Quantitative Hyperbolic Geometry Toolkit: Examples

By F1 the endpoint geodesic has the same image as the given segment. Each point in either image belongs to the other image, hence has distance zero from it. Both suprema defining their Hausdorff distance are zero by F2, including for a constant segment.

If the excursion begins at parameter $s$, its return occurs at $t=s+2a$ and $q(s)=q(t)$. The lower quasi-geodesic inequality in F2 gives $0=d(q(s),q(t))\ge 2a\mathrm{/}\lambda -\epsilon$. Multiplying by $\lambda \mathrm{/}2>0$ yields $a\le \lambda \epsilon \mathrm{/}2$. Thus for instance a $(2,3)$-quasi-geodesic cannot have such an excursion of length outwards exceeding $3$; for $\epsilon =0$ every such excursion has $a=0$. This is a necessary bound, not an assertion that every path meeting it is quasi-geodesic.

By F1, for every $s\le t$ in $I$ the path between them is the unique geodesic, of length $t-s$. Hence $d(q(s),q(t))=t-s$. The two $(1,0)$ inequalities are both this equality, and restricting to $t-s\le k$ proves locality for each $k$. In particular this supplies a concrete zero-slim instance of F2 for any positive radius, such as $k=1$.

For example, in the free group on $a,b$ the reduced path labelled $ab{a}^{-1}b$ has length $4$ and endpoint word length $4$. Its subpath between parameters $1\mathrm{/}2$ and $13\mathrm{/}4$ has distance $13\mathrm{/}4-1\mathrm{/}2=11\mathrm{/}4$, even though both endpoints lie inside edges. A one-letter path has endpoint distance $1$, and a zero-length restriction has distance $0$. These are the same equality from step 1.1, without additive error.

For vertices, $d_S(gx,gy)=\mathrm{\mid }(gx{)}^{-1}(gy){\mathrm{\mid }}_S=\mathrm{\mid }{x}^{-1}y{\mathrm{\mid }}_S=d_S(x,y)$. Left multiplication carries each right-labelled edge $x\to xs$ to $gx\to (gx)s$ and extends linearly as an isometry on it. Every finite edge route and its translated route have equal length; translating back by $g^{-1}$ shows equality of the path distances, including interior-edge points.

For points $x,y$ in the tree, their paths from $e$ have a common initial segment of length $c$ and then separate, so $d(x,y)=d(e,x)+d(e,y)-2c$ by F1. Thus $(x\mathrm{\mid }y{)}_e=c$. A Gromov sequence therefore eventually shares, for each integer $r$, a common prefix of length $r$, and its radii tend to infinity because $(x_n\mathrm{\mid }{x}_n{)}_e=d(e,x_n)$. These eventual prefixes are uniquely determined and compatible, so they define one infinite reduced word. Conversely the vertices along an infinite reduced word have products $\min \{n,m\}$ and are a Gromov sequence. Two such sequences are equivalent exactly when all their eventual prefixes agree. This identifies the entire boundary with infinite reduced words, without a representative selection theorem.

For two different infinite words whose common prefix has length $r$, any representatives of their classes eventually lie beyond that common prefix in the respective different branches. Their mixed products then equal $r$ by step 1.2. Equal words give infinite product. Hence F2's boundary topology is exactly the prefix topology: requiring common prefix length greater than $R$ gives $U_R$. For the empty alphabet there are no infinite reduced words and the boundary is empty.

In the concatenation of the reduced finite word $g$ with an infinite reduced word $w$, cancellations occur only at their join. Each cancellation removes one letter of $g$, so at most $\mathrm{\mid }g\mathrm{\mid }$ letters from the start of $w$ disappear. After that finite process the remaining infinite word is reduced. The same cancellation applies to every sufficiently long finite prefix of $w$, so the translated vertices converge to the resulting end under the identification in step 1.2. For example, $g=ab$ sends $w=b^{-1}{a}^{-1}bbb\cdots$ to $bbb\cdots$, cancelling precisely two letters at the join.

By step 1.1, $(gx\mathrm{\mid }gy{)}_e=(x\mathrm{\mid }y{)}_{g^{-1}}$. The basepoint estimate in F2 bounds its difference from $(x\mathrm{\mid }y{)}_e$ by $\mathrm{\mid }g\mathrm{\mid }$. Passing to tail products and then the boundary gives $B_e(g\xi ,g\eta )\ge B_e(\xi ,\eta )-\mathrm{\mid }g\mathrm{\mid }$. Thus sharing a prefix longer than $R+\mathrm{\mid }g\mathrm{\mid }$ forces the images to share one longer than $R$, proving continuity by step 2.1. Translation by $g^{-1}$ supplies the inverse, by its equality on every finite vertex and the identification of ends. It is continuous by the same bound. The empty word gives the identity, and no infinite sequence of choices occurs in the finite cancellation rule.

