, if $\beta _k$ is the geodesic joining the endpoints of $c_k$ , we have $d_{\mathrm {Haus}}(c_k,\beta _k)\le H + 4\delta $ .

Since $\Gamma $ is proper, and the geodesics $\beta _k$ all pass through the $(H+4\delta )$ –ball about $s_0$ , they subconverge to a biinfinite geodesic $\gamma $ . Notice that all the segments $[s_i,s_{i+1}]$ lie in the $(H+4\delta )$ –neighborhood of $\gamma $ . We will show this $\gamma $ satisfies the conclusions of the lemma.

If $s\in S$ , then there is a $\frac {R}{4}$ –coarse path joining $s_0$ to s, that is to say there exist points $s_0=p_0 ,\ p_1,\ \ldots ,\ p_k = s$ in S satisfying.

$$\begin{align*}d_{\Gamma}(p_i,p_{i+1})\le \frac{R}{4},\ \forall i. \end{align*}$$

We first claim that for each $p_i$ , there is some $s_{j(i)}$ with $d_{\Gamma }(s_{j(i)},p_i)\le \frac {R}{4}+2 H$ . Clearly this is true for $p_0$ . Arguing by induction, we see that $d_{\Gamma }(p_{i+1},s_{j(i)})$ is at most $\frac {R}{2}+2 H$ . In particular, there is a point q on $\gamma _{s_{j(i)}}$ within H of $p_{i+1}$ . Recalling that $\gamma _{s_{j(i)}}(0)$ lies within H of $s_{j(i)}$ , we see that $q = \gamma _{s_{j(i)}}(t)$ for some t with

$$\begin{align*}|t| \le \frac{R}{2} + 3H<\frac{3}{4}R.\end{align*}$$

Thus, for some $t' \in \{-\frac {R}{2}, 0, \frac {R}{2}\}$ , we have $|t-t'|\le \frac {R}{4}$ . Since the points $\gamma _{s_{j(i)}}(\pm \frac {R}{2})$ are within H of $s_{j(i)\pm 1}$ , there is some $s_{j(i+1)}\in \{s_{j(i)-1},s_{j(i)},s_{j(i)+1}\}$ so that $d_{\Gamma }(p_{i+1},s_{j(i+1)}) \le 2H + \frac {R}{4}$ , as desired.

Now let $s_j = s_{j(k)}$ , so we have $d_{\Gamma }(s,s_j)\le \frac {R}{4} + 2H$ , and let q be a closest point to s on $\gamma _j$ . Without loss of generality, we suppose that $q = \gamma _j(t)$ for $t\ge 0$ . Any quadrilateral with corners $s_j, s_{j+1},\gamma _j(0),\gamma _j(\frac {R}{2})$ is $2\delta $ –thin, so there is a point r on $[s_j,s_{j+1}]$ within $H+2\delta $ of q. This point r is within $H+4\delta $ of some point z on $\gamma $ . Adding up the constants, we have

$$ \begin{align*} d_{\Gamma}(s,z) & \le d_{\Gamma}(s,q) + d_{\Gamma}(q,r) + d_{\Gamma}(r,z)\\ & \le H + H + 2\delta + H + 4\delta = 3H + 6\delta. \end{align*} $$

This shows

(7) $$ \begin{align} S \subset N_{3H + 6\delta}(\gamma). \end{align} $$

Conversely, let $x\in \gamma $ . Then $x\in \beta _k$ for some k, and so for some i, there is a point $y\in [s_i,s_{i+1}]$ with $d_{\Gamma }(x,y)\le H + 4\delta $ . This point is within $H+2\delta $ of a point on $\gamma _i$ , which is within H of a point of S, so we have

