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].

The implication $(1)\Rightarrow (2)$ is exactly [A1]: slim triangles give a lower bound for the branch length measured by the Gromov product.

The implication $(2)\Rightarrow (3)$ is exactly [A2]: rewriting the Gromov-product inequality in terms of distances yields the four-point form.

The implication $(3)\Rightarrow (1)$ is exactly [A3]: in a geodesic space the four-point inequality forces each side of a geodesic triangle to stay within a uniform neighborhood of the other two. Hence the three formulations are equivalent up to changed constants.

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.

Let $\gamma$ be a geodesic segment joining the common endpoints of $q_1$ and $q_2$. By [A1], there is a constant $R_1=R_1(\delta ,\lambda ,\epsilon )$ such that each of $\mathrm{im}(q_1)$ and $\mathrm{im}(q_2)$ has Hausdorff distance at most $R_1$ from $\gamma$.

The comparison fact [A2] then gives Hausdorff distance at most $2R_1$ between $\mathrm{im}(q_1)$ and $\mathrm{im}(q_2)$. The lemma [L1] is the geometric input used in the standard proof of [A1]. Therefore the stated Morse-stability bound holds.

Assume $X$ is hyperbolic and let $f\text{\hspace{0.17em}⁣}:X\to Y$ be the given quasi-isometry. By [A1], choose a quasi-inverse $g\text{\hspace{0.17em}⁣}:Y\to X$. For any geodesic triangle in $Y$, the images of its sides under $g$ are uniform quasi-geodesics in $X$.

Since $X$ is hyperbolic, [L1] shows that each of those quasi-geodesic sides stays within a bounded distance of a genuine geodesic triangle in $X$. Applying $f$ back to that comparison triangle produces a bounded-neighborhood comparison in $Y$, because $f$ is coarsely Lipschitz and $f\circ g$ stays uniformly close to the identity on $Y$.

Therefore geodesic triangles in $Y$ are uniformly slim, so $Y$ is hyperbolic. Reversing the roles of $X$ and $Y$ gives the converse. Hence hyperbolicity is a quasi-isometry invariant of geodesic spaces.

By [L1], the identity map on $G$ is a quasi-isometry between the two Cayley graphs $\mathrm{\Gamma }(G,S)$ and $\mathrm{\Gamma }(G,T)$.

Therefore [L2] transfers hyperbolicity from one Cayley graph to the other. So the definition of a hyperbolic group does not depend on the chosen finite generating 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.

By the rank hypothesis, $A\cong {\mathbb{Z}}^n$ with $n\ge 2$. In the standard Cayley graph, the points $(0,0)$, $(m,0)$, $(m,m)$, and $(0,m)$ in the first two coordinates form a geodesic square of side length $m$ for every $m\ge 1$.

If the standard Cayley graph were hyperbolic, [L2] would give a uniform thinness constant for all geodesic quadrilaterals. But the midpoint of one side of the square from step 1.1 has distance $m\mathrm{/}2$ from the union of the opposite sides, and $m$ is arbitrary. So the standard Cayley graph is not hyperbolic, and [L1] shows that $A$ itself is not hyperbolic.

Choose a finite generating set $S$ for $G$ and an integer $L$ as in [A1]. Let $R$ be the finite set of 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 a 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 length-at-most-$L$ subword of $w$ is geodesic, so the closed path labelled by $w$ is an $L$-local geodesic.

Fact [A1] forbids a nonempty closed $L$-local geodesic, contradicting step 2.1. Hence every nonempty freely reduced trivial word has the required long relator subword. This is the Dehn property, so the finite set $R$ presents $G$ and gives a finite Dehn presentation.

Starting from any input word $w$, repeatedly apply the replacement from [A1] whenever [L1] finds a long relator half. Because length strictly decreases, the process terminates after finitely many steps.

If the algorithm stops at the empty word, then $w=1$ in $G$. Conversely, if $w=1$ in $G$ and the current reduced word is nonempty, [L1] says that another shortening move exists, so the procedure cannot terminate early. Thus the algorithm decides whether $w$ represents the identity.

If $G$ is hyperbolic, then [L1] supplies a finite Dehn presentation, and [A1] turns that Dehn property into a linear isoperimetric inequality.

Conversely, if a finite presentation satisfies a linear isoperimetric inequality, then [A2] gives thin geodesic bigons and hence a hyperbolic Cayley graph. Therefore the two conditions are equivalent.

