Hyperbolic Spaces and Hyperbolic Groups

Let $\mathrm{△}xyz$ be a geodesic triangle in $T$. By [A1], the three geodesic segments $[x,y]$, $[y,z]$, and $[z,x]$ are unique, so their union is a tripod with a single branch point.

In a tripod, each side is contained in the union of the other two sides. Thus every geodesic triangle in $T$ is $0$-slim, so $T$ is $0$-hyperbolic. The final sentence follows from [L1].

If triangles are $\delta$-slim, [F1] proves condition (2) with the same constant $3\delta$ at every basepoint.

Now assume (2) and fix just one point $o\in X$. Let $\kappa ={\delta }_o^{\mathrm{\prime }}$. For any four points $a,b,c,d$, write $p_{uv}=(u\mathrm{\mid }v{)}_o$ and $r_u=d(o,u)$. On these four points define $q_{uv}$ to be the maximum, over all simple edge paths from $u$ to $v$ in the complete graph, of the least $p$-value of an edge on the path; put $q_{uu}=r_u$. There are finitely many paths. The one-edge path gives $q_{uv}\ge p_{uv}$. Along a two-edge path the assumed product inequality gives $p_{uv}\ge \min (p_{us},p_{sv})-\kappa$, and along a three-edge path it gives $p_{uv}\ge \min (p_{us},p_{st},p_{tv})-2\kappa$. Hence $0\le q_{uv}-p_{uv}\le 2\kappa$. Also $q_{uv}\le \min (r_u,r_v)$, since the first and last edges of every path satisfy these respective bounds.

Concatenate paths attaining $q_{uv}$ and $q_{vw}$ and erase any loops. Erasing loops cannot lower the minimum edge value. Thus $q_{uw}\ge \min (q_{uv},q_{vw})$: $q$ is an exact ultrametric similarity on these four labels. For completeness, its positive threshold relations $u{\sim }_{t}v⟺q_{uv}\ge t$ are nested equivalence relations (on labels with $r_u\ge t$). Make a finite rooted tree from these nested clusters, with each leaf $u$ at height $r_u$ and each common ancestor of $u,v$ at height $q_{uv}$. Nonnegative edge lengths follow from the bound in step 1.2. The tree distance between leaves is $D(u,v)=r_u+r_v-2q_{uv}$. Removing the finite subtree spanned by four leaves at its central edge or central vertex shows that the largest two of its three opposite-pair distance sums are equal: each uses the central edge twice, while the third uses it zero times; zero-length edges and repeated leaves follow by the same calculation.

The original metric satisfies $d(u,v)=r_u+r_v-2p_{uv}$, so $0\le d(u,v)-D(u,v)\le 4\kappa$. Each opposite-pair sum therefore differs from its tree counterpart by a number in $[0,8\kappa ]$. Since the two largest tree sums are equal, the largest and second-largest original sums differ by at most $8\kappa$: the two original sums corresponding to those equal tree sums both lie in one interval of length $8\kappa$, while the remaining original sum can only increase the second-largest if it becomes larger. This is the four-point condition with constant $4\kappa$ (additive error $8\kappa$). The bound uses the one fixed basepoint $o$, so condition (2)'s per-basepoint quantifier causes no uniformity gap.

Finally (3) is the four-point condition with constant $\kappa ={\delta }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2$. By [F2] every triangle is $4\kappa =2{\delta }^{\mathrm{\prime }\mathrm{\prime }}$-slim. This proves (1) and closes the cycle.

By [A1], each point of the side $[a,b]$ lies either within distance $\delta$ of $[a,c]$ or within distance $\delta$ of $[b,c]$ by applying [L1] to triangle $abc$.

If such a point lies near $[a,c]$, then applying [L1] to triangle $acd$ shows that the nearby point on $[a,c]$ lies within distance $\delta$ of $[a,d]\cup [d,c]$. Hence every point of $[a,b]$ lies within distance $2\delta$ of $[b,c]\cup [c,d]\cup [d,a]$. Cyclic symmetry gives the same bound for each side, so the quadrilateral is $2\delta$-thin.