(8) $$ \begin{align} \gamma \subset N_{3H+6\delta} (S). \end{align} $$

Together, (7) and (8) imply the first statement of the lemma, that is to say the bound on Hausdorff distance. It remains to show the statement about Gromov products. Breaking symmetry, we consider just the ray $\gamma _s|[0,\infty )$ . Let $y'$ be a point on $\gamma _s|[0,\infty )$ at distance R from s. Let $s'\in S$ be a point within H of $y'$ , and let $z'$ be a point on $\gamma $ within $3H + 6\delta $ of $s'$ . Let $\alpha $ be a ray starting at s with limit point $\gamma _s^{+\infty }$ , and let $\beta $ be a ray starting at s with limit point $\gamma ^{+\infty }$ . There are points y on $\alpha $ and z on $\beta $ which are within $\delta $ of $y'$ , $z'$ , respectively. We have $d_{\Gamma }(s,y)\ge R-\delta $ , $d_{\Gamma }(s,z) \ge R - (4H + 7\delta )$ and $d_{\Gamma }(y,z) \le 3H + 8\delta $ , so

$$\begin{align*}({y}\,|\,{z})_{s} \ge \frac{1}{2}(R - \delta + R-(4H + 7\delta) - (4 H + 8\delta)) = R-(4H + 8\delta). \end{align*}$$

Lemma 2.1 allows us to conclude $({\gamma _s^{+\infty }}\,|\,{\gamma ^{+\infty }})_{s} \ge R - (4H + 10\delta )$ , as desired.

Proof. The group G already acts properly and cocompactly on X, and $N_\kappa (F)$ is open and G–invariant, so G still acts properly and cocompactly on $X_\kappa $ . The only points of $\Xi $ with nontrivial stabilizer are in F, which has been removed.

Proof. Let $D'$ be a compact subset of $\Xi $ containing $B_{ \epsilon '}(x) \times B_{ \epsilon '}(y) \times B_{ \epsilon '}(z)$ for all $(x, y, z) \in D$ . Let F be the set of elements $g\in G - \{\mathbf {e}\}$ so that g fixes some triple in $D'$ . By Corollary 5.7, the fixed point sets of torsion elements form a null sequence in $\partial G$ . For each $g\in F$ , we have $\mathrm {diam}(\mathrm {fix}(g))\ge \mathrm {minsep}(D')> 0$ , so F is finite. Let $C = \bigcup \{\mathrm {fix}(g) \mid g\in F\}$ be the union of these fixed sets in $\partial G$ . Since C is a finite union of closed sets with empty interior, C is closed with empty interior.

By Lemma 5.9, we can choose a positive $\lambda < \epsilon "$ , such that $\partial G-N_\lambda (C)$ is $\epsilon "$ –dense in $\partial G$ . Choose $\kappa $ small enough so that $N_{2\kappa }(F) \cap D'$ is a subset of the neighborhood of F of radius $\lambda $ in the product visual metric on $(\partial G)^3$ .

To verify this $\kappa $ works, choose $(a, \theta , c) \in D$ . Since $\partial G-N_\lambda (C)$ is $\epsilon "$ –dense, we can find $c'\in \partial G - N_\lambda (C)$ with $d_{\mathrm {vis}}(c, c')< \epsilon "$ . We claim that $\{a\} \times B_{\epsilon '}(\theta ) \times \{c'\}$ is contained in $\Xi - N_{2\kappa }(F)$ , and hence in the interior of $X_\kappa $ . Indeed, for any $b \in B_{\epsilon '}(\theta )$ , we have $(a,b,c') \in D'$ . Suppose $(x,y,z) \in F$ is a closest point of F to $(a,b,c')$ in the product metric. If the visual distance between each coordinate of $(x,y,z)$ and $(a,b,c)$ were less than $\lambda $ , which is less than $\epsilon '$ , then we would have $(x,y,z) \in D' \cap F$ . By our definition of C, this means that x, y, and z all belong to C. But the minimum distance from $c'$ to a point of C is greater than $\lambda $ by construction, so we conclude $(a,b,c')$ lies outside the $\lambda $ –neighborhood of F in the product metric, and hence outside $N_{2\kappa }(F)$ .

Proof

Fix a round metric $d_{\mathrm {rnd}}$ on $\partial G = S^n$ . The collection

$$\begin{align*}C_m:= \left \{(a, b, c) \in \Xi \mid d_{\mathrm{rnd}}(a,b)\geq \tfrac{1}{m},\ d_{\mathrm{rnd}}(a, c) \geq \tfrac{1}{m},\ d_{\mathrm{rnd}}(b,c) \geq \tfrac{1}{m}\right \}\end{align*}$$

forms an exhaustion of $\Xi $ by compact sets, so we may choose m sufficiently large so that $K \subset C_m$ and the diameter of $S^n$ in the round metric is larger than $\frac {10}{m}$ .

