Bi-Lipschitz embeddings of $SRA$-free spaces into Euclidean spaces

TL;DR

This paper proves that a broad class of metric spaces, including many geometric and algebraic structures, can be embedded into Euclidean spaces using bi-Lipschitz maps, with applications to Alexandrov spaces.

Contribution

It establishes bi-Lipschitz embedding results for $SRA$-free spaces, including a conjectured embedding for Alexandrov space balls, and introduces an extension theorem for bi-Lipschitz maps.

Findings

$SRA$-free spaces admit bi-Lipschitz embeddings into Euclidean spaces.

A quantitative embedding theorem for balls in finite-dimensional Alexandrov spaces.

An extension theorem for bi-Lipschitz maps that may be of independent interest.

Abstract

$SRA$-free spaces is a wide class of metric spaces including finite dimensional Alexandrov spaces of non-negative curvature, complete Berwald spaces of nonnegative flag curvature, Cayley Graphs of virtually abelian groups and doubling metric spaces of non-positive Busemann curvature with extendable geodesics. This class also includes arbitrary big balls in complete, locally compact $CAT(k)$-spaces $(k \in \mathbb R)$ with locally extendable geodesics, finite-dimensional Alexandrov spaces of curvature $\ge k$ with $k \in R$ and complete Finsler manifolds satisfying the doubling condition. We show that $SRA$-free spaces allow bi-Lipschitz embeddings in Euclidean spaces. As a corollary we obtain a quantitative bi-Lipschitz embedding theorem for balls in finite dimensional Alexandrov spaces of curvature bounded from below conjectured by S. Eriksson-Bique. The main tool of the proof is an extension theorem for bi-Lipschitz maps into Euclidean spaces. This extension theorem is close in nature with the embedding theorem of J. Seo and may be of independent interest.

Equations

$$d(x,y)>\max \{d(x,z)+\alpha d(z,y),\alpha d(x,z)+d(z,y)\}.$$

$$f^{\prime }(x)=(f(x),0),$$

$$\mathrm{Lip}(\varphi )=\underset{x_1\ne x_2\in X}{\sup }\frac{d_Y(\varphi (x_1),\varphi (x_2))}{d_X(x_1,x_2)}\in [0,\infty ],$$

$$s{d}_X(x_1,x_2)\le d_Y(\varphi (x_1),\varphi (x_2))\le Ds{d}_X(x_1,x_2)$$

$$sd(y_1,y_2)\le ||\varphi (y_1)-\varphi (y_2)||\le Dsd(y_1,y_2),$$

$$d(x_1,x_2)>K\nu \max \{d(x_1,Y),d(x_2,Y)\}$$

$$d(x_1,x_2)>5D\sqrt{n}\max \{d(x_1,Y),d(x_2,Y)\}$$

$$||{\varphi }_1(x_1)-{\varphi }_1(x_2)||\ge ||{\varphi }_1(y_1)-{\varphi }_1(y_2)||-||{\varphi }_1(y_1)-{\varphi }_1(x_1)||-||{\varphi }_1(y_2)-{\varphi }_1(x_2)||.$$

$$||{\varphi }_1(y_1)-{\varphi }_1(y_2)||=||\varphi (y_1)-\varphi (y_2)||\ge sd(y_1,y_2).$$

$$||{\varphi }_1(y_1)-{\varphi }_1(x_1)||\le Ds\nu d(y_1,x_1)\le Ds\sqrt{n}\delta ,$$

$$||{\varphi }_1(y_2)-{\varphi }_1(x_2)||\le Ds\nu d(y_2,x_2)\le Ds\sqrt{n}\delta .$$

$$||{\varphi }_1(x_1)-{\varphi }_1(x_2)||\ge sd(y_1,y_2)-2Ds\nu \delta .\ge$$

$$\geq s\big{(}d(x_{1},x_{2})-d(x_{1},y_{1})-d(x_{2},y_{2})\big{)}-2Ds\nu{\delta}\geq$$

$$\geq s\big{(}d(x_{1},x_{2})-2{\delta}-2D\nu{\delta}\big{)}.$$

$$||{\varphi }_1(x_1)-{\varphi }_1(x_2)||\ge sd(x_1,x_2)(1-\frac{2}{K\nu }-\frac{2D\nu }{K\nu }).$$

$$||{\varphi }_1(x_1)-{\varphi }_1(x_2)||\ge sd(x_1,x_2)(1-\frac{2}{K\nu }-\frac{2D}{K})\ge \frac{s}{5}d(x_1,x_2).$$

$$||P_x(z)||\le ||P_x(x)||+||P_x(z)-P_x(x)||\le 2^k+(D{2}^{k-1})\le 2^{k+1}D.$$

$$d(x,y)\ge 10\cdot 2^k\zeta .$$

$$\overline{a}=\{b\in \mathbb{Z}||b=a\text{ mod }M\}.$$

$$Q_{\tilde{k},\tilde{j}}(z)=\sum _{k\in \tilde{k}}\sum _{x\in N_k,\chi (x)=\tilde{j}}{P}_x(z).$$

$$\Phi(z)=\Big{(}Q_{\widetilde{k},\widetilde{j}}(z)\Big{)}_{\widetilde{k}=\overline{1},\dots,\overline{M},1\leq\widetilde{j}\leq j}.$$

$$d(z_1,z_2)\ge \frac{2}{5}{2}^{\max \{k_1,k_2\}}.$$

$$Q_{\tilde{k},\tilde{j}}=\sum _{k\in \tilde{k}}\sum _{x\in N_k,\chi (x)=\tilde{j}}{P}_x$$

$$||Q_{\tilde{k},\tilde{j}}(z_1)-Q_{\tilde{k},\tilde{j}}(z_2)|||=||P_{x_1}(z_1)-P_{x_1}(z_2)||\le 40D\sqrt{n+1}d(z_1,z_2)$$