Choose one geodesic $\gamma$ joining the common endpoints. By [F1], each image $Q_i=\mathrm{im}(q_i)$ has Hausdorff distance at most $M$ from $\gamma$. This means both that every point of $Q_i$ is within distance $M$ of $\gamma$ and that every point of $\gamma$ is within distance $M$ of $Q_i$, with infimum distances understood as in [F1].

Fix $x\in Q_1$ and $h>0$. Choose $z\in \gamma$ with $d(x,z)<M+h$, then $y\in Q_2$ with $d(z,y)<M+h$. Thus $d(x,Q_2)\le 2M+2h$; letting $h$ decrease to zero gives $d(x,Q_2)\le 2M$. Reverse the roles of $Q_1,Q_2$ for the other inclusion. Therefore their Hausdorff distance is at most $2M=184{\lambda }^2(\epsilon +3\delta )$, as claimed.

Let $f:X\to Y$ be the given quasi-isometry. If $Y$ is $\delta$-slim, [F1] first extracts uniform quasi-isometric embedding constants for $f$ and then gives an explicit slimness constant for $X$. The extraction uses the supplied coarse inverse and both bounded composite errors; the transport uses the two Hausdorff inclusions of Morse stability.

If instead $X$ is hyperbolic, [F1] gives a controlled quasi-isometric inverse $g:Y\to X$. Applying the same transport assertion to $g$ makes $Y$ hyperbolic. In the empty-space case the quasi-isometry convention forces both spaces empty and the claim is vacuous. Thus hyperbolicity is invariant in both directions.

By [L1], the identity map on the vertex sets is bilipschitz for the two word metrics. Extend it over each edge of $\mathrm{\Gamma }(G,S)$ by a chosen shortest $T$-path for that edge label, and conversely for $T$-edges. Because the generating sets are finite, these paths can be fixed by finitely many choices. The resulting maps are quasi-isometries of the geometric Cayley graphs: every point is within $1\mathrm{/}2$ of a vertex, and the vertex metrics have the bilipschitz bounds from [L1].

By [L2], under [A1] hyperbolicity transfers from one geometric Cayley graph to the other. Since $S,T$ were arbitrary finite generating sets, the definition does not depend on the set.

If $G$ is finite, then its Cayley graph has finite diameter. Every geodesic triangle in a finite-diameter space is $\mathrm{diam}(\mathrm{\Gamma }(G,S))$-slim, so $G$ is hyperbolic.

If $F(X)$ is free, then [L2] says that its Cayley graph is a tree, and [L1] therefore makes it $0$-hyperbolic. Hence $F(X)$ is hyperbolic.

If the basis is infinite, every finite set of group elements uses only finitely many basis coordinates and cannot generate $A$. Thus [L0] excludes hyperbolicity. Otherwise identify $A$ with ${\mathbb{Z}}^n$, $n\ge 2$, and fix any finite generating set. Replace it by its nonzero symmetric closure $S$, which leaves the geometric Cayley graph unchanged by [L1]. It spans ${\mathbb{R}}^n$.

Choose $s\in S$ of maximal Euclidean norm. The linear functional $f(x)=⟨s,x⟩\mathrm{/}\mathrm{\parallel }s{\mathrm{\parallel }}^2$ satisfies $f(s)=1$ and $\mathrm{\mid }f(a)\mathrm{\mid }\le 1$ for every $a\in S$, by Cauchy–Schwarz and maximality. Since $S$ spans a space of dimension at least two, choose $v\in S$ independent of $s$ and put $w=v-⟨v,s⟩s\mathrm{/}\mathrm{\parallel }s{\mathrm{\parallel }}^2$. Then $w\mathrm{\ne }0$, $⟨w,s⟩=0$ and $⟨w,v⟩>0$. Choose $t\in S$ maximizing $⟨w,t⟩$. Symmetry gives a positive maximum and $\mathrm{\mid }⟨w,a⟩\mathrm{\mid }\le ⟨w,t⟩$ on $S$. Hence $g(x)=⟨w,x⟩\mathrm{/}⟨w,t⟩$ has $g(t)=1$, $g(s)=0$ and $\mathrm{\mid }g(a)\mathrm{\mid }\le 1$ on $S$. In particular $s,t$ are independent.

