# Does closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?

As the title says. Let $$A^n$$ be an $$n$$-dimensional closed Alexandrov space. Does it admit a bi-Lipschitz embedding into Euclidean space $$\mathbb R^N$$ for sufficiently large $$N$$?

I know there are some spaces that do not admit such an embedding; for example, a theorem by Pansu says that:

The Heisenberg group equipped with the Carnot-Caratheodory distance does not biLipschitz embed into $$\mathbb R^n$$, for any $$n$$.

• The Heisenberg group doesn't embed isometrically with respect to Euclidean distance, but Enrico Le Donne proved that there are Nash-type embeddings that preserve the length of every curve! See arxiv.org/abs/1005.1623 . – Robert Young Apr 12 '13 at 16:04

Theorem 2. For $$n \in \mathbb{N}$$, $$k < 0$$ and $$R > 0$$ there exist $$D > 0$$ and $$N \in \mathbb{N}$$ satisfying the following. For every $$n$$-dimensional Alexandrov space of curvature $$\ge k$$ and every $$x \in X$$ there exists an embedding $$\phi:B_R(x) \rightarrow \mathbb{N}^N$$ which bi-Lipschitz distortion does not exceed $$D$$.
For $$n \in N$$ there exist $$D_0 > 0$$ and $$N_0 \in \mathbb{N}$$ such that every $$n$$-dimensional Alexandrov space of non-negative curvature allows an embedding into $$\mathbb{N}^{N_0}$$ which bi-Lipschitz distortion does not exceed $$D_0$$.