Are hyperbolic spaces actually better for embedding trees than Euclidean spaces?

Question

There is a folklore in the empirical computer-science literature that, given a tree $(X,d)$, one can find a bi-Lipschitz embedding into a hyperbolic space $\mathbb{H}^n$ and that $n$ is "much smaller" than the smallest dimension of a Euclidean space in which $(X,d)$ can be bi-Lipschitz embedded with similar distortion.

Question A: Is there any theoretical grounding to this claim? Namely, can one prove that $(X,d)$ (where $\# X = n\in\mathbb{N}_+$) admits a bi-Lipschitz embedding into some $\mathbb{H}^n$ with:

• distortion strictly less that $O(\log(n))$
• $n<O(\log^2n)$?

! Edit - (Following Discussion of YCor, WillSawin, and TomTheQuant): What can be said if $s=1$ in Equation (1)?

Question B (Converse): For every $n\in \mathbb{N}_+$ and every $D>0$ does there exist a finite metric space $(X,d)$, which don't admit a bi-Lipschitz (resp. possibly uniform embedding) into $\mathbb{H}^n$ with distortion at-most $D$?

Relevant Definition (For completeness)

A bi-Lipschitz embedding $f:X\rightarrow \mathbb{H}^n$ of a metric space $(X,d)$ into $\mathbb{H}^n$ with distortion $D>0$ is a Lipschitz homeomorphism $f:X\rightarrow \mathbb{H}^n$ Lipschitz inverse $f^{-1}$ such that there is some $s>0$ satisfying $$ sd(x_1,x_2) \leq d_{\mathbb{H}^n}(f(x_1),f(x_2)) \leq sDd(x_1,x_2) \qquad (1) $$ for every $x_1,x_2\in X$. Here, $d_{\mathbb{H}^n}$ is the usual geodesic distance on the $n$-dimensional hyperbolic space.

Some Relevant posts:

Hyperbolic embeddings

• References to embedding into hyperbolic spaces
• Representability of finite metric spaces

Flat Embeddings

• Problem with embedding expanders into "flat" spaces
• Characterizing finite metric spaces which embed into Euclidean space

Uniform Embeddings

• Notes on coarse and uniform embeddings

Answer 1

I am not sure if the following paper answers your question. The abstract suggests so, but it is written in a computer science style that is less transparent to me in terms of stating a precise theorem. Also: (a) I am not an expert, (b) I am confused by the way that $n$ seems to play two different roles in your question, and (c) I doubt if it is the earliest answer to your question, if indeed it does answer it.

Low Distortion Delaunay Embedding of Trees in Hyperbolic Plane by Rik Sarkar

Abstract. This paper considers the problem of embedding trees into the hyperbolic plane. We show that any tree can be realized as the Delaunay graph of its embedded vertices. Particularly, a weighted tree can be embedded such that the weight on each edge is realized as the hyperbolic distance between its embedded vertices. Thus the embedding preserves the metric information of the tree along with its topology. Further, the distance distortion between non adjacent vertices can be made arbitrarily small – less than a $(1+\epsilon)$ factor for any given $\epsilon$. Existing results on low distortion of embedding discrete metrics into trees carry over to hyperbolic metric through this result. The Delaunay character implies useful properties such as guaranteed greedy routing and realization as minimum spanning trees.

(bolding is mine.)

Answer 2

Here is a trivial example for question B (in the $s=1$ case):

The discrete metric space on $N$ points with distance 1 between every two distinct points has the minimal distortion of an embedding into $\mathbb H^n$ going to $\infty$ as $N$ goes to $\infty$ with respect to $n$.

Indeed, for an embedding of distortion $D$, any two points in the embedding must have distance at least $1$, so the balls of radius $1/2$ around these points must be disjoint. But every two points have distance at most $D$, so the balls of radius $1/2$ around these points must be contained in the ball of radius $D+1/2$ around one point.

Thus, for an embedding to exist, the volume of the ball of radius $D+1/2$ must be at least $N$ times the volume of the ball of radius $1/2$.

Thus $D$ must go to $\infty$ if $N$ goes to $\infty$ with fixed $n$.