# Dose 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 the Euclidiean space $\mathbb R^N$ for sufficiently large $N$?

I know there are some spaces does not admit such 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 $n$.

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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

See Distance embedding (27.5) in our book

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Dear Anton, are chapters 13 and 15 available? – Suvrit Apr 12 '13 at 23:14
@Suvrit, everything which is nearly ready is there, check it time to time, maybe more sections will appear. If you really-really need some sections I could send it with a "read on your own risk" notice --- send me an e-mail. – Anton Petrunin Apr 13 '13 at 18:23
Thanks Anton. I'll email if you if this need arises (I'm mostly interested in the theory of first-order approximation in metric spaces, and in particular in CAT(0) spaces and slight perturbations of these spaces. – Suvrit Apr 13 '13 at 22:00