Since D is compact, only finitely many elements of G fix a triple of points that meets D. Let $C \subset S^n$ denote the union of these finitely many fixed sets; it is a closed subset of $\partial G$ with empty interior.

By Lemma 5.9, there exists $\lambda < \frac {1}{m}$ , such that $S^n - N_\lambda (C)$ is $\frac {1}{m}$ -dense in $S^n$ (again, using the round metric). Consider a shortest length geodesic path with respect to the round metric on $S^n$ between a and $a'$ . If this path does not meet the closed $\frac {1}{m}$ -ball about b, then call this path $a_t$ . Otherwise, modify this path by removing the segment that intersects the closed $\frac {1}{m}$ -ball, replacing it with a path that lies on the boundary of this ball, and call the resulting path $a_t$ (we are here using our Assumption 4.4 that $n \ge 2$ ). In either case, $a_t$ lies in the union of a radius $\frac {1}{m}$ -ball and a geodesic segment.

Since the diameter of $S^n$ in the chosen round metric is larger than $\frac {10}{m}$ , we may find a point $\hat {c} \in S^n$ that avoids both the $\frac {2}{m}$ -neighborhood of this path $a_t$ and the $\frac {2}{m}$ -neighborhood of b. Since $S^n - N_\lambda (C)$ is $\frac {1}{m}$ –dense, we may move this point a distance of at most $\frac {1}{m}$ to find a point $c" \notin N_\lambda (C)$ . Thus, we have $(a_t, b, c") \in D$ for all t.

Finally, choose $\kappa $ small enough so that any element of a triple in $N_\kappa (F) \cap D$ lies within the $\lambda $ -neighborhood (in the round metric) of some point of C. Note that this choice of $\kappa $ only depends on D and $\lambda $ , which both depended only on the set K. The point $c"$ was chosen so that $c" \notin N_\lambda (C)$ . Thus, we conclude that $(a_t, b, c") \notin N_\kappa (F)$ holds for all t, as desired.

Notation 6.1. Let $H = \max \{2\delta ,Q(3\delta )\}+1$ , where $Q(\cdot )$ is from Lemma 2.5.

Proof. We let $D_0$ be a compact subset of $\Xi $ containing $\pi ^{-1}(B_1(\mathbf {e}))$ and define $D_{\frac {1}{2}}$ as in Item (2). We fix any positive $\epsilon _1$ satisfying Item (3). Let $K \subset \Xi $ be a compact set large enough to contain the union of $\pi ^{-1}(B_R(\mathbf {e}))$ , where R is in equation (⋆⋆), as well as the union of the sets $B_{\epsilon _1}(x)\times B_{\epsilon _1}(y)\times B_{\epsilon _1}(z)$ for every $(x,y,z)\in D_{\frac {1}{2}}$ . Applying Lemma 5.13 to K gives us a constant $\kappa _0$ and set $D_1$ containing K so that the $\kappa _0$ path property holds for $D_1$ ; note that by definition, the $\kappa $ path property remains true for any $\kappa < \kappa _0$ , since $N_\kappa (F) \subset N_{\kappa _0}(F)$ . Then take any $\epsilon _2$ and $D_2$ as described.

Proof. That the first condition can be met is a consequence of compactness. Indeed, by Lemma 2.6, the $10 \delta $ -neighborhood of the set $\pi (D_2)$ is bounded in $\Gamma $ , hence contained in some compact ball centered at $\mathbf {e}$ . Thus, there is a positive lower bound on the visual distance between the endpoints of any geodesic passing through that ball.

We now address the second condition. Since $S^n$ is compact, the visual and round metrics are uniformly equivalent. Thus, there is a $\mu>0$ so that the round ball of radius $2\mu $ about p is contained in $B_{\epsilon _2}(p)$ for every $p\in S^n$ , so we can take $B_p$ to be the closed round $\mu $ –ball. There is then some $\epsilon>0$ so that $B_\epsilon (p)\subset B_p$ for every p.

Convention 6.5. Fix some $\rho \in \mathcal U(V) $ . Since $\rho $ is fixed, we henceforth drop it from the notation, writing $f = f^\rho $ . Since $\kappa $ is also fixed, we also drop it from our notation when convenient, writing X for $X_\kappa $ .