$$d(z_1,z_2)\ge \frac{2}{5}{2}^{\max \{k_1,k_2\}}.$$

$$||P_{x_i}(z_i)||\le 22D\sqrt{n+1}{2}^{k_i},$$

$$||Q_{\tilde{k},\tilde{j}}(z_1)-Q_{\tilde{k},\tilde{j}}(z_2)||=||P_{x_1}(z_1)-P_{x_2}(z_2)||\le 44D\sqrt{n+1}{2}^{\max \{k_1,k_2\}}.$$

$$||Q_{\tilde{k},\tilde{j}}(z_1)-Q_{\tilde{k},\tilde{j}}(z_2)||\le 110D\sqrt{n+1}d(z_1,z_2).$$

$$||\Phi (z_1)-\Phi (z_2)||\ge Kd(z_1,z_2).$$

$$||P_x(z_1)-P_x(z_2)||\ge d(z_1,z_2).$$

Full text

We show that $\operatorname{SRA}$-free spaces allow bi-Lipschitz embeddings in Euclidean spaces. As a corollary we obtain a quantitative bi-Lipschitz embedding theorem for balls in finite dimensional Alexandrov spaces of curvature bounded from below conjectured by S. Eriksson-Bique.

The main tool of the proof is an extension theorem for bi-Lipschitz maps into Euclidean spaces. This extension theorem is close in nature with the embedding theorem of J. Seo and may be of independent interest.

1. Introduction

111Acknowledgements: I thank my advisor Sergey V. Ivanov for all his ideas, advice and continuous support. The mission of investigating connections between $\operatorname{SRA}({\alpha})$-condition and bi-Lipschitz embeddability into Euclidean spaces was given to me by Alexander Lytchak. I’m very grateful for that. I express my gratitude to Nina Lebedeva for multiple useful discussions. Research is supported by ”Native towns”, a social investment program of PJSC ”Gazprom Neft”.

Definition 1**.**

Let $k\in{\mathbb{N}}$ and $0<{\alpha}<1$. We say that a metric space $X$ is free of $k$-point $\operatorname{SRA}({\alpha})$-subspaces if for every $k$-point subset $Y\subset X$ there exist $x,z,y\in Y$ such that

[TABLE]

We say that a metric space is an $\operatorname{SRA}({\alpha})$-free space if it is free of $k$-point $\operatorname{SRA}({\alpha})$-subspaces for some $k\in{\mathbb{N}}$.

The condition of being $\operatorname{SRA}({\alpha})$-free is a strengthening of the doubling condition (see [8, Theorem 6]). The reason why $\operatorname{SRA}({\alpha})$-free spaces were studied in previous works [8, 13] is the following. By showing that a space is an $\operatorname{SRA}({\alpha})$-free space we are showing that:

• •

this space does not contain large ${\alpha}$-snowflakes as isometric subspaces (see [7, Proof of Theorem 1.1]).

• •

bounded self-contracted curves are rectifiable in this space (see [13, Theorem 1]).

The aim of this paper is to give a new application for the concept by proving that every $\operatorname{SRA}({\alpha})$-free space allows a bi-Lipschitz embedding into some Euclidean space.

In the present paper ${\mathbb{R}}^{n}$ is always considered as metric spaces with Euclidean metric.

Theorem 1**.**

For $k\in{\mathbb{N}}$ and $0<{\alpha}<1$ there exist $N\in{\mathbb{N}}$ and $D\geq 1$ satisfying the following. For every metric space $X$ which is free of $k$-point $\operatorname{SRA}({\alpha})$-subspaces there exists a bi-Lipschitz embedding $\Phi:X\rightarrow{\mathbb{R}}^{N}$ which bi-Lipschitz distortion does not exceed $D$.

The class of $\operatorname{SRA}({\alpha})$-free spaces is huge, it includes

• •

finite dimensional normed spaces (see [8, Proposition 15] and [8, Theorem 2]),

• •

finite dimensional Alexandrov spaces of non-negative curvature (see [8, Proposition 14] and [8, Theorem 2]),

• •

complete Berwald spaces of nonnegative flag curvature (see [8, Proposition 17]) and [8, Theorem 2]),

• •

Cayley Graphs of virtually abelian groups (see [8, Proposition 18] and [8, Theorem 2]),

• •

doubling metric spaces of non-positive Busemann curvature with extendable geodesics (see [8, Proposition 12] and [8, Theorem 2]),

• •

globally $\operatorname{ATB}({\epsilon})$-spaces (see Definition 10 and [8, Theorem 2])).

It also includes arbitrary big balls in following classes of spaces:

• •

complete, locally compact $\operatorname{CAT}(k)$-spaces $(k\in{\mathbb{R}})$ with locally extendable geodesics (see [8, Proposition 12], [8, Theorem 2]) and [13, Theorem 5],

• •

finite-dimensional Alexandrov spaces of curvature $\geq k$ with $k\in R$ (see [8, Proposition 14], [8, Theorem 2]) and [13, Theorem 6],

• •

complete locally compact Finsler manifolds (see [8, Proposition 16], [8, Theorem 2]) and [13, Theorem 5],

• •

proper locally $\operatorname{ATB}({\epsilon})$-spaces (see Definition 10, [8, Theorem 2]) and [13, Theorem 5]).

Thus, Theorem 1 generalizes several previously known results on bi-Lipschitz embeddings into Euclidean spaces, see [3, Theorem 1–3, Theorem 1–4], [6, Therorem 4.5], and its generalization [13, Theorem 4]. On the other hand Theorem 1 does not provide a criterion for embeddability into Euclidean spaces i.e., there are metric spaces which allow bi-Lipschitz embeddings into Euclidean spaces which are not $\operatorname{SRA}({\alpha})$-free spaces for any $0<{\alpha}<1$.

As an application of Theorem 1 we prove the following embedding result for Alexandrov spaces of curvature bounded from below, conjectured by S. Eriksson-Bique [3, Conjecture 1–9].

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 $\geq k$ and every $x\in X$ there exists an embedding $\phi:B_{R}(x)\rightarrow{\mathbb{R}}^{N}$ 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 ${\mathbb{R}}^{N_{0}}$ which bi-Lipschitz distortion does not exceed $D_{0}$.