Fix integers $p,q\ge N$ and set $A=g^p$, $B=h^q$. For letters $s\in \{A,A^{-1},B,B^{-1}\}$ write $D_s$ for the corresponding one of the four domains. The assumed inclusions say precisely $s(Z\setminus D_{s^{-1}})\subseteq D_s$. Every $D_s$ is nonempty, and distinct letter domains are disjoint.

Let $w=s_1\cdots s_l$ be any nonempty reduced word in these formal letters. Choose a letter $t$ different from both $s_1$ and $s_l^{-1}$; among four letters at most two are excluded. Choose $z\in D_t$. Apply the word to $z$ from right to left. Since $t\mathrm{\ne }{s}_l^{-1}$, the first application sends $z$ into $D_{s_l}$. Inductively, a point in $D_{s_{j+1}}$ lies outside $D_{s_j^{-1}}$ because reducedness means $s_{j+1}\mathrm{\ne }{s}_j^{-1}$ and the domains are disjoint. Thus applying $s_j$ sends it into $D_{s_j}$. Finally $wz\in D_{s_1}$, disjoint from the initial domain $D_t$. Therefore $wz\mathrm{\ne }z$, so the group element represented by $w$ is not the identity. This works also for $l=1$.

By F1 there is a homomorphism from the reduced-word free group on two formal generators to the ambient group sending them to $A,B$. Its image is $⟨A,B⟩$: every product of $A,B$ and their inverses is an image, and such products form that subgroup. Step 2.1 shows its kernel has no nonempty reduced word; by F1 the empty word is the only remaining element. The map is therefore injective and identifies its image with the rank-two free group. For instance the reduced commutator word $AB{A}^{-1}{B}^{-1}$ is nonidentity by exactly the same domain test, rather than by an assumed independence theorem. Only one point from one specified nonempty domain was needed for each word, so AC is not used.

Define $d(o,(j,r))=r$ and $d((j,r),(k,s))=\mathrm{\mid }r-s\mathrm{\mid }$ if $j=k$, and $r+s$ otherwise. These are path lengths on the given branches: points on one branch are joined by its interval; points on different branches are joined through the root. Subdivide each branch at positive integers. The resulting unit-edge graph is connected, and has no cycles since removing any open edge separates its outer tail from the root. F1 therefore verifies that this path metric is geodesic with unique geodesics and $0$-slim triangles.

The closed unit ball has an open cover, in its relative topology, consisting of $B(o,3\mathrm{/}4)$ and all $B((j,1),1\mathrm{/}2)$. A point of radius below $3\mathrm{/}4$ is in the first set; a point on branch $j$ with radius at least $3\mathrm{/}4$ and at most $1$ is in the $j$th set. No finite subfamily covers the ball: choose an index $j$ absent from its finitely many outer balls. The point $(j,1)$ is outside $B(o,3\mathrm{/}4)$ and has distance $2$ from every other outer centre. Thus the closed unit ball is noncompact and the space is not proper.

Direct calculation gives $((j,r)\mathrm{\mid }(k,s){)}_o=\min \{r,s\}$ for $j=k$, and $0$ for $j\mathrm{\ne }k$; products involving $o$ are zero. If $(x_n)$ is Gromov, the threshold $0$ in F2 forces its entire sufficiently late tail onto one fixed branch, and its diagonal products force the radii to tend to infinity. Conversely any sequence eventually on one branch with radii tending to infinity has jointly diverging products. Two such sequences are equivalent exactly when their branches agree. Thus each positive integer $j$ determines exactly one boundary point, represented explicitly by $x_n=(j,n)$, and these are all the boundary points.

For two distinct branch classes, every representative pair eventually has mixed product zero, so their supremal boundary product is zero. A class has infinite product with itself. Hence $U_0(\xi )=\{\xi \}$ for every class $\xi$, and F3 gives the discrete topology. The collection of all singleton classes is an open cover with no finite subcover, since there are infinitely many positive integers. The boundary is therefore countably infinite and noncompact, despite the geodesic and zero-hyperbolicity conclusions of step 1.1. This is the required failed conclusion.