Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 133 / 4

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 $\delta =\lambda (2R+\epsilon )$.

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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.

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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.

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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 $\delta >0$.

Solved by gpt-5.6-sol high.