Fix a hyperbolicity constant ${\delta }_0>0$ for ${\mathbb{H}}^2$. For a prescribed $\delta >0$, scale its distance byAll distances in every geodesic triangle, including its thinness constant, scale by $\delta /{\delta }_0$. Henceis $\delta$-hyperbolic. Multiplication of a metric by a fixed positive constant is a bilipschitz equivalence and hence a quasi-isometry. Therefore $X_{\delta }$ is quasi-isometric to ${\mathbb{H}}^2$ and cannot be a quasi-tree. This supplies an example for every $\delta >0$.

The hyperbolic plane ${\mathbb{H}}^2$ is a geodesic Gromov-hyperbolic metric space for some universal constant ${\delta }_0$. It is not a quasi-tree. Indeed, for every $C>0$, choose two points on opposite sides of a large closed metric ball centred at the midpoint of their joining geodesic. The complement of that ball in ${\mathbb{H}}^2$ is path connected, so the endpoints can be joined by a continuous path that stays more than $C$ from the midpoint. Thus ${\mathbb{H}}^2$ fails the bottleneck property, whereas part (b) shows that every quasi-tree satisfies it.

Retain the $(\lambda ,\epsilon )$-quasi-isometry $f:X\to T$ and the Morse lemma for quasi-geodesics constant $R$. Given $x,y\in X$, let $m$ be the midpoint of a geodesic $[x,y]$. There is a point $z$ on the tree geodesic $[f(x),f(y)]$ with

For any continuous path $\alpha$ from $x$ to $y$, choose a partition fine enough that consecutive $\alpha (t_i)$ are at distance at most one. Consecutive images under $f$ are then at distance at mostRemoving $z$ separates $f(x)$ from $f(y)$ in the tree along their geodesic, so this finite $D$-chain must contain an element within $D$ of $z$. For the corresponding $t_i$,The lower quasi-isometry inequality yieldsThis constant depends only on the chosen quasi-isometry, so every quasi-tree has the bottleneck property.

Let $f:X\to T$ be a $(\lambda ,\epsilon )$-quasi-isometry to a tree. The image under $f$ of every geodesic segment in $X$ is a $(\lambda ,\epsilon )$-quasigeodesic in $T$. By the Morse lemma for quasi-geodesics, it lies within a constant $R=R(\lambda ,\epsilon )$ of the tree geodesic with the same endpoints.

Consider a geodesic triangle in $X$. A point $p$ on one side maps within $R$ of the corresponding side of the comparison triangle in $T$. Every geodesic triangle in a tree is $0$-thin, so that comparison side is contained in the other two sides. Those two tree sides are in turn within $R$ of the images of the other two sides of the original triangle. Hence some point $q$ on one of those sides satisfiesThe lower quasi-isometry inequality givesThus $X$ is Gromov-hyperbolic metric space with $\delta =\lambda (2R+\epsilon )$.

Let $T^{Z(G)}$ be the common fixed subtree of the centre, which is nonempty by hypothesis and is $G$-invariant because $Z(G)$ is central. Restrict the action to this subtree. There $x$ and $y$ act pointwise trivially, so the action factors throughIn particular, the induced tree isometries satisfy

Assume, as usual for a combinatorial tree action, that edge inversions have been removed by barycentric subdivision. The finite-order elements $a,b,c$ are then elliptic. Moreoverso each pairwise product is elliptic. Serre lemma for tree actions implies that the fixed subtrees of each pair intersect. Convex subtrees of a tree have the Helly property, soSince $a,b,c$ generate $G/⟨x,y⟩$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.

A hyperbolic isometry of a tree has a unique invariant axis of a tree isometry, on which it acts by a nonzero translation. Let $L$ be the axis of $x$. By part (c), $x$ is central, so for every $g\in G$,Thus $gL$ is another axis of $x$. Uniqueness gives $gL=L$, and hence the whole group $G$ preserves the line $L$.

The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab{)}^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus

Quotienting by $⟨x,y⟩$ givesA non-elementary Fuchsian group has trivial center: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in ${\mathrm{PSL}}_2(\mathbb{R})$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that

Letbe the orientation-preserving $(2,3,7)$ hyperbolic triangle group. DefineEvery defining relator of $G$ maps to the identity, and $A,B,C$ lie in the image, so this is a surjective group homomorphism. Since$\Delta (2,3,7)$ is a non-elementary Fuchsian group.