The main tool for the proof of Theorem 1 is the following extension theorem for bi-Lipschitz maps into Euclidean spaces.

Theorem 3**.**

Let ${\lambda}\in{\mathbb{N}}$ and $X$ be a metric space with the doubling constant ${\lambda}$. Let $Y\subset X$ and $\phi:Y\rightarrow{\mathbb{R}}^{n}$ be a bi-Lipschitz embedding into some Euclidean space. Suppose that there exist $D,\overline{n}>0$ and $0<{\theta}\leq 1$ such that for every $x\in X$ there exists an embedding of $B_{{\theta}d(x,Y)}(x)$ into ${\mathbb{R}}^{\overline{n}}$ which bi-Lipschitz distortion does not exceed $D$. Then there exist $\widetilde{\phi}:X\rightarrow{\mathbb{R}}^{n}\times{\mathbb{R}}^{N}={\mathbb{R}}^{n+N}$ such that

1. (1)

$N$* and $\operatorname{dist}(\overline{\phi})$ are bounded by some functions of $n$, $\overline{n}$, $\operatorname{dist}(\phi)$, $D$, ${\theta}$ and ${\lambda}$.* 2. (2)

for every $y\in Y$ we have $\overline{\phi}(y)=(\phi(y),0).$

The first remark about Theorem 3 we should do is that the second part of the conclusion (2) is not that important. Let us denote by Theorem 3-emb a version of Theorem 3 where (2) is skipped. Theorem 3-emb implies Theorem 3 if combined with the following proposition.

Proposition 2**.**

([9, Theorem 1.2], see also [1, Theorem 5.5]) Let $X\subset{\mathbb{R}}^{n}$ and $f:X\rightarrow{\mathbb{R}}^{m}$ be a bi-Lipschitz map with distortion at most $D$. There exists $f^{\prime}:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}^{m+n}$ with the distortion at most $3D$ and such that

[TABLE]

for every $x\in X$.

Our second remark is that Theorem 3 seems to address the same phenomena as the embedding result by J. Seo [12, Theorem 1.1]. It also seems very plausible that by applying [11, Lemma 2.5] and some standard arguments one can show that Theorem 3-emb is equivalent to Romney’s version of Seo’s result [11, Theorem 2.3].

We have not verify carefully the validity of this approach and give a direct proof of Theorem 3 instead.

2. Proof of Theorem 3

For a map between metric spaces $\phi:X\rightarrow Y$ we denote by $\operatorname{Lip}(\phi)$ the Lipschitz constant of $\phi$ i.e.,

[TABLE]

and by $\operatorname{dist}(\phi)$ its (bi-Lipschitz) distortion, which the infimum over those $D\in[1,\infty]$ such that there exists $s>0$ satisfying

[TABLE]

for all $x_{1},x_{2}\in X$.

A metric space is said to have the doubling constant ${\lambda}\in{\mathbb{N}}$ if for every $x\in X$ and $R>0$ there exists $x_{1},\dots,x_{\lambda}\in X$ such that $B_{R}(x)\subset\cup_{i=1}^{\lambda}B_{R/2}(x_{i})$.

We denote the standard Euclidean norm by $||\cdot||$.

It is well known that partial Lipschitz maps to Euclidean spaces allow extensions to whole spaces.

Proposition 3** (McShane).**

Let $X$ be a metric space, $Y\subset X$, $L>0$, $f:Y\rightarrow{\mathbb{R}}^{n}$ be a $L$-Lipschitz map then there exists $F:X\rightarrow{\mathbb{R}}^{n}$ such that $F$ is $\sqrt{n}L$-Lipschitz and $F|_{Y}=f$.

Suppose that we start from a bi-Lipschitz function $f$ and apply the previous proposition. The resulting map is not necessarily bi-Lipschitz. But if we have two points $x_{1},x_{2}\in X$ such that both of them are close to $Y$ and they are far from each other then $F$ will not distort distance between them too much. The following lemma formalizes this statement.

Lemma 4**.**

Let $X,Z$ be metric spaces, $Y\subset X$, $\phi:Y\rightarrow Z$ be a map having bi-Lipschitz distortion $dist(\phi)=D\in[1,\infty)$ for some $s>0$ i.e.,

[TABLE]

for every $y_{1},y_{2}\in Y$. Let $\nu\geq 1$ and $\phi_{1}:X\rightarrow Z$ be a $Ds\nu$-Lipschitz extension of $\phi$. Then

1. (1)

for every $K\geq 0$ if $x_{1},x_{2}\in X$ satisfy

[TABLE]

then $||\phi_{1}(x_{1})-\phi_{1}(x_{2})||\geq s(1-\frac{2}{K\nu}-\frac{2D}{K})d(x_{1},x_{2}),$ 2. (2)

in particular if $\nu=\sqrt{n}$ and $x_{1},x_{2}\in X$ satisfy

[TABLE]

then $||\phi_{1}(x_{1})-\phi_{1}(x_{2})||\geq\frac{s}{5}d(x_{1},x_{2}).$

Proof of (1).

Indeed, fix $x_{1},x_{2}\in X$ such that $d(x_{1},x_{2})>K\nu\max\{d(x_{1},Y),d(x_{2},Y)\}$. Fix ${\delta}$ such that $\max\{d(x_{1},Y),d(x_{2},Y)\}<{\delta}<\frac{d(x_{1},x_{2})}{K\nu}$ and $y_{1},y_{2}\in Y$ be such that $d(x_{1},y_{1}),d(x_{2},y_{2})<{\delta}$. By the triangle inequality we have

[TABLE]

By the left part of (2.1) we have

[TABLE]

And since $\phi_{1}$ is $Ds\nu$-Lipschitz we have

[TABLE]

[TABLE]

Substituting last three inequalities into (2.2) provides

[TABLE]

[TABLE]

[TABLE]

Since ${\delta}<\frac{d(x_{1},x_{2})}{K\nu}$ we have

[TABLE]

And we are done. ∎

Proof of (2).