Notation 6.7. For $\theta \in \partial G$ , we let $Y_\theta = f(X \times \{\theta \}) \subset X \times \partial G$ . For $a, c, \theta \in \partial G$ , we let $f_{a,c,\theta }$ denote the map $x \mapsto f_\theta (a, x, c)$ . This map is defined on all points $x \in \partial G$ , such that $(a, x, c) \in X$ and it is continuous on its domain of definition.

Proof. By Item (2) of Definition 6.4, there exists a point $c' \in B_\epsilon (c)$ , such that $\{a\}\times B_{\epsilon _2}(\theta )\times \{c'\} \subset X$ . Thus, $f_{a,c',\theta }$ is defined on $B_{\epsilon _2}(\theta )$ and Definition 6.4, Item (4) implies that $f_{a,c',\theta }(B_{\epsilon _2}(\theta )) \subset B_\epsilon (\theta )$ . Lemma 6.3 states that there exists a closed, contractible set $B_\theta $ , such that

$$\begin{align*}B_{\epsilon}(\theta) \subset B_\theta \subset B_{\epsilon_2}(\theta) .\end{align*}$$

Thus,

$$\begin{align*}f_{a,c',\theta}(B_\theta) \subset f_{a,c',\theta}(B_{\epsilon_2}(\theta))\subset B_\epsilon(\theta) \subset B_\theta.\end{align*}$$

By the Lefschetz fixed point theorem, $f_{a,c',\theta }$ has a fixed point $b_\theta \in B_{\epsilon }(\theta )$ , which means exactly that $((a, b_\theta , c'), b_\theta ) \in L_a \cap Y_\theta $ . Recall that $\epsilon < \epsilon _2$ , so by our choice of sets and constants as in Lemma 6.2, we have also $(a, b_\theta , c') \in D_2$ .

Notation 6.9. Going forward, we define $S(a,\theta ) := \bar \pi (L_a\cap Y_\theta ) \subset \Gamma $ .

Proof. For the first inclusion, suppose that p lies in $S(a,\theta )\cap B_R(\mathbf {e})$ . Then, since $D_1$ contains $\pi ^{-1}(B_R(\mathbf {e}))$ , we have $p \in \overline {\pi }(f((D_1 \cap X)\times \{\theta \})\cap L_a)$ . In particular, there exist $b,c\in \partial G$ , such that all the following hold simultaneously:

1. 1. $p\in \pi (a,b,c)$ ,

2. 2. $(a,b,c)\in D_1 \cap X$ ,

3. 3. $f_\theta (a,b,c) = b$ .

Both $(a,b,c)$ and $(a,b_0,c_0)$ lie in $(D_1 \cap X) \subset (D_2 \cap X)$ . By item (4) in Definition 6.4, we conclude that $b = f_\theta (a,b,c)$ and $b_0 = f_\theta (a,b_0,c_0)$ both lie in $B_\epsilon (\theta )$ . In particular, $d_{\mathrm {vis}}(b,b_0) < 2\epsilon $ , so any geodesic from b to $b_0$ misses $B_R(\mathbf {e})$ by at least $10\delta $ (Lemma 6.3.(ϵ1)). By $\delta $ –slimness of ideal triangles, this implies that if $[a,b]$ is any geodesic joining a to b, then $[a,b]\cap B_{R+8\delta }(\mathbf {e})$ lies in the $\delta $ –neighborhood of $\gamma $ (the geodesic joining a to $b_0$ ). By definition of $\pi $ , the point p lies within $\delta $ of a point on $[a,b]$ which is at most $R+\delta $ from $\mathbf {e}$ , so lies within $2\delta $ of $\gamma $ . This shows the first inclusion.

For the second inclusion, suppose that $p\in \gamma \cap B_R(\mathbf {e})$ . By assumption (δ5) on $\delta $ , there is some c so that $p\in \pi (a,b_0,c)$ . Since $D_1$ contains $\pi ^{-1}(B_R(\mathbf {e}))$ , we have $(a,b_0,c)\in D_1$ .