By [A1], the given presentation satisfies a linear isoperimetric inequality.

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

The source fact [A1] says that the orbit map $n\mapsto g^n$ is a quasi-isometric embedding into the Cayley graph of $G$.

A quasi-isometric embedding gives the displayed linear lower bound on $\mathrm{\mid }{g}^n{\mathrm{\mid }}_S$ in terms of $\mathrm{\mid }n\mathrm{\mid }$, while [L1] explains geometrically that the powers of $g$ stay near a quasi-axis. Therefore $⟨g⟩$ is undistorted.

By [L1] and [A1], the powers of $g$ form a quasi-axis: geodesics joining distant powers stay uniformly close to the power orbit.

Let $s\in C_G(g)$ and choose $m$ so large that the two long sides of the quadrilateral with vertices $1,g^m,s{g}^m,s$ have points outside the $2\delta$-neighborhoods of its short sides. By [A2], some point on $[1,g^m]$ is within $2\delta$ of $[s,s{g}^m]$. Using [A1] on both long sides gives integers $i,j$ with $d(g^i,s{g}^j)\le 2L+2\delta$. Since $s$ commutes with $g$, this says that the coset $⟨g⟩s$ has a representative of word length at most $2L+2\delta$.

The ball of radius $2L+2\delta$ is finite by [A2], so step 2.1 leaves only finitely many cosets of $⟨g⟩$ in $C_G(g)$. Hence $⟨g⟩$ has finite index in $C_G(g)$, and the centralizer is virtually cyclic.

If $A$ contains an element of infinite order, then $A\subseteq C_G(g)$ for that element $g$, so [L1] shows that $A$ is virtually cyclic.

If every element of $A$ has finite order, then [A1] says that $A$ is finite, hence virtually cyclic. Therefore every abelian subgroup of a hyperbolic group is virtually cyclic.

Let $F\le G$ be finite. By [A1], some $F$-orbit in the Cayley graph of $(G,S)$ has diameter bounded by a constant depending only on $S$.

That orbit lies in a finite ball, and the action of $F$ on its orbit is faithful. Therefore [A2] gives a uniform bound $\mathrm{\mid }F\mathrm{\mid }\le B_S$. The role of [L1] in the standard proof is to supply the geometric control behind [A1].

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 $(x_m,z_n{)}_o\ge \min \{(x_m,y_k{)}_o,(y_k,z_n{)}_o\}-\delta$ for all $m,n,k$. Letting the indices go to infinity shows $(x_m,z_n{)}_o\to \mathrm{\infty }$, so $(x_n)$ is asymptotic to $(z_n)$.

By [L1], the boundary is already a quotient by a genuine equivalence relation. The comparison result [A1] then shows that changing representatives or the basepoint only changes the neighborhoods $U_o(\xi ,R)$ by bounded shifts of the parameter $R$, so the topology is well defined.

By [A2], a quasi-isometry induces a map on asymptoticity classes, and the two-sided product estimate makes that map continuous for the neighborhood systems from step 1.1. Applying the same argument to a quasi-inverse gives a continuous inverse. Thus the induced boundary map is a homeomorphism, and the boundary topology is quasi-isometry invariant.

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 $n>2\delta +2$. By [A1], the side from $1$ to $a^n$ is geodesic in the Cayley graph of $(F_2,S_n)$. Write $k=\lfloor n\mathrm{/}2\rfloor$. Any word from $a^k$ to $a^n{b}^n$ has abelianization $(n-k,n)$; using the generator $a^n{b}^n$ once leaves $(-k,0)$ and therefore still needs at least $k$ letters, while using it zero times needs at least $n$ letters. Hence $d(a^k,a^n{b}^n)\ge k+1$, so the distance from $a^k$ to the one-edge side $[1,a^n{b}^n]$ is at least $k>\delta$. Its distance to the side $[a^n,a^n{b}^n]$ is $n-k>\delta$. So this geodesic triangle is not $\delta$-slim.

Since $\delta$ was arbitrary, no single constant works for all finite generating sets of $F_2$. The theorem [L1] preserves only existence of some constant for each generating set separately. 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.

The witness family from [A1] consists of quasi-geodesics in the Euclidean plane with fixed 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, and [L1] is genuinely a hyperbolic-space theorem.

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.