We substitute $\nu=\sqrt{n}$ and $K=5D$ into the last inequality of the previous proof. This provides

[TABLE]

∎

Lemma 5**.**

Let $X$ be a metric space with the doubling constant ${\lambda}\in{\mathbb{N}}$, ${\Gamma}\geq 1$, $f\geq 0$ be a ${\Gamma}$-Lipschitz function on X. Suppose that there are $n\in{\mathbb{N}}$ and $D\geq 1$ such that for every $x\in X$ there exists an embedding of $B_{f(x)}(x)$ in ${\mathbb{R}}^{n}$ with distortion less then $D$. Then for every ${\zeta}\geq 1$ there exist $N=N(n,D,{\lambda},{\zeta},{\Gamma})$ and $\Phi:X\rightarrow{\mathbb{R}}^{N}$ such that

1. (1)

if $f(x)=0$ then $\Phi(x)=0$, 2. (2)

$\operatorname{Lip}(\Phi)$* is bounded by some function of $n$, $D$, ${\lambda}$, ${\zeta}$ and ${\Gamma}$,* 3. (3)

if $x_{1},x_{2}\in X$ satisfy $d(x_{1},x_{2})\leq{\zeta}\max\{f(x_{1}),f(x_{2})\}$ then $||\Phi(x_{1})-\Phi(x_{2})||\geq Kd(x_{1},x_{2})$, for $K=\frac{9{\Gamma}}{40{\zeta}}$.

Proof.

We will only deal with the case ${\Gamma}=1$ and the general case follows by rescalings of metrics. Most of ideas of the following proof came from the proof of Assouad’s embedding theorem (see [2, 4, 10]).

Step 1: maps $P_{x}$. On this step we introduce maps $P_{x}$ which will be used as building blocks for $\Phi$.

A subset of a metric space is said to be $r$-separated if every pair of distinct points in that set is of distance $\geq r$. For every $k\in{\mathbb{Z}}$ we fix $N_{k}$ to be a maximal $\frac{2^{k}}{10}$-separated subset in $\{x\in X|2^{k}\leq f(x)\leq 2^{k+2}\}$. In the later text we assume that for $k\neq l\in{\mathbb{Z}}$ we have $N_{k}\cap N_{l}=\emptyset$. The proof of the general case is the same but requires more messy notation.

For $x\in N_{k}$ we construct $P_{x}:X\rightarrow{\mathbb{R}}^{n+1}$ such that

1. (1)

${P_{x}}(B_{2^{k-1}}(x))\subset\{(y_{1},\dots,y_{n+1})\in{\mathbb{R}}^{n+1}||y_{1}+\dots+y_{n+1}=2^{k}\sqrt{n+1}\}$, 2. (2)

$P_{x}(x)=\frac{2^{k}}{\sqrt{n+1}}(1,\dots,1),$ 3. (3)

${P_{x}}|_{B_{2^{k-1}}(x)}$ has bi-Lipshitz distortion $D$ for $s=1$, 4. (4)

$P_{x}(X\setminus B_{\frac{11}{10}2^{k-1}}(x))=0$. 5. (5)

$P_{x}$ is $40D\sqrt{n+1}$-Lipschitz.

Construction. First we define $P_{x}$ only on $B_{2^{k-1}}(x)$ and in a way that it satisfies (1-3) which can be done by assumptions of the lemma. Then we define $P_{x}$ on $X\setminus B_{\frac{11}{10}2^{k-1}}(x)$ by zero. For every $z\in B_{2^{k-1}}(x)$ we have

[TABLE]

This implies that $P_{x}$ is $40D$-Lipschitz on $B_{2^{k-1}}(x)\cup(X\setminus B_{\frac{11}{10}2^{k-1}}(x))$. By Proposition 3 $P_{x}$ can be extended to the whole $X$ in a way that the resulting map is $40D\sqrt{n+1}$-Lipschitz.

Step 2: coloring $\chi$. On this step we introduce coloring $\chi$ of $\cup_{k=-\infty}^{\infty}N_{k}$. It will assist us in our mission of constructing a finite dimensional map $\Phi$ from the countable set of maps $\{P_{x}\}_{x\in\cup_{k=-\infty}^{\infty}N_{k}}$.

For $a,R>0$ and a metric space having doubling constant ${\lambda}$ we have that the size of an $a$-separated subset of a ball of radius $R$ does not exceed ${\lambda}^{2+\log_{2}(\frac{R}{a})}$. Thus if we take $j={\lambda}^{2+\log_{2}(100{\zeta})}$ then for every $x\in X$ the size of $\{y\in N_{k}|d(x,y)<10\cdot 2^{k}{\zeta}\}$ is less or equal to $j$. We fix a coloring $\chi:\cup_{k=-\infty}^{\infty}N_{k}\rightarrow\{1,\dots,j\}$ such that for every $k\in{\mathbb{Z}}$ and every $x,y\in N_{k}$ equality $\chi(x)=\chi(y)$ implies

[TABLE]

Step 3: Construction of $\Phi$. We will start from the set of maps $\{P_{x}\}_{x\in\cup_{k=-\infty}^{\infty}N_{k}}$. We are going to split them into finite amount of groups in such a manner that supports of elements of the same group are far from each other (see Claim 6). From each group we will get a map $Q_{\widetilde{k},\widetilde{j}}$ by simply taking the sum of all elements. Finally, we will bundle all those maps $Q_{\widetilde{k},\widetilde{j}}$ together to get $\Phi$. More precisely we do the following.

Let $M=\max\{8,\log_{2}(440D\sqrt{n+1})\}$. For in $a\in Z$ we denote

[TABLE]

For $\widetilde{k}=\overline{1},\dots,\overline{M}$ and $1\leq\widetilde{j}\leq j$ we define $Q_{\widetilde{k},\widetilde{j}}:X\rightarrow{\mathbb{R}}^{n+1}$ by

[TABLE]

We claim that a map $\Phi:X\rightarrow{\mathbb{R}}^{N}={\mathbb{R}}^{(n+1)Mj}$ defined by

[TABLE]

satisfies conditions (1 - 3).

Step 3: proof that (1) is satisfied.