By Item (2) of Definition 6.4, there exists $c' \in B_\epsilon (c)$ , such that

$$\begin{align*}\{a\}\times B_{\epsilon_2}(b_0)\times\{c'\} \subset (X \cap D_2).\end{align*}$$

Let $B = B_\theta $ be the contractible set from Lemma 6.3.(ϵ2). This set contains $B_\epsilon (\theta )$ , so in particular, $b_0\in B$ . We have $\{a\}\times B\times \{c'\}\subset X \cap D_2$ , so as in the proof of Lemma 6.8, $f_\theta $ induces a map $f_{a,c',\theta }\colon \thinspace B\to B_{\epsilon }(\theta )\subset B$ , and $f_{a,c',\theta }$ fixes some $b'\in B_\epsilon (\theta )\subset B_{2\epsilon }(b_0)$ . The point $((a,b',c'),b')$ is therefore in $L_a\cap Y_\theta $ , so $\pi (a,b',c')$ is a subset of $S(a,\theta )$ . Let $q\in \pi (a,b',c')$ .

See Figure 1.

Figure 1

$p \in \pi (a,b,c)$ is close to any point q of $\pi (a, b', c')$ .

We claim q is close to p. To see this, we first show p is within $2\delta $ of any geodesic from a to $b'$ . Fix such a geodesic $[a,b']$ , and geodesics $[b_0, b']$ and $[b_0, a]$ . Since $p \in B_R(\mathbf {e})$ , it lies distance at least $10 \delta $ from any point on $[b_0, b]$ . Also, we have that p lies within $\delta $ of some point on $[b_0,a]$ , so by $\delta $ -thinness, p is distance at most $2\delta $ from a point on $[a,b']$ .

Repeating this argument with $c'$ in place of $b'$ shows that p is within $2\delta $ of any geodesic from a to $c'$ . A slight modification shows that p is within $3\delta $ of any geodesic $[b',c']$ , as follows: Considering a quadrilateral with sides $[b_0, c]$ , $[c, c']$ , $[b', c']$ , and $[b_0, b']$ , we know that p is within $\delta $ of some point on $[b_0, c]$ and this must lie within $2\delta $ of a point on $[b', c']$ since the other two sides are each distance at least $10\delta $ from p. Thus, $p \in \pi _{3\delta }(a,b',c')$ , and of course q lies in this set as well.

By Lemma 2.5, $\pi _{3\delta }(a,b,c)$ has diameter at most $Q(3\delta )$ . In particular, $d_{\Gamma }(p,q)\le Q(3\delta )$ . Since $q\in S(a,\theta )$ , this shows the second inclusion.

Proof. We first verify the hypotheses of Lemma 2.12 for the set $S = S(a,\theta )\cap G$ (recall that G is canonically identified with the vertices of $\Gamma $ ). By Corollary 5.5, $\pi $ is surjective, so for each $s\in \mathcal {S}$ , we can choose some $b_s, c_s$ so that $\sigma (a,b_s,c_s) \in L_a\cap Y_\theta $ and $s\in \pi (a,b_s,c_s)$ . Since $\pi ^{-1}(\mathbf {e})\subset D_0$ , the point $\sigma (a,b_s,c_s)$ lies in $\sigma (s\cdot D_0)\cap L_a\cap Y_\theta $ . Left translating by $s^{-1}$ , we have

$$\begin{align*}(s^{-1}a,s^{-1}b_s,s^{-1}c_s)\in \sigma(D_0)\cap L_{s^{-1}a}\cap Y_{\rho(s^{-1})\theta}.\end{align*}$$

Let $\hat {\gamma }_s$ be any biinfinite geodesic from $s^{-1}a$ to $s^{-1}b_s$ . Lemma 6.10 implies

$$\begin{align*}s^{-1}S\cap B_R(\mathbf{e})\subset N_H(\hat{\gamma}_s),\mbox{ and } \hat{\gamma}_s\cap B_R(\mathbf{e})\subset N_H(S).\end{align*}$$

Setting $\gamma _s$ equal to $s \cdot \hat {\gamma }_s$ , we find that

$$\begin{align*}S\cap B_R(s)\subset N_H(\gamma_s),\mbox{ and } \gamma_s\cap B_R(s)\subset N_H(S).\end{align*}$$

Now fix any $\frac {R}{4}$ –connected component $S_0$ of S so that we may apply Lemma 2.12 (we will see later that S is $\frac {R}{4}$ –connected, so in fact $S_0 = S$ ). The paragraph above shows that all the hypotheses of Lemma 2.12 hold for $S_0$ , so we conclude that there is a biinfinite geodesic $\gamma $ with $d_{\mathrm {Haus}}(\gamma ,S_0) \le 3H + 6\delta $ .

