Fix a positive integer $n$, let $\mathbb{H}^n$ be the $n$-dimensional hyperbolic space, $r>0$, $x\in \mathbb{H}^n$ and consider the closed (compact) geodesic ball $B_{\mathbb{H}^n}(x,r)$. Are there known estimates on the minimum distortion of a bi-Lipschitz embedding of $B_{\mathbb{H}^n}(x,r)$ into $\mathcal{W}_2(\mathbb{R})$.

Let me just note that a bi-Lipschitz embedding into $\mathcal{W}_2(\mathbb{R})$ must exist by this paper.


3 Answers 3


For global distortion.

Choose a 4 point set --- vertices of an equilateral triangle in $B(x,r)_{\mathbb{H}^n}$ and its center and observe that it cannot be emebdded isometrically in nonnegatively curved space in the sense of Alexandrov, which includes $\mathcal{W}_2(\mathbb{R})$.

In is straightforward to get some explicit bounds this way, most likely they are far from being optimal.

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    $\begingroup$ Just apply the (3+1)-point comparison, you will get that for any $\varepsilon>0$ and all sufficiently large $R$ the map cannot be $[\sqrt3,2-\varepsilon]$-bilipschitz. $\endgroup$ Apr 3, 2022 at 16:48
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    $\begingroup$ By the way, if you apply (n+1)-point comparison you could get that distortion should be at least $\sqrt2-\varepsilon$ for large $R$. Most likely the bound should approach infinity, but I do not see an argument to prove it. $\endgroup$ Apr 3, 2022 at 16:55
  • $\begingroup$ May I ask you to add the details for this, I'm really not an expert in this type of argument :/ $\endgroup$ Apr 8, 2022 at 17:33
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    $\begingroup$ @Carlos_Petterson, I guess it is about (3+1)-point comparison --- check our book arxiv.org/abs/1903.08539 $\endgroup$ Apr 8, 2022 at 18:08

$\mathcal{W}_2(\mathbb{R})$ is isometric to a convex subset of a Hilbert space (and embedding a hyperbolic space into Euclidean/Hilbert one has a lot of history).

The argument can find an argument in S.S. Vallender "Calculation of the Wasserstein distance between probability distributions on the line". I will do an informal sketch.

I want to see probability measures in $\mathbb{R}$ as piles of sand with sand grains enumerated from left to right and indexed by $[0,1]$. For a measure $\mu$ define function $\Phi_{\mu}:[0,1]\rightarrow \mathbb{R}$ by $\Phi_\mu(t)$ be a position of $t$-th grain in $\mathbb{R}$. I claim that for every $\mu, \nu$ $$\mathcal{W}_2(\mu,\nu) = ||\Phi(\mu)-\Phi(\nu)||_{L_2}.$$

Before proving this let say how the optimal transport looks on $\mathbb{R}$.

Lemma 1: for the measures $\mu$ and $\nu$ the optimal transport plan between them sends $t$-th sand grain of $\mu$ into $t$-th sand grain of $\nu$.

I give the proof of the lemma for the case when $\mu = \frac{1}{n}(\delta(\mu_1) + \dots + \delta(\mu_n)),$ for some real $\mu_1 < \dots < \mu_n$ and $\nu = \frac{1}{n}(\delta(\nu_1) + \dots + \delta(\nu_n))$, for some real $\nu_1 < \dots < \nu_n$. I need to show that optimal transport moves $\frac{1}{n} \delta(\mu_1)$ into $\frac{1}{n}\delta(\nu_1)$ and then the rest will follow by induction. Okay, lets assume its not and (a part of) $\frac{1}{n} \delta(\mu_1)$ is moved to $\frac{1}{n} \delta(\nu_k)$ and (a part of) $\frac{1}{n}\delta(\mu_j)$ is moved to $\frac{1}{n} \delta(\nu_1)$ for some $j,k > 1$. I claim that if we change our transport by moving (the part of) $\frac{1}{n} \delta(\mu_1)$ into $\frac{1}{n}\delta(\nu_1)$ and (the part of) $\frac{1}{n}\delta(\mu_j)$ into $\frac{1}{n} \delta(\nu_k)$ it will get cheaper. Which follows from the following trivial lemma.

Lemma 2: Suppose that $a,b,c > 0$ then $a^2 + (b + c)^2 < (a + c)^2 + b^2$ iff $b < a$.

Now it's really easy to proof our main statement $$\mathcal{W}_2(\mu,\nu) = ||\Phi(\mu)-\Phi(\nu)||_{L_2}.$$
for the case when $\mu = \frac{1}{n}(\delta(\mu_1) + \dots + \delta(\mu_n)),$ for some real $\mu_1 < \dots < \mu_n$ and $\nu = \frac{1}{n}(\delta(\nu_1) + \dots + \delta(\nu_n))$, for some real $\nu_1 < \dots < \nu_n$. Indeed, squares of both sides are equal to $$\frac{1}{n}\sum_{i=1}^{n}|\mu_i - \nu_i|^2.$$

PS: note that measures of the type $\mu = \frac{1}{n}(\delta(\mu_1) + \dots + \delta(\mu_n)),$ for some real $\mu_1 < \dots < \mu_n$ are dense in Wasserstein space see Proposition 2.10 from Karl-Theodor Sturm. "On the geometry of metric measure spaces.". So all the above can be formalized with some suffering.


It admits an isometric embedding.

The Nash--Kuiper theorem provides an isometric embedding of $\mathbb{H}^n\to \mathbb{R}^{n+1}$; moreover, its image might be in a bounded set. Now observe that $\mathcal{W}_2(\mathbb{R})$ contains a subset isometric to a ball in $\mathbb{R}^{n+1}$.

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    $\begingroup$ The Nash-Kuiper theorem provides an isometric embedding (in the Riemannian sense) and not in the metric sense (as YCor pointed out to me a couple of days ago). So then $f:\mathbb{H}^n \rightarrow \mathbb{R}^{2n+1}$ is only locally Lipschitz with constant $\max_{x\in ...}\,\|\nabla f(x)\|\neq 1$. $\endgroup$ Apr 2, 2022 at 12:15
  • $\begingroup$ @Carlos_Petterson then you should say "global" somewhere in your question, otherwise it is unclear. $\endgroup$ Apr 2, 2022 at 15:14

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