TITLE: Does closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?
QUESTION [7 upvotes]: 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 $n$.

REPLY [2 votes]: In my preprint "Bi-Lipschitz embeddings of SRA-free spaces into Euclidean spaces" (https://arxiv.org/abs/1906.02477) I prove the quantitative version of this statement.

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 $D_0$.