We next claim that one of the endpoints of $\gamma $ is a. We argue by contradiction. Using $\delta $ –slimness of ideal triangles, there is a point $p\in \gamma $ so that p is within $2\delta $ of any geodesic joining a to any endpoint of $\gamma $ . Since $S_0$ is Hausdorff distance at most $3H + 6\delta $ from $\gamma $ , there is some $s\in S_0$ so that $d_{\Gamma }(s,p)\le 3H + 6\delta $ . Now $\gamma _s$ has an endpoint at a, and we have $\gamma _s\cap B_R(s)\subset N_H(S)$ . From the inclusions $S\cap B_R(s)\subset N_H(\gamma _s),\mbox { and } \gamma _s\cap B_R(s)\subset N_H(S)$ and the inequality $6H < \frac {R}{4}$ , we conclude that $N_H(\gamma _s) \cap S \subset S_0$ . Choosing $x\in \gamma _s$ at distance $\frac {R}{2}$ from s, in the direction of a, we find some point $s'\in S_0$ with $d_{\Gamma }(x,s')\le H$ . See Figure 2 for a schematic. By repeated applications of the triangle inequality, one can easily show that this $s'$ is further than $3H + 6\delta $ from $\gamma $ , a contradiction.

Figure 2

$\gamma $ and $\gamma _s$ should both be close to $S_0$ on the shaded region $B_R(s)$ , giving a contradiction if a is not an endpoint of $\gamma $ .

We now argue that $S = S_0$ . To see this, suppose that $S_1$ were some other component. We may apply the same argument to $S_1$ to produce a biinfinite geodesic $\gamma _1$ . The geodesics $\gamma $ and $\gamma _1$ share an endpoint a, and so contain points within $\delta $ of one another. This implies that $S_0$ and $S_1$ contain points within $6H + 13\delta < \frac {R}{4}$ of one another, so they cannot be different $\frac {R}{4}$ –connected components.

We have established the first conclusion, since $d_{\mathrm {Haus}}(S,S(a,\theta ))\le 1$ .

Now we suppose that $s\in S$ is in $\bar \pi (\sigma (D_0)\cap L_a\cap Y_\theta )$ . By Lemma 6.8, we may take s to be in $\bar \pi ((a,b,c),b)$ for some $b\in B_\epsilon (\theta )$ . We can therefore take $\gamma _s$ in the first part of this argument to be a geodesic joining a to b. The second conclusion of Lemma 2.12 implies that $({e^+}\,|\,{b})_{s}\ge R-(4H + 10\delta )$ , where $e^+$ is the endpoint of $\gamma $ which is not equal to a. We thus have

$$ \begin{align*} ({e^+}\,|\,{b})_{\mathbf{e}} & \ge ({e^+}\,|\,{b})_{s} - d_{\Gamma}(\mathbf{e},s)\\ & \ge ({e^+}\,|\,{b})_{s} - \mathrm{diam} (\pi(D_0)) \\ & \ge R - (4H + 10\delta) .\end{align*} $$

The first condition on $\epsilon $ in Lemma 6.3 implies that $({b}\,|\,{\theta })_{\mathbf {e}}\ge R$ . We have

$$ \begin{align*} ({e^+}\,|\,{\theta})_{\mathbf{e}}& \ge \min\{ ({e^+}\,|\,{b})_{\mathbf{e}}, ({b}\,|\,{\theta})_{\mathbf{e}}\} - \delta\\ & \ge R - (4H + 11\delta), \end{align*} $$

establishing the last claim of the proposition.

Proof of Lemma 7.2

If $B_r(\mathbf {e})\cap S(a_0,\theta _0)$ is empty, there is nothing to show, so we suppose $B_r(\mathbf {e})\cap S(a_0,\theta _0)$ is nonempty. Let $K\subset X$ be the closure of $\pi ^{-1}(B_r(\mathbf {e}))$ .