Extend these linear functions from graph vertices affinely over each edge. Their slopes have absolute value at most one, so they are 1-Lipschitz for the graph path metric. Thus a path of $m$ successive $s$-edges, or $m$ successive $t$-edges, has endpoints at distance exactly $m$: the path supplies the upper bound and $f$, respectively $g$, supplies the lower bound. Translates and reversals are likewise geodesics. Therefore the four such paths through $0,ms,ms+mt,mt$ form a geodesic quadrilateral.

Choose linear functionals $\alpha ,\beta$ on ${\mathbb{R}}^n$ with $\alpha (s)=1$, $\alpha (t)=0$, $\beta (s)=0$ and $\beta (t)=1$; solving the nonsingular two-vector Gram system constructs them. Let $C={\max }_{a\in S}\max \{\mathrm{\mid }\alpha (a)\mathrm{\mid },\mathrm{\mid }\beta (a)\mathrm{\mid }\}\ge 1$. Their affine extensions to graph edges are $C$-Lipschitz. At the midpoint of the side $0$ to $ms$, their values are $(m\mathrm{/}2,0)$. On each of the other three sides, either $\alpha =0$, $\alpha =m$, or $\beta =m$. Consequently every point on those sides is at distance at least $m\mathrm{/}(2C)$ from that midpoint. Taking arbitrarily large $m$ contradicts [L2] for every proposed hyperbolicity constant. Since the finite generating set was arbitrary, [L0] excludes hyperbolicity of $A$.

Choose a finite generating set $S$ for $G$ and a positive slimness constant $\delta$ for its Cayley graph. Choose an integer $L>\max \{6\delta ,12\delta \}$. By [F1], any $L$-local geodesic arc is a $(3,4\delta )$-quasi-geodesic. If such an arc has the same initial and terminal vertex and positive length $N$, the quasi-geodesic inequality gives $0\ge N\mathrm{/}3-4\delta$, hence $N\le 12\delta <L$; then the whole arc lies within the local-geodesic radius and would have to be geodesic, impossible between equal endpoints. Let $R$ be the finite set of nonempty freely reduced words over $S^{\pm 1}$ of length at most $2L$ that represent the identity. Finiteness follows from [A2], and $⟨S\mid R⟩$ presents $G$ once the Dehn property below is proved.

Suppose a nonempty freely reduced trivial word $w$ contains no subword longer than half of a cyclic conjugate of a member of $R^{\pm 1}$. If an ordinary subword of $w$ of length at most $L$ were nongeodesic, choose one of minimal length and call it $u$, and choose a shorter geodesic word $v$ with the same endpoints. Minimality of $u$ implies that $u$ and $v$ share neither an initial nor a terminal edge: deleting such a common edge would give a shorter nongeodesic subword. Hence the loop word $u{v}^{-1}$ is freely and cyclically reduced. It belongs to $R$, has length $\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }<2\mathrm{\mid }u\mathrm{\mid }\le 2L$, and contains $u$ as more than half of a cyclic conjugate, a contradiction. Thus every ordinary length-at-most-$L$ subword of $w$ is geodesic. Read $w$ as the parameterized open path from the identity vertex back to itself; all its short consecutive segments are geodesic, so this open path is $L$-local geodesic. No condition is imposed across a cyclic junction of $w$.

Step 1.1 forbids a nonempty $L$-local geodesic arc with equal endpoints, contradicting step 2.1. Hence every nonempty freely reduced trivial word has the required long relator subword. Replacing that subword by the shorter complementary piece of its relator, then freely reducing, strictly decreases word length while preserving its value in $G$. Finite induction reduces every trivial word to the empty word using relations from $R$, so $R$ presents $G$ and has the Dehn property.

Fix the finite alphabet $S^{\pm 1}$ and finite relator list $R$ supplied by [L1]. Freely reduce the input word. At each stage enumerate its finitely many subwords and the finitely many cyclic conjugates of relators in $R^{\pm 1}$; if a subword is longer than half of one of those relators, replace it by the inverse of the complementary piece and freely reduce again. Choose the first match in a fixed finite ordering. Each replacement preserves the group element and strictly decreases length, so this effective procedure terminates.

