For fixed n≥ 3 and c∈(0,1), let GT(n,c) be the class of geodesic metric trees of valence at most n whose branch points are uniformly relatively separated with constant c. Does there exist a geodesic metric tree M∈GT(n,c) such that every T∈GT(n,c) admits a bi-Lipschitz embedding into M; that is, for every T∈GT(n,c), does there exist an injective map f:T→ M and a constant L_T≥ 1 such that L_T^-1d_T(x,y)≤ d_M(f(x),f(y))≤ L_Td_T(x,y) for all x,y∈ T?

Universal element for doubling geodesic trees

For fixed $n\ge 3$ and $c\in(0,1)$ , let $\mathcal{GT}(n,c)$ be the class of geodesic metric trees of valence at most $n$ whose branch points are uniformly relatively separated with constant $c$ . Does there exist a geodesic metric tree $M\in\mathcal{GT}(n,c)$ such that every $T\in\mathcal{GT}(n,c)$ admits a bi-Lipschitz embedding into $M$ ; that is, for every $T\in\mathcal{GT}(n,c)$ , does there exist an injective map $f:T\to M$ and a constant $L_T\ge 1$ such that $L_T^{-1}d_T(x,y)\le d_M(f(x),f(y))\le L_Td_T(x,y)$ for all $x,y\in T$ ?

References

Primary source

arXiv

Additional references

On universal elements for doubling geodesic trees — arXiv

Posted 4 weeks ago · edited 4 weeks ago

Progress summary

Refreshed 4 weeks ago

Claimed solved

An unrefereed preprint claims that no single separated geodesic tree can contain all such trees with controlled distortion, resolving the question negatively.

A November 2024 paper posed whether one tree could receive bi-Lipschitz embeddings from every member of the relevant class, and whether such a universal tree must resemble the regular tree $\mathbb{T}^{n}$ .

September 2026 negative answer

A preprint announced on September 8, 2026 claims a construction answering Question 1.11 negatively. It presents a non-dimensional obstruction to universal bi-Lipschitz embeddings, but the result is unrefereed and unverified.

Current status (as of September 2026): the universal-element question has a claimed negative answer, but that claim has not been independently verified.

Sources

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