Recall from Notation 6.7 that $f_{a,c,\theta }$ denotes the map $x \mapsto f_\theta (a, x, c)$ , and recall that $D_0\subset D_1$ has the following properties

1. 1. $X \subset GD_0$

2. 2. For any $(a,b,c)\in D_0$ , the set $B_{\epsilon _1}(a) \times B_{\epsilon _1}(b) \times B_{\epsilon _1}(c)$ is contained in $D_1$ (see Lemma 6.2.(4)).

Recall also that $\epsilon _1> \epsilon _2 > 2 \epsilon $ .

We cover $\sigma (K) \cap (L_{a_0} \cap Y_{\theta _0})$ with translates of $D_0$ , as follows. Since K is compact, there are finitely many elements $g_1, g_2, \ldots g_k \in G$ so that

$$\begin{align*}\sigma(K) \cap (L_{a_0} \cap Y_{\theta_0}) \subset \bigcup_{i=1}^k \sigma(g_i D_0). \end{align*}$$

By deleting elements from the list if necessary, we may assume

$$\begin{align*}\sigma(g_i D_0) \cap \left(L_{a_0} \cap Y_{\theta_0} \right) \neq \emptyset\end{align*}$$

for each i.

Our next goal is to show that, for each i, the projection of the larger translate $g_iD_1$ to the Cayley graph contains a point of $S(a, \theta )$ , provided that $(a, \theta )$ is chosen close enough to $(a_0,\theta _0)$ . Here, “close enough” depends on the set K and hence on the constant r.

Translating back to $D_0$ , for each i, we have $\sigma (D_0) \cap \left ( L_{g_i^{-1}a_0} \cap Y_{\rho (g_i)^{-1}\theta _0} \right ) \neq \emptyset $ . Let $b_i, c_i \in \partial G$ be such that

$$\begin{align*}\left( ( g_i^{-1}a_0, b_i, c_i ), b_i \right) \in \sigma(D_0) \cap \left( L_{g_i^{-1} a_0} \cap Y_{\rho(g_i)^{-1}\theta_0} \right). \end{align*}$$

Because $ \left ( g_i^{-1}a_0, b_i, c_i \right ) \in D_0 \subset D_2$ , Definition 6.4, Item (4) implies that

$$\begin{align*}b_i = f_{\rho(g_i)^{-1}\theta_0}(g_i^{-1}a_0, b_i, c_i) \in B_{\epsilon}(\rho(g_i)^{-1}\theta_0). \end{align*}$$

Hence, $d_{\mathrm {vis}}(b_i,\rho (g_i)^{-1}\theta _0) < \epsilon $ , and so $(g_i^{-1}a_0, \rho (g_i)^{-1}\theta _0, c_i ) \in D_1$ .

Let ${p_i} = \rho (g_i)^{-1}\theta _0$ . By Item (2) of Definition 6.4, there is some s $c^{\prime }_i \in B_\epsilon (c_i)$ so that $ \{a_i\} \times B_{\epsilon _2}({p_i}) \times \{c^{\prime }_i\}$ lies in the interior of X. Furthermore, by Lemma 6.3, Item (ϵ2), there is a closed contractible set $B_{p_i} \subset \partial G$ with $B_\epsilon ({p_i}) \subset B_{p_i} \subset B_{\epsilon _2}({p_i})$ , and

$$\begin{align*}f_{g_i^{-1} a_0, c^{\prime}_i, {p_i}}(B_{\epsilon_2}({p_i})) \subset B_\epsilon({p_i}),\end{align*}$$

so $f_{g_i^{-1} a_0, c^{\prime }_i, {p_i}}$ has a fixed point in $B_\epsilon ({p_i})$ . The property of taking the compact set $B_{p_i}$ into the open ball $B_\epsilon ({p_i})$ is open (in the compact-open topology on continuous maps), so also holds for any map sufficiently close to $f_{g_i^{-1} a_0, c^{\prime }_i, {p_i}}$ , provided the map is defined on $B_{p_i}$ . Recall that the domain of definition of $f_{x,y,z}$ is the set $\{ w \mid (x, w, z) \in X \}$ . Thus, if a function $f_{x,y,z}$ is defined on a set $\{x\} \times B_{\epsilon _2}({p_i}) \times \{z\}$ contained in the interior of X, and $B_{\epsilon _2}({p_i}) \supset B_{p_i}$ , then for all sufficiently close $x', y', z'$ , the function $f_{x',y', z'}$ will be defined on $B_{p_i}$ as well. Additionally, the functions $f_{x,y,z}$ vary continuously in the arguments $(x,y,z)$ . Thus, we may take a neighborhood $N_i$ of $(a_0, \theta _0)$ , such that for each $(a, \theta ) \in N_i$ ,