If the algorithm stops at the empty word, then $w=1$ in $G$. Conversely, if $w=1$ in $G$ and the current freely reduced word is nonempty, [L1] supplies another enumerated shortening move, so the procedure cannot stop there. Thus it decides whether $w$ represents the identity.

If $G$ is hyperbolic, [F1] gives a finite presentation with the Dehn reduction property. For every nonempty null word, one reduction replaces a subword longer than half a defining relator by the complementary shorter subword, using one conjugate of that relator; the resulting freely reduced word is strictly shorter. Iterate. There are at most $\mathrm{\mid }w\mathrm{\mid }$ reductions before the empty word, so reversing them gives an expression of $w$ as at most $\mathrm{\mid }w\mathrm{\mid }$ conjugated relators. By [F3] it has a van Kampen diagram with at most $\mathrm{\mid }w\mathrm{\mid }$ cells. The empty word has an empty diagram. Thus the displayed linear inequality holds, with $C=1$ (or any larger positive constant).

Conversely suppose a finite presentation has diagrams with at most $C\mathrm{\mid }w\mathrm{\mid }$ cells for every null word. The boundary word of any finite disc diagram is a product of conjugates of its face relators: choose a spanning tree of its edges, cut along that tree, and peel cells from the exterior; each peel contributes one conjugated relator and the cut-tree traversals cancel freely. Repeated vertices and edges are treated by their separate directed occurrences. Thus the algebraic relator area of $w$ is at most $C\mathrm{\mid }w\mathrm{\mid }$. If the given bound is real, take $K=\max \{0,C\}$; the area is integral, so the same inequality holds. Since the relator set is finite, its lengths have a finite bound $L$.

By [F2] and [A1], the bound in step 1.2 makes the labelled geometric Cayley graph uniformly slim. Hence $G$ is hyperbolic. Together with step 1.1 this proves the equivalence.

By [L0], the given presentation satisfies a linear isoperimetric inequality.

Therefore [L1] applies under [A1], and the presented group is hyperbolic.

By the definition of a hyperbolic group, some finite generating set $T$ has a hyperbolic geometric Cayley graph. Apply [L1] with $T$: for some $C_T>0$, $\mathrm{\mid }n\mathrm{\mid }\le C_T\mathrm{\mid }{g}^n{\mathrm{\mid }}_T$ for every integer $n$. The proof of [L1] is choice-free.

By [L2] there is a finite $K>0$ with $\mathrm{\mid }h{\mathrm{\mid }}_T\le K\mathrm{\mid }h{\mathrm{\mid }}_S$ for all $h\in G$. Thus $\mathrm{\mid }n\mathrm{\mid }\le C_{T}K\mathrm{\mid }{g}^n{\mathrm{\mid }}_S$. Take $A=C_{T}K$ and any $B>0$. This proves undistortion for the stated arbitrary finite $S$ without importing a choice-dependent hyperbolicity transfer theorem.

The standing hyperbolic-group convention supplies a finite generating set whose geodesic Cayley realization is $\delta$-slim. By [L1], the power orbit of $g$ is quasi-isometrically embedded; fix its constants $\lambda ,c$. The hypotheses of [L2] are therefore met.

Apply [L2]. It gives finitely many cosets of $⟨g⟩$ in $C_G(g)$, with a representative of each in one finite word ball. Thus $⟨g⟩$ is a cyclic subgroup of finite index in the centralizer, as claimed.

If $A$ contains an element $g$ of infinite order, then $A\subseteq C_G(g)$. By [L1], the cyclic subgroup $⟨g⟩$ has finite index in $C_G(g)$. Thus $A\cap ⟨g⟩$ has finite index in $A$ and is cyclic as a subgroup of $⟨g⟩$. Hence $A$ is virtually cyclic.

If every element of $A$ has finite order, each finitely generated subgroup of $A$ is finite: for generators of orders $n_1,\dots ,n_k$, commutativity makes it a quotient of the finite group ${\prod }_i\mathbb{Z}\mathrm{/}{n}_i\mathbb{Z}$. By [L2] it has at most $B$ elements. Were $A$ to contain $B+1$ distinct elements, their finitely generated subgroup would contradict this bound. So $A$ is finite, hence virtually cyclic. The two cases prove the claim.