Fix $x\in N_{k}$, we have $f(x)\geq 2^{k}$ and $f$ is $1$-Lipschitz thus $d(x,\{p\in X|f(p)=0\})\geq 2^{k}$. Since $P_{x}(X\setminus B_{\frac{11}{10}2^{k-1}}(x))=0$ we conclude that $P_{x}(\{p\in X|f(p)=0\})=0$. Thus $\Phi(\{p\in X|f(p)=0\})=0$ too.

Step 4: Claim 6. The following claim says that supports of $P_{x_{1}}$, $P_{x_{2}}$ which are summands in the same $Q_{\widetilde{k},\widetilde{j}}$ are far from each other. This claim will we useful for proving both (2) and (3).

Claim 6**.**

Let $k_{1},k_{2}\in{\mathbb{Z}}$ be such that $k_{1}=k_{2}\mod M$, $x_{1}\in N_{k_{1}}$, $x_{2}\in N_{k_{2}}$, $z_{1}\in\operatorname{Spt}(P_{x_{1}})$, $z_{2}\in\operatorname{Spt}(P_{x_{2}})$. If $\chi(x_{1})=\chi(x_{2})$ and $x_{1}\neq x_{2}$ then

[TABLE]

Proof.

The case $k_{1}=k_{2}$ follows from (2.3). In the case $k_{1}\neq k_{2}$ we can assume $k_{1}\geq k_{2}+M$ without loss of generality. We have that $f(x_{1})\geq 2^{k_{1}}$. Since $f$ is $1$-Lipschitz and $\operatorname{Spt}(P_{x_{1}})\in\overline{B_{\frac{11}{10}2^{k_{1}-1}}(x_{1})}$ we conclude that $f(z_{1})\geq\frac{9}{10}2^{k_{1}-1}$. The similar argument shows that $f(z_{2})\leq 2^{k_{2}+3}\leq 2^{k_{1}-(M-3)}$. Thus $f(z_{1})-f(z_{2})\geq\frac{2}{5}2^{k_{1}}$. Once again we apply that $f$ is $1$-Lipschitz and obtain $d(z_{1},z_{2})\geq\frac{2}{5}2^{k_{1}}$. ∎

Step 5: proof of (2). We are going to show that

[TABLE]

is Lipschitz for every $\widetilde{k}=\overline{1},\dots,\overline{M}$ and $1\leq\widetilde{j}\leq j$. By Claim 6 we know that supports of functions $P_{x}$ in the sum do not intersect each other. Fix $z_{1},z_{2}\in X$. If there exists $k\in\widetilde{k}$ and $x_{1}\in N_{k}$ such that $\chi(x_{1})=\widetilde{j}$ and $z_{1},z_{2}\in\operatorname{Spt}(P_{x_{1}})$ then we have

[TABLE]

by (5). The same is true if at least one of points $z_{1},z_{2}\in X$ is outside of supports of all functions $P_{x}$.

The remaining case is if there exist $k_{1},k_{2}\in\widetilde{k}$ and different $x_{1}\in N_{k_{1}},x_{2}\in N_{k_{2}}$ such that $\chi(x_{1})=\chi(x_{2})=\widetilde{j}$ and $z_{1}\in\operatorname{Spt}(P_{x_{1}})$, $z_{2}\in\operatorname{Spt}(P_{x_{2}})$. In this case by the claim we have

[TABLE]

On the other hand since each $P_{x_{i}}$ is $40D\sqrt{n+1}$-Lipschitz and $\operatorname{Spt}(P_{x_{i}})\subset\overline{B_{\frac{11}{10}2^{k_{i}-1}}(x)}$ we have

[TABLE]

for $i=1,2.$ Thus,

[TABLE]

Combining this with (2.4) provides

[TABLE]

Thus we conclude that $Q_{\widetilde{k},\widetilde{j}}$ is $110D\sqrt{n+1}$-Lipschitz for every $\widetilde{k}=\overline{1},\dots,\overline{M}$ and $1\leq\widetilde{j}\leq j$. And $\Phi$ is $110D\sqrt{(n+1)Mj}$-Lipschitz.

Step 6: proof of (3). Let $z_{1},z_{2}\in X$ be such that $d(z_{1},z_{2})\leq{\zeta}\max\{f(z_{1}),f(z_{2})\}$. We have to show that

[TABLE]

Without loss of generality we assume that $f(z_{1})\geq f(z_{2})$. We fix $k\in{\mathbb{N}}$ such that $f(z_{1})\in[2^{k},2^{k+1})$.

The first case is $d(z_{1},z_{2})<\frac{4}{5}2^{k-1}$. By the construction of $N_{k}$ there exists $x\in N_{k}$ such that $d(z_{1},x)<\frac{1}{5}2^{k-1}$. Then we also have $d(z_{2},x)<2^{k-1}$ by the triangle inequality. Thus by the property (3) from the construction of $P_{x}$ we have

[TABLE]

By the Claim 6 we conclude that

[TABLE]

And thus

[TABLE]

Now lets consider the case $d(z_{1},z_{2})\geq\frac{4}{5}2^{k-1}$. By the construction of $N_{k-1}$ there exists $x\in N_{k-1}$ such that $d(z_{1},x)<\frac{1}{5}2^{k-2}$. By the triangle inequality we conclude that $d(z_{2},x)\geq\frac{7}{5}2^{k-2}$. By the property (1) from the construction of $P_{x}$ we have

[TABLE]

And thus

[TABLE]

Now we are going to show that

[TABLE]

By the Claim 6 we know that at most one $P_{x^{\prime}}(z_{2})$ in the previous sum is non zero. The case when all summands are zero is trivial. Thus we assume that non-zero summand exist and denote by $k^{\prime}$ and $x^{\prime}$ corresponding indexes. First lets note that $k^{\prime}\leq k$. Indeed, suppose that $k^{\prime}\geq k+M$. Then $f(x^{\prime})\geq 2^{k^{\prime}-1}$ and since $f$ is $1$-Lipschitz and $\operatorname{Spt}(P_{x^{\prime}})\subset\overline{B_{\frac{11}{10}2^{k^{\prime}-2}}(x^{\prime})}$ we conclude that $f(z_{2})\geq\frac{9}{10}2^{k^{\prime}-2}>2^{k+1}\geq f(z_{1})$. Which contracts the assumption that $f(z_{2})\leq f(z_{1})$.