1. 1. the map $f_{g_i^{-1} a, c^{\prime }_i, \rho (g_i)^{-1}\theta }$ is defined on $B_{p_i}$

2. 2. the map $f_{g_i^{-1} a, c^{\prime }_i, \rho (g_i)^{-1}\theta }$ has a fixed point contained in $B_\epsilon ({p_i})$ ; and

3. 3. $d_{\mathrm {vis}}(g_i^{-1} a,g_i^{-1} a_0)< \epsilon _1$ .

Set $N = \bigcap _{i=1}^k N_i$ . Thus, for any $(a, \theta ) \in N$ , each of the sets $L_{g_i^{-1}(a)} \cap Y_{\rho (g_i)^{-1}\theta }$ contains a point $\sigma (g_i^{-1} a, z_i, c^{\prime }_i)$ , where $z_i \in B_\epsilon ({p_i})$ . In particular, we have

$$ \begin{align*} & d_{\mathrm{vis}}(g_i^{-1} a, g_i^{-1} a_0)< \epsilon_1 \text{ and } \\ & d_{\mathrm{vis}}(z_i,b_i)\le 2\epsilon <\epsilon_1. \end{align*} $$

Since $(g_i^{-1} a_0,b_i,c_i)\in D_0$ , this means that $(g_i^{-1} a, z_i, c^{\prime }_i)\in D_1$ . Multiplying on the left by $g_i$ , we obtain $(a,g_iz_i,g_ic^{\prime }_i) \in g_iD_1$ and $\sigma (a,g_iz_i,g_ic^{\prime }_i)\in L_{a}\cap Y_{\theta }$ , so the intersection

$$\begin{align*}\sigma(g_i D_1) \cap (L_a\cap Y_\theta) \end{align*}$$

is nonempty for each of the elements $g_i$ . Projecting to the Cayley graph, we have $B_r(\mathbf {e})\cap S(a_0,\theta _0)$ contained in the $\mathrm {diam}(\pi (D_1))$ –neighborhood of $S(a,\theta )$ , which proves the lemma.

Proof of Proposition 7.1 from Lemma 7.2

Suppose $(a_0, \theta _0) \in \partial G \times \partial G$ is such that there exists c with $(a_0, \theta _0, c) \in D_1$ . Then by Lemma 6.8, $L_{a_0} \cap Y_{\theta _0}$ contains a point of $\sigma (D_2)$ , so is nonempty. Lemma 7.2 states that, given $r>0$ , there is a neighborhood N of $(a_0,\theta _0)$ so that if $(a , \theta ) \in N$ , then $B_r(\mathbf {e})\cap S(a_0,\theta _0)$ lies in the $\mathrm {diam}(\pi (D_1))$ –neighborhood of $S(a,\theta )$ . By Proposition 6.11, both $S(a_0,\theta _0)$ and $S(a,\theta )$ are Hausdorff distance at most $3H+6\delta +1$ from some biinfinite geodesic, so these geodesics will $2(3H + 6\delta + 1)+ \mathrm {diam}(\pi (D_1))$ fellow-travel each other over a compact set, which can be taken as large as we wish by taking r large. This gives continuity.

Proof. Equivariance of each map in the composition implies that we have the equivariance property

$$\begin{align*}\mathcal{E}^+(g x) = \mathcal{E}^+(g a(x),\rho(g)\theta(x)) .\end{align*}$$

It therefore suffices to check continuity of the composition above on the set $D_0 \cap X$ containing a fundamental domain for the action of G on X.

By definition, the section $\sigma $ is continuous. For the second map in (13), note that $\theta (x)$ is simply projection onto the second coordinate of $f^{-1}(\sigma (x)) \in X \times \partial G$ . Since $f^{-1}$ is a homeomorphism of $X \times \partial G$ , its projection $\theta (x)$ is continuous. Finally, Proposition 7.1 says that $(a, \theta (a,b,c)) \mapsto e^+(a, \theta )$ is continuous for $(a,b,c) \in D_1$ .