Let $F\le G$ be finite and put $M=F\cdot e$, a finite set of vertices of $X$. For a vertex $x$ define $R(x)={\max }_{m\in M}d(x,m)$. The nonempty set of integer values $R(x)$ has a least value $R$, attained at some vertex $x$. Left translation by each $f\in F$ preserves $M$ and distances, so $R(fx)=R(x)=R$. Thus the center set $C=\{v\in G:R(v)=R\}$ is $F$-invariant and contains the orbit $Fx$.

Let $x,y\in C$, write $D=d(x,y)$, and let $z$ be the midpoint of a geodesic $[x,y]$. For any $m\in M$, slimness of the triangle with vertices $x,y,m$ gives a point $p$ on $[x,m]$ or $[y,m]$ with $d(z,p)\le \delta$. In the first case, $d(x,p)\ge D\mathrm{/}2-\delta$ and $d(x,m)\le R$, so $d(z,m)\le R-D\mathrm{/}2+2\delta$; the second case is symmetric. Choose a vertex $v$ of the edge containing $z$, with $d(v,z)\le 1\mathrm{/}2$. Then $R(v)\le R-D\mathrm{/}2+2\delta +1\mathrm{/}2$. Minimality of $R$ forces $D\le 4\delta +1$. Hence every two vertices of $C$, and in particular of $Fx$, are at distance at most $4\delta +1$.

The map $f\mapsto fx$ is injective by [L2]. The orbit $Fx$ lies in the vertex ball about $x$ of radius $N=\lceil 4\delta +1\rceil$, whose cardinality is the fixed finite number $B=\mathrm{\mid }{B}_X(e,N)\cap G\mathrm{\mid }$ by Cayley vertex transitivity. Thus $\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }Fx\mathrm{\mid }\le B$ for every finite $F\le G$. Taking $B_S=B$ proves the assertion for the given $S$.

By [A1], choose independent infinite-order elements $g,h\in G$ with disjoint attracting and repelling neighborhoods on the boundary.

Choose $N$ large enough for all four north--south inclusions in [A1]. Write $D_{g^N}=U_g^{+}$, $D_{g^{-N}}=U_g^{-}$, $D_{h^N}=U_h^{+}$, and $D_{h^{-N}}=U_h^{-}$. If $w=s_1\cdots s_k$ is a nonempty reduced word in $g^{\pm N},h^{\pm N}$, choose a letter $t$ distinct from both $s_1$ and $s_k^{-1}$ and a point $x\in D_t$. Acting from right to left, [A1] gives successively. [A1, step 1.1, choose] $s_j\cdots s_{k}x\in D_{s_j}(j=k,k-1,\dots ,1),$ because reducedness says $s_{j+1}\mathrm{\ne }{s}_j^{-1}$. Thus $wx\in D_{s_1}$, while $x\in D_t$, and these domains are disjoint. Hence $wx\mathrm{\ne }x$, so $w$ is not trivial. Therefore $g^N$ and $h^N$ freely generate a free subgroup of rank $2$.

By [A1], every Gromov sequence is asymptotic to itself, and if $(x_n)$ is asymptotic to $(y_n)$ then $(y_n)$ is asymptotic to $(x_n)$.

Suppose $(x_n)$ is asymptotic to $(y_n)$ and $(y_n)$ is asymptotic to $(z_n)$. By [L1], there is $\delta \ge 0$ with $(x_m,z_n{)}_o\ge \min \{(x_m,y_k{)}_o,(y_k,z_n{)}_o\}-3\delta$ for all $m,n,k$. Given $R>0$, choose $N$ so both mixed products exceed $R+3\delta$ whenever both of their indices are at least $N$. For every $m,n\ge N$, taking $k=N$ in the inequality gives $(x_m,z_n{)}_o>R$. Hence $(x_m,z_n{)}_o\to \mathrm{\infty }$, so $(x_n)$ is asymptotic to $(z_n)$.