Next we are going to show that $k^{\prime}\neq k$. Once again we argue by contradiction and suppose that $k^{\prime}=k$. Combining $d(z_{2},x)\geq\frac{7}{5}2^{k-2}$ and the property (4) of $P_{x}$ we obtain that

[TABLE]

Thus $x^{\prime}\neq x$. Then by the construction of $N_{k-1}$ we have that

[TABLE]

Since $z_{1}\in\operatorname{Spt}(P_{x})\subset\overline{B_{\frac{11}{10}2^{k-2}}(x)}$ and $z_{2}\in\operatorname{Spt}(P_{x^{\prime}})\subset\overline{B_{\frac{11}{10}2^{k-2}}(x^{\prime})}$ the previous inequality implies

[TABLE]

which contradicts the assumption that

[TABLE]

We conclude that the only possible case is that $k^{\prime}\leq k-M$. Thus we have

[TABLE]

[TABLE]

[TABLE]

where the last inequality follows from the definition of $M$. Thus we have that

[TABLE]

Which implies

[TABLE]

In combination with

[TABLE]

the previous inequality implies

[TABLE]

∎

Proof of Theorem 3.

Suppose that we are in conditions of Theorem 3. Without loss of generality we can assume that

[TABLE]

Let $\phi_{1}$ be the extension of $\phi$ provided by Proposition 3. We define $f:X\rightarrow[0,\infty)$ by

[TABLE]

And we take ${\zeta}=\frac{5\operatorname{dist}(\phi)\sqrt{n}}{{\theta}}$. Let $\Phi:X\rightarrow{\mathbb{R}}^{N}$ be the map provided by Lemma 5 for those $f,{\zeta}$ and ${\Gamma}=1$. We claim that the map $\widetilde{\phi}=(\phi_{1},\Phi)$ satisfies the conclusion of the Theorem 3.

Indeed, (2) from the conclusion of Theorem 3 follows from (1) from the conclusion of Lemma 5. Once again $N$ comes from Lemma 5 with the required upper bound which depends only on $n$, $\overline{n}$, $\operatorname{dist}(\phi)$, $D$, ${\theta}$ and ${\lambda}$. Note that both $\phi_{1}$ are $\Phi$ are Lipschitz and their Lipschitz constants depend only $n$, $\overline{n}$, $\operatorname{dist}(\phi)$, $D$, ${\theta}$ and ${\lambda}$. Thus, the same is true for $\widetilde{\phi}$.

Finally we have to provide the inequality

[TABLE]

for every $x_{1},x_{2}\in X$.

In the case

[TABLE]

we have

[TABLE]

where the last inequality follows from Lemma 5(2).

In the case

[TABLE]

we have

[TABLE]

where the last inequality comes from Lemma 5(3). ∎

3. Proof of Theorem 1

Definition 7** (Small rough angle condition; $\operatorname{SRA}({\alpha})$).**

Let $(X,d)$ be a metric space and $0<{\alpha}<1$. We say that $X$ satisfies the $\operatorname{SRA}({\alpha})$-condition if, for every $x,y,z\in X$, we have

[TABLE]

Lemma 8**.**

Let $0<{\alpha}<1$ and $X$ be a metric space such that it is free of $k$-point $\operatorname{SRA}({\alpha})$-subspaces. Suppose that $\{x_{1},\dots,x_{k-1}\}\subset X$ taken with induced metrics satisfies $\operatorname{SRA}(\frac{{\alpha}}{2})$-condition. Then $B_{\frac{{\alpha}R}{6}}(x_{1})\subset X$ could be embedded into ${\mathbb{R}}^{k-2}$ with distortion $\frac{\sqrt{k-2}}{{\alpha}}$, where $R=\min_{1\leq i<j\leq k-1}{d(x_{i},x_{j})}$.

Proof.

We claim that a map $\phi:X\rightarrow{\mathbb{R}}^{k-2}$ given by

[TABLE]

does the trick. Clearly $\phi$ is a $\sqrt{k-2}$-Lipschitz map. It suffices to show that for every $y_{1},y_{2}\in B_{\frac{{\alpha}R}{6}}(x_{1})$ we have

[TABLE]

Fix $y_{1},y_{2}$. We claim that there exists $1\leq m\leq k-2$ such that

[TABLE]

By contradiction suppose that for every $1\leq m\leq k-2$

[TABLE]

We claim that in this case $\{y_{1},y_{2},x_{2},\dots,x_{k-1}\}$ satisfies $\operatorname{SRA}({\alpha})$-condition. We have to check that (3.1) is satisfied for every $x,y,z\in\{y_{1},y_{2},x_{2},\dots,x_{k-1}\}$. We define a function $\overline{\cdot}:\{y_{1},y_{2},x_{2},\dots,x_{k-1}\}\rightarrow\{x_{1},x_{2},\dots,x_{k-1}\}$ by

[TABLE]

[TABLE]

Case A: $x,y,z\subset\{x_{2},\dots,x_{k-1}\}$. This case is trivial since $\{x_{1},\dots,x_{k-1}\}\subset X$ satisfies $\operatorname{SRA}(\frac{{\alpha}}{2})$-condition.

Case B: $\{x,y,z\}=\{y_{1},x_{i},x_{j}\}$. Since $\{x_{1},\dots,x_{k-1}\}\subset X$ satisfies $\operatorname{SRA}(\frac{{\alpha}}{2})$-condition. We have

[TABLE]

Since $y_{1},y_{2}\in B_{\frac{{\alpha}R}{6}}(x_{1})$ we have

[TABLE]

On the other hand since $R=\min_{1\leq i<j\leq k-1}{d(x_{i},x_{j})}$ we have

[TABLE]

[TABLE]

Once again we use that $y_{1},y_{2}\in B_{\frac{{\alpha}R}{6}}(x_{1})$

[TABLE]

[TABLE]

Adding together (3.3 - 3.6) provides

[TABLE]

Case C: $\{x,y,z\}=\{y_{1},y_{2},x_{i}\}$. If $z=x_{i}$ from $y_{1},y_{2}\in B_{\frac{{\alpha}R}{6}}(x_{1})$ we have that $d(x,y)\leq 2\frac{{\alpha}R}{6}$ and $d(x,z)\geq R-\frac{{\alpha}R}{6}$. And (3.1) follows.

