# What is the structure preserved by strong equivalence of metrics?

Let $$X$$ be a set. Then we can define at least three equivalence relations on the set of metrics on $$X$$. We say that two metrics $$d_1$$ and $$d_2$$ are topologically equivalent if the identity maps $$i:(X,d_1)\rightarrow(X,d_2)$$ and $$i^{-1}:(X,d_2)\rightarrow(X,d_1)$$ are continuous. We say that $$d_1$$ and $$d_2$$ are uniformly equivalent if $$i$$ and $$i^{-1}$$ are uniformly continuous. And we say that $$d_1$$ and $$d_2$$ are strongly equivalent if there exists constants $$\alpha,\beta>0$$ such that $$\alpha d_1(x,y)\leq d_2(x,y)\leq\beta d_1(x,y)$$ for all $$x,y\in X$$.

We can take equivalence classes of metrics under each of these equivalence relations. Now two metrics are topologically equivalent if and only if they induce the same topology on $$X$$, so we can identify equivalence classes under topological equivalence with topologies on $$X$$. And two metrics are uniformly equivalent if and only if they induce the same uniformity on $$X$$, so we can identify equivalence classes under uniform equivalence with uniformities on $$X$$. But my question is, what structures can we identify equivalence classes under strong equivalence with? To put it another way, two metrics are strongly equivalent if and only if they have the same ... what?

One thing worth noting is that if two metrics are strongly equivalent, then they’re both uniformly equivalent and they have the same bounded sets. Or in fancier language, they induce both the same uniformity and the same bornology. But the converse is not true; uniformly equivalent metrics with the same bounded sets need not be strongly equivalent. So that means there must be some structure on top of the uniformity and bornology that is preserved by strong equivalence.

Note: This was posted on Mathematics Stack Exchange with no answer.

• Crossposted from MSE. It is best to link between your posts to avoid duplication of effort. – Mike Miller Dec 30 '18 at 2:30
• I'm not sure there's a well-established terminology, but these are metrics up to bilipschitz equivalence. Keywords: "bilipschitz class", "Lipschitz category". – YCor Dec 30 '18 at 6:12
• I don't know of anything that's not somewhat contrived besides what @YCor and Tim Campion have said, but in case it could be of use, I posted a lengthy summary of results involving these three types of equivalences in this 4 October 2006 sci.math post archived at Math Forum. (Note: I didn't see your earlier question in MSE. I've long since given up trying to even glance at all question titles appearing there, and now I just look over the first page of hits whenever I'm there, and sometimes search under the tag "reference request".) – Dave L Renfro Dec 30 '18 at 17:52
• See mathoverflow.net/questions/291563/… for a related question (with some answers). – Taras Banakh Jan 5 '19 at 21:40
• A bi-lipschitz bijection may also be called a lipeomorphism. So if two spaces are related like this, we could say that they are lipeomorphic. – Gerald Edgar Jan 7 '19 at 20:51

Metrics are strongly equivalent if the identity mapping $$Id:(X,d_1)\to (X,d_2)$$ is bi-Lipschitz. They preserve the class of Lipschitz mappings.

Roughly speaking classical topology deals with notions that are invariant under homeomorphisms. In particular, changing metric to the one that gives equivalent topology does not make much harm. But the situation chances rapidly if we look at Lipschitz mappings and invariance under bi-Lipschitz mappings. This kind of topology leads to so called analysis on metric spaces.

If, in addition to a metric structure one assumes that the metric space is equipped with the so called doubling measure, then one can even develop analysins of first order derivatives on such spaces, including the theory of Sobolev spaces, quasiconformal mappings and differentiablity of Lipschitz functions.

This is quite new area of research developed since nineties and it already has its subject in MSC:

30L View Publications (2010-now) Analysis on metric spaces

Good introductory references include:

J. Heinonen, P. Koskela, N. Shanmugalingam, J. T. Tyson, Sobolev spaces on metric measure spaces. An approach based on upper gradients. New Mathematical Monographs, 27. Cambridge University Press, Cambridge, 2015.

J. Heinonen, Nonsmooth calculus. Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 2, 163–232.

J. Heinonen, Lectures on Lipschitz analysis Report. University of Jyväskylä Department of Mathematics and Statistics, 100. University of Jyväskylä, Jyväskylä, 2005.

• I would say that to get differentiability results it is not always necessary to have a doubling measure, it can be sufficient to have a suitable notion of a negligible set which can play the role of sets of measure zero in the Rademacher theorem. See Chapter 6 in the book by Benyamini and Lindenstrauss mentioned in my answer. – August Cleaner Jan 6 '19 at 2:57
• @AugustCleaner I am not sure if what I had in mind is in that book since I am talking about the Rademacher theorem for Lipschitz functions defined on metric spaces. This is a celebrated result of Cheeger from 1999. – Piotr Hajlasz Jan 6 '19 at 3:00

First of all, there are two aspects to Lipschitz maps between metric spaces: Local (having implications such as differentiability properties, etc) and global. The two aspects have different non-metric formalizations. Below, I will be mostly addressing the local aspect. If you are interested in the global aspect (appearing under the umbrella of "Coarse Geometry"), read section 2.1 of

Roe, John, Lectures on coarse geometry, University Lecture Series 31. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3332-4/pbk). vii, 175 p. (2003). ZBL1042.53027.