The boundary product in The boundary topology defined by Gromov products is the supremal product of [F1]. Its comparison with every representative product proves independence of representatives; the basepoint inequality gives cofinal threshold neighbourhoods at any two basepoints.

The definition's open-set criterion is exactly the one proved in [F1], including its treatment of threshold sets as neighbourhoods that need not be open. Hence it is a topology and is Hausdorff.

By [F2] under [A1], the quasi-isometry induces a continuous map of these boundary topologies. Its controlled quasi-inverse induces a continuous inverse because the bounded-distance composites induce identity maps. This proves the claimed homeomorphism; properness is included in the statement but not required by [F1] or [F2].

By [L1], the free group $F_2$ is hyperbolic.

The group $F_2$ is countable, but [A1] says that ${\mathbb{H}}^2$ is uncountable. So $F_2$ cannot contain the hyperbolic plane as a subgroup or even as an underlying set, yet it is hyperbolic by step 1.1. Therefore the claim is false.

The generating set $\{a,b\}$ gives a tree Cayley graph, so $F_2$ is $0$-hyperbolic for that choice.

Let $\delta \ge 0$, and choose an even $n=2k$ with $k>\delta$. The paths labelled $a^n$ from $1$ to $a^n$ and $b^n$ from $a^n$ to $t$ have length $n$. They are geodesic: for every integer $r$, [A1] gives respective lower bounds $\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge n$ and $\mathrm{\mid }nr\mathrm{\mid }+\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge n$. The edge labelled $t$ joins $1$ to $t$, so these paths form a geodesic triangle.

Its midpoint vertex $a^k$ is distance $k$ from $1$. For any word from $a^k$ to $t$, [A1] gives $\mathrm{\ell }\ge \mathrm{\mid }k-nr\mathrm{\mid }+\mathrm{\mid }n-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }\ge k+1$ for every integer $r$: the first term is at least $k$, and either $r\mathrm{\ne }0$ or the second term is positive. Thus $a^k$ is at distance at least $k$ from the one-edge side $[1,t]$, including its interior. Every vertex $a^n{b}^j$ on the other long side, $0\le j\le n$, is also at distance at least $k$ from $a^k$, since the first term of the lower bound $\mathrm{\mid }k-nr\mathrm{\mid }+\mathrm{\mid }j-nr\mathrm{\mid }+\mathrm{\mid }r\mathrm{\mid }$ is at least $k$ for every integer $r$. The same bound holds for points inside its edges. Therefore this geodesic triangle is not $\delta$-slim.

Since $\delta$ was arbitrary, no single constant works for all finite generating sets of $F_2$. Therefore the claim is false.

The group ${\mathbb{Z}}^2$ is abelian.

The group ${\mathbb{Z}}^2$ has rank $2$, so [L1] shows that it is not hyperbolic. Therefore the claim is false.

Parametrize each broken path by arclength. For two points on the same side of the path, the subpath length equals their Euclidean distance. For points on adjacent sides, the subpath has two perpendicular legs, so its length is at most $\sqrt{2}$ times their distance. For points on the two vertical sides, their distance is at least $n$ while their subpath length is at most $3n$. Thus every path is a $(3,0)$-quasi-geodesic under the definition, with constants independent of $n$. The midpoint $(n\mathrm{/}2,n)$ of the top side is distance $n$ from the straight segment between the endpoints.

Their distance from the corresponding geodesic segments is unbounded as $n\to \mathrm{\infty }$, so no uniform fellow-traveling constant exists. Hence the global claim is false.

A quotient set consists of equivalence classes, so it is defined only when the proposed relation is an equivalence relation. The sequence model on this page therefore depends essentially on [L1].

Consequently the proposed quotient cannot be licensed merely by writing down the relation: reflexivity, symmetry, and transitivity must be checked. The properness hypothesis used elsewhere on this page is a scope choice for this construction, not a claim that every possible boundary model requires properness.

By the external geometric result [F1], ${\mathbb{H}}^2$ is geodesic and all its geodesic triangles are $\delta$-slim for a single $\delta \ge 0$.

Therefore [L1] shows that ${\mathbb{H}}^2$ is hyperbolic.