Finally the last case is $z=y_{1}$, $x=x_{i}$, $y=y_{2}$. We have to show that

[TABLE]

It suffices to show that

[TABLE]

Which follows directly from (3.2). ∎

Proof of Theorem 1.

First we need the following proposition.

Proposition 9** ([8], Theorem 6).**

For $0<{\alpha}<1$, $k\in{\mathbb{N}}$ there exist $L=L(k,\alpha)$ such that every metric space $(X,d)$, which is free of $k$-point $\operatorname{SRA}({\alpha})$-subspaces, satisfies the doubling condition with the constant $L$.

We are going to prove the theorem via induction by $k$. The base is that the theorem is true for $k=3$.

Proof of the base..

Case A: $\operatorname{diam}(X)<\infty$. Fix a maximal $\frac{\operatorname{diam}(X)}{10}$-separated set $\{x_{1},\dots,x_{n}\}$ in $X$. By Proposition 9 we have that $n$ is bounded from above by some $C({\alpha})$. We claim that a map $\phi:X\rightarrow{\mathbb{R}}^{n}$ given by

[TABLE]

has distortion less or equal to $\sqrt{C({\alpha})}\max\{10,\frac{1}{{\alpha}}\}$. Clearly $\operatorname{Lip}(\phi)\leq\sqrt{n}\leq\sqrt{C({\alpha})}$. Thus it suffices to show that for every $y_{1},y_{2}\in X$ there exists $1\leq i^{\prime}\leq n$ such that

[TABLE]

Since $\{x_{1},\dots,x_{n}\}$ is a maximal $\frac{\operatorname{diam}(X)}{10}$-separated set there exists $x_{i}$ such that $d(x_{i},y_{1})\leq\frac{\operatorname{diam}(X)}{10}$.

If we have $d(y_{1},y_{2})\geq\frac{3\operatorname{diam}(X)}{10}$ then we can take $x_{i^{\prime}}=x_{i}$. Indeed, by the triangle inequality we have $d(x_{i},y_{2})\geq\frac{2\operatorname{diam}(X)}{10}$. Which implies that

[TABLE]

And (3.7) follows.

If we have $d(y_{1},y_{2})\leq\frac{3\operatorname{diam}(X)}{10}$ then there exists $x\in X$ such that $d(x_{i},x)\geq\frac{\operatorname{diam}{X}}{2}$. Once again we apply that $\{x_{1},\dots,x_{n}\}$ is a maximal $\frac{\operatorname{diam}(X)}{10}$-separated set and obtain that there exists $x_{j}$ such that $d(x_{j},x_{i})\geq\frac{\operatorname{diam}{X}}{2}-\frac{\operatorname{diam}(X)}{10}$. In this case we can take $x_{i^{\prime}}=x_{j}$. We claim that

[TABLE]

By contradiction suppose that

[TABLE]

From (3.9) we have

[TABLE]

[TABLE]

On the other hand we know that a subspace $\{x_{i},y_{1},y_{2}\}$ does not satisfy $\operatorname{SRA}({\alpha})$-condition. Thus,

[TABLE]

By the triangle inequality we have

[TABLE]

We conclude that

[TABLE]

Which contradicts with our assumption.

Case B: $\operatorname{diam}(X)=\infty$. This case follows from the following standard limiting argument. Fix a point $x\in X$. From Case A we have that there exist $N=N({\alpha}),D=D({\alpha})$ such that for every $m\in{\mathbb{N}}$ there exist an embedding $\phi_{n}:B_{m}(x)\rightarrow{\mathbb{R}}^{N}$ which distortion does not exceed $D$ and such that $\phi(x)=0$.

Since $X$ satisfies the doubling condition it is also separable. We fix a dense countable subset $Y\subset X$. By the diagonal argument there exists a map $\phi:Y\rightarrow{\mathbb{R}}^{N}$ having distortion less or equal to $D$. The continuous extension of $\phi$ provides the required embedding. ∎

From now we have to deal with the inductive step. We will show that theorem holds for $k\in{\mathbb{N}}$ and $0<{\alpha}<1$ under assumption that it holds for $k-1$ and $\frac{{\alpha}}{2}$.

Proof.

For a point $x\in X$ we define $R(x)\in[0,\infty]$ as the infimum of $R>0$ such that every $Y\subset X$ satisfying the $\operatorname{SRA}(\frac{{\alpha}}{2})$-condition and also such that $x\in Y$, $||Y||=k-1$ we have

[TABLE]

We define $X^{\prime}\subset X$ as a maximal subset of $X$ such that for every $x\neq y\in X^{\prime}$,

[TABLE]

Claim: $X^{\prime}$ is free of $(k-1)$-point $\operatorname{SRA}(\frac{{\alpha}}{2})$-subspaces.

Proof.

This follows from definitions of $R$ and $X^{\prime}$ in a tautological fashion. Indeed, suppose that $(k-1)$-point $Y\subset X^{\prime}$ satisfies the $\operatorname{SRA}(\frac{{\alpha}}{2})$-condition. Let $x\in Y$ be such that

[TABLE]

From the definition of $R(x)$ we conclude that there exist $y_{1}\neq y_{2}\in Y$ such that

[TABLE]

Thus we have

[TABLE]

Which is a contradiction. ∎

By the inductive assumption there exist $N^{\prime}=N^{\prime}(k-1,\frac{{\alpha}}{2})$, $D^{\prime}=D^{\prime}(k-1,\frac{{\alpha}}{2})$ and a map $\phi:X^{\prime}\rightarrow{\mathbb{R}}^{{\mathbb{N}}^{\prime}}$ which bi-Lipschitz distortion does not exceed $D$. We are going to construct the required embedding by extending $\phi$ via Theorem 3.

