Gromov hyperbolicity constant vs. Gromov-Hausdorff distance to a tree

Question

Let $X$ be a compact, geodesic metric space which is Gromov hyperbolic with a constant $\delta>0$. To fix scaling, let us also assume that $X$ has diameter $1$.

To fix a definition of Gromov hyperbolicity, I will use the first definition here under ``Definitions using triangles''.

I understand that mostly one does not care about the precise value of $\delta$ because we think of this as an asymptotic notion, and that all compact metric spaces are Gromov hyperbolic for $\delta$ larger than their diameters. Nonetheless, in this question I would like to care about the size of $\delta$.

Gromov hyperbolicity is often motivated as saying that a metric space looks ``tree-like'' (at least at large scales), since trees are Gromov hyperbolic with $\delta=0$. So:

Question 1: Given $\epsilon>0$, is there $\delta>0$ such that if $X$ is $\delta$-hyperbolic, then $X$ is within Gromov-Hausdorff distance $\epsilon$ from a compact metric tree?

Here a metric tree means a geodesic metric space in which each pair of points is joined by a unique arc.

I am nervous that someone might answer the above question, if the answer is indeed yes, by an appeal to ultralimits. Although this would be fine for question 1, I would be happier with a more constructive understanding. So I also ask:

Question 2: If the answer to question 1 is yes, is there an explicit estimate for the rate of decay for $\delta$ given $\epsilon$? Equivalently, is there an upper bound, tending to zero with $\delta$, for the maximal Gromov-Hausdorff distance to a tree among all metric spaces $X$ with the above properties?

I apologize if these questions are simple or well-known. If the latter, a reference would be great.

Answer 1

Just to remove this question from the un-answered list. First of all, if $d_{GH}(X,Y)\le \epsilon$ then there is a $(1,2\epsilon)$-quasi-isometry $X\to Y$. See for instance

Burago, D.; Burago, Yu.; Ivanov, S., A course in metric geometry, Graduate Studies in Mathematics. 33. Providence, RI: American Mathematical Society. 2001.

If a geodesic metric space $X$ is $(L,A)$-quasi-isometric to a tree then it satisfies the following bottleneck property:

For every pair of points $x, y\in X$ and every mid-point $m$ of this pair, every path in $X$ connecting $x$ and $y$ passes through the open ball $B(m, L(A/2 +D) +A)$ centered at $m$.

Here $D=D(L,A)$ is some function which I can write for you if you really want. Incidentally, the bottleneck property also characterizes geodesic metric spaces quasi-isometric to trees. For a proof, see Theorem 4.6 in

J. Manning, Geometry of pseudocharacters. Geom. Topol. 9 (2005), 1147–1185.

On the other hand, if $X_R$ is the closed ball $\bar{B}(m, R)$ of radius $R$ in the hyperbolic plane and $x, y$ are boundary points of $\bar{B}(m,R)$ at the distance $2R$ from each other (with the mid-point $m$), then either arc on the boundary circle of $\bar{B}(m,R)$ connecting the points $x, y$ avoids $B(m,R)$. Hence, assuming that $R>(A +D) +2A$, the GH-distance from $X_R$ to any tree is $>A$. Here $D=D(1,2A)$. Of course, all the spaces $X_R$ are $\delta$-hyperbolic, where $\delta$ is the hyperbolicity constant of the hyperbolic plane.

Thus, your question 1 has negative answer. However, the following might be true, I am not sure:

Conjecture. There exists a separable geodesic Gromov-hyperbolic spaces $U$ (the "universal hyperbolic space") and a function $\epsilon(\delta)$ such that every compact $\delta$-hyperbolic hyperbolic space $X$ is within Gromov-Hausdorff distance $\epsilon(\delta)$ from a subset of $U$.

See also

M. Bonk, O. Schramm, Embeddings of Gromov hyperbolic spaces. Geom. Funct. Anal. 10 (2000), no. 2, 266–306.

Answer 2

A straightforward estimates show that any $n$-point $\delta$-hyperbolic space lies on distance at most $n{\cdot}\delta$ from a $0$-hyperbolic space. The following statement improves this bound to $\log n\cdot \delta$. I learned it from Rostislav Matveyev, who attributed it to Étienne Ghys.

Let $d$ be a $\delta$-hyperbolic metric on an $n$-points set $F$. Then there is a $0$-hyperbolic metric $d'$ on $F$ such that $$d \leqslant d' \leqslant d+\mathrm{const}\cdot\log n\cdot \delta.$$

To prove it, you choose a base point and define new Gromov's product $(\ |\ )'$ using the old one $(\ |\ )$ $$(x|y)'\mathrel{:=}\max \{\,\min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}\,\},$$ where the maximum is taken for all chains of points $x=x_0,\dots, x_m=y$. Plus you need to apply the following lemma:

For any sequence of points $x_0,\dots,x_m$ in a $\delta$-hyperbolic space, we have $$(x_0|x_m)\geqslant \min \{\,(x_0|x_1),\dots,(x_{m-1}|x_m)\,\}-\mathrm{const}\cdot\log m\cdot \delta.$$

The proof is recursive application of the inequality for Gromov's product for triples $$(x_i|x_k)\geqslant \min\{\,(x_{i}|x_j),(x_{j}|x_k)\,\}- \delta.$$ You start with $(x_0|x_m)$ and choose roughly the middle index each time. This way you get $\lceil\log_2 n\rceil$ iterations; each iteration gives you extra $\delta$.

Postscript. Actually, it is 6.1.B in Gromov's "Hyperbolic groups".