Write the generators additively in the abelianization. The relators giveandAfter substitution, the last relation becomesThe commutator relators disappear automatically, sobecause $\gcd (5,4)=1$. Explicitly, an isomorphism to $\mathbb{Z}$ sends

Use the amalgam from part (c), with edge groupAn odd power of $a$ belongs to $A\setminus C$, because $C$ is the index-two translation subgroup of the Klein bottle group $A$. Similarly, an odd power of $c$ belongs to $B\setminus C$. Thusis a reduced alternating word whose syllables lie in $A\setminus C$ and $B\setminus C$. The normal form theorem for an amalgamated free product says that every nonempty reduced alternating word is nonidentity. The displayed element is therefore nontrivial for every $n\ge 1$.

Introduce $d=c^2$ and $e=a^2$. The two vertex groupsare Klein bottle groups. In $A$, the subgroup $⟨a^2,d⟩$ is ${\mathbb{Z}}^2$ of index two; in $B$, the subgroup $⟨e,c^2⟩$ is also ${\mathbb{Z}}^2$ of index two. IdentifyinggivesEliminating $d,e$ from this amalgamated free product recovers exactly the two given relators.

The Bass-Serre tree is bipartite with vertex sets $G/A$ and $G/B$ and edge set $G/⟨a^2,c^2⟩$. Both edge-group inclusions have index two, so every vertex has degree two. The tree is therefore a bi-infinite line, with $A$- and $B$-type vertices alternating.

The relation isso $K$ is the Klein bottle group. Let it act on ${\mathbb{R}}^2$ byThese are Euclidean isometries and satisfy $ab{a}^{-1}=b^{-1}$. Every element has a normal form $b^m{a}^n$. The orbit of $(0,0)$ is discrete, and a rectangle of finite size meets every orbit, so the action is proper and cocompact.

The squareis a vertical translation, while $b$ is a horizontal translation. They commute, andas an isometry only when $m=n=0$. Hence $⟨a^2,b⟩\cong {\mathbb{Z}}^2$. The normal form shows that every element lies in either $⟨a^2,b⟩$ or $a⟨a^2,b⟩$, so this subgroup has index two in $K$.

Choose a finite generating set $h_1,\dots ,h_r$ for $H$. Under the standard classification of connected covering spaces, each $h_i$ is represented by a based combinatorial loop ${\gamma }_i$ in $Y$. Let $Z$ be the union of the images of these finitely many finite edge paths. Then $Z$ is a finite connected subgraph containing $y_0$.

Part (b) makes the inclusion-induced mapinjective. Its image contains every $h_i$, because every ${\gamma }_i$ lies in $Z$, and therefore contains the subgroup they generate, namely all of $H$. The map is consequently an isomorphism.

Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.

The universal cover $\tilde{Y}$ of a connected graph is a tree. The lift $\tilde{\beta }$ is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\tilde{\beta }$ is injective unless $\beta$ is constant.

Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\tilde{Y}$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving thatis an injective group homomorphism.

Suppose for a contradiction that $Y\times \mathbb{R}$ carries a complete Ricci-flat metric. Since $Y$ is closed, the two subsets $Y\times (A,\infty )$ and $Y\times (-\infty ,-A)$ are different unbounded components outside the compact set $Y\times [-A,A]$. Thus $Y\times \mathbb{R}$ is disconnected at infinity and, by part (a), contains a line in a Riemannian manifold.

Its Ricci curvature is zero, so the Cheeger-Gromoll splitting theorem gives an isometryThe product Ricci tensor shows that $N$ is a complete three-dimensional Ricci-flat manifold. By the allowed fact, $N$ is flat, and hence so is $Y\times \mathbb{R}$.

The universal cover of a complete flat manifold is complete, simply connected, and has zero sectional curvature. The Hadamard-Cartan theorem therefore identifies it diffeomorphically with ${\mathbb{R}}^4$, so it is contractible. On the other hand, the universal cover of the product iswhich deformation retracts onto $\tilde{Y}$. It is contractible only if $\tilde{Y}$ is contractible, contrary to the hypothesis. Hence no such complete Ricci-flat metric exists.