Fix a point $x\in X\setminus X^{\prime}$. Note that that the only possible reason why we cannot add $x$ to $X^{\prime}$ is that there exists $x^{\prime}\in X^{\prime}$ such that $d(x,x^{\prime})<R(x).$ Thus we have

[TABLE]

From Lemma 8 it follows that $B_{\frac{aR(x)}{6}}(x)$ allows an embedding into ${\mathbb{R}}^{k-2}$ with distortion $\frac{\sqrt{k-2}}{{\alpha}}$. Combined with (3.10) this provides all required ingredients for Theorem 3. So the induction step follows. ∎

∎

4. Proof of Theorem 2

It was shown in [8] that balls in Alexandrov spaces are $\operatorname{SRA}({\alpha})$-free for every $0<{\alpha}<1$ via the $\operatorname{ATB}$-condition. To prove Theorem 2 we are going to reformulate this result in a slightly stronger form. To do this we take from [8]

• •

the definition of $\operatorname{ATB}({\epsilon})$-condition,

• •

the theorem saying that metric spaces satisfying $\operatorname{ATB}({\epsilon})$-condition are $\operatorname{SRA}({\alpha})$-free,

• •

and the theorem providing $\operatorname{ATB}({\epsilon})$-condition for Alexandrov spaces.

Definition 10** (Angular total boundedness; $\operatorname{ATB}({\epsilon})$).**

Let $(X,d)$ be a metric space and $0<{\epsilon}<\pi/2$. We say that a point $p\in X$ satisfies the $\operatorname{ATB}({\epsilon})$-condition if there exist some $L\in{\mathbb{N}}$ and $R>0$ such that, for every $y_{1},\dots,y_{L}\in B_{R}(p)\setminus\{p\}$, we can find $i\neq j$ satisfying

[TABLE]

where $\widetilde{\angle}$ denotes the planar comparison angle. We say that $X$ is a globally $\operatorname{ATB}({\epsilon})$-space if there exists $L\in{\mathbb{N}}$ such that every $p\in X$ satisfies the $\operatorname{ATB}({\epsilon})$-condition with constants $L$ and $R=\infty$. We say that $X$ is a locally $\operatorname{ATB}({\epsilon})$-space if, for any $x\in X$, there exist $L\in{\mathbb{N}}$ and $R>0$ such that every $p\in B_{R}(x)$ satisfies the $\operatorname{ATB}({\epsilon})$-condition with constants $L$ and $R$.

Proposition 11** ([8], Theorem 2).**

For every $L\in{\mathbb{N}}$ there exists $N(L)\in{\mathbb{N}}$ satisfying the following. For every $0<{\alpha}<1$, ${\epsilon}:=\arccos({\alpha})/2>0$, $r>0$, every metric space $(X,d)$ and every $x\in X$ such that every $p\in B_{r}(x)$ satisfies the $\operatorname{ATB}({\epsilon})$-condition with constants $L$ and $2r$ we have that $B_{r}(x)$ is free of $N(L)$-point $\operatorname{SRA}({\alpha})$-subspaces.

The $\operatorname{ATB}({\epsilon})$-condition for Alexandrov spaces is provided by [8, Proposition 14]. Here we formulate this proposition in a stronger form then in [8], but the same proof works.

Proposition 12** ($\operatorname{ATB}$ of Alexandrov spaces).**

For every $k<0$, $0<{\epsilon}<\pi/2$ and $n\in{\mathbb{N}}$ there exist $L\in{\mathbb{N}}$ and $r>0$ satisfying the following. For every $n$-dimensional Alexandrov space $X$ of curvature $\geq k$, every $x\in X$ satisfies the $\operatorname{ATB}({\epsilon})$-condition for $L$ and $r$.

For every $n\in{\mathbb{N}}$, $0<{\epsilon}<\pi/2$ there exists $L_{0}\in N$ such that every point in every $n$-dimensional non-negatively curved Alexandrov space satisfies $\operatorname{ATB}({\epsilon})$-condition for $L_{0}$ and $R_{0}=\infty$.

Proof of Theorem 2.

We only give a proof for the case $k<0$, the proof for the case $k=0$ uses the same arguments and it is even simpler.

We fix $n\in{\mathbb{N}}$, $k<0$ and $R>0$. Next we fix an arbitrary $0<{\alpha}<1$ and set ${\epsilon}:=\arccos({\alpha})/2>0$. (For the proof of Theorem 2 we can use the $\operatorname{SRA}$-free condition for any $0<{\alpha}<1$).

By Proposition 12 we have that there exist $L(k,{\epsilon})\in{\mathbb{N}}$ and $r(k,{\epsilon})>0$ such that every point in $X$ satisfies the $\operatorname{ATB}({\epsilon})$-condition for $L$ and $r$. By Proposition there exists $N(k,{\epsilon})$ such that any open ball of radius $\frac{r}{2}$ in $X$ is free of $N$-point $\operatorname{SRA}({\alpha})$-subspaces. Since $X$ is an Alexandrov space there exists ${\lambda}(n,k,R)$ such that any ball of radius $R$ satisfies doubling condition with a constant ${\lambda}$. It is easy to see that two last statements imply that there exist $\widetilde{N}=\widetilde{N}(n,k,R,{\epsilon})\in{\mathbb{N}}$ such that any ball of radius $R$ in $X$ is free of $\widetilde{N}$-point $\operatorname{SRA}({\alpha})$-subspaces (see [13, Theorem 6] for details). Thus the existence of the required embedding follows from Theorem 1. ∎

5. Open problems

Question 13** (see [5]).**

Let $n$ be a natural number. Is it true that there exist $N(n),D(n)>0$ such that any complete $n$-dimensional Riemannian manifold of nonnegative Ricci curvature can be embedded into $N$-dimensional Euclidean space with bi-Lipschitz distortion less then $D$?

See also a similar question by S. Eriksson-Bique [3, Conjecture 1–11].