The key difference is that instead of uniformities in my answer Roe uses "coarse structures" ("coarsivities"). The definition is almost the same but instead of been closed under taking supersets, coarsivities are closed under taking subsets. Large-scale Lipschitz maps are then defined as "bornologous maps," see also here.

There are some interesting relationships between micro and macro aspects of Lipschitz maps, but I will not go into this.

One can define filtered uniformities for more general posets than $${\mathbb Z}$$ and non-Hausdorff uniformities, which will lead to the notion of Lipschitz maps between nonmetrizable spaces. I will not do this.

Definition 1. A Hausdorff uniformity $${\mathcal U}$$ on a set $$X$$ will be called $${\mathbb Z}$$-filtered if it comes equipped with a map $${\mathbb Z}\times {\mathcal U}\to {\mathcal U}, (a,U)\mapsto U_a$$, such that

1. $$a\le b\Rightarrow U_a\subset U_b,$$ and $$U_0=U$$ for all $$U\in {\mathcal U}$$.

2. For some $$U\in{\mathcal U}$$, the subset $$=\{U_n: n\in {\mathbb Z}\}$$ is a basis of $${\mathcal U}$$. (I will refer to such $$U$$'s as "generators" of $${\mathcal U}$$.)

Example. Suppose that $${\mathcal U}={\mathcal U}_d$$ is the metric uniformity defined by a metric $$d$$ on $$X$$: A basis $${\mathcal B}$$ of $${\mathcal U}_d$$ is given by subsets of the form $$U(\epsilon)=\{(x,y)\in X^2: d(x,y)<\epsilon\}$$, $$0<\epsilon<\infty$$. Then $${\mathcal U}$$ consists of all subsets of $$X^2$$ containing elements of $${\mathcal B}$$. For $$n\in {\mathbb Z}$$, $$U=U(\epsilon)$$, define $$U_n:= \{(x,y)\in X^2: d(x,y)< 2^n \epsilon\}.$$ Thus, for each $$U\in {\mathcal B}$$, $$$$ is a basis of $${\mathcal U}$$.

For each $$U\in {\mathcal U}$$ which is not in $${\mathcal B}$$, I will take $$U_n:= U, n\in {\mathbb Z}.$$ (I am sure somebody can come up with a more imaginative choice here.)

In what follows, I will simply say that a uniformity is filtered rather than $${\mathbb Z}$$-filtered. Thus, every metric uniformity is filtered.

Lemma 1. Each filtered uniformity $${\mathcal U}$$ is metrizable.

Proof. The proof is quite standard. Pick a generator $$U$$ of $${\mathcal U}$$. For each pair $$x, y\in X$$ define $$\phi_U(x,y)$$ as $$\inf \{2^n: n\in {\mathbb Z}, (x,y)\in U_n\}.$$ Then set $$d_U(x,y)=\inf \sum_i \phi_U(x_i, x_{i+1})$$ where the infimum is taken over all chains $$x=x_1, x_2...,x_{k-1}, x_k=y$$. Then $$d_U$$ is a metric (recall that $${\mathcal U}$$ is assumed to be Hausdorff) and its uniformity is equivalent to $${\mathcal U}$$. qed

I will use the notation $$d_U$$ for the metric on $$X$$ given by this construction.

Suppose that $${\mathcal U}, {\mathcal V}$$ are filtered uniformities on sets $$X, Y$$ and $$f: X\to Y$$ is a map. Then $$f$$ induces the map $$f: X^2\to Y^2$$, $$f(x_1, x_2)= (f(x_1), f(x_2)).$$

Definition 2. A map $$f: X\to Y$$ (between two sets equipped with filtered uniformities) is said to be Lipschitz (in the sense of uniformities) if for each generator $$V\in {\mathcal V}$$, there exists a generator $$U\in {\mathcal U}$$ such that for all sufficiently small $$n\in {\mathbb Z}$$, $$f(U_n)\subset V_n.$$

Equivalently, one can require this property for all generators $$V$$; also, equivalently, one can require this for all sufficiently small generators $$U$$.

Definition 3. Two uniformities $${\mathcal U}, {\mathcal V}$$ on the same set $$X$$ are Lipschitz-equivalent (in the sense of uniformities) iff the identity map $$(X, {\mathcal U})\to (X, {\mathcal V})$$ is bi-Lipschitz (i.e. is Lipschitz and its inverse is also Lipschitz). A bi-Lipschitz structure on a set $$X$$ is the equivalence class of a filtered uniformity on $$X$$.

I will omit the proofs of the lemmata below:

Lemma 2. If $${\mathcal U}, {\mathcal V}$$ are metric uniformities on metric spaces $$X, Y$$, a map $$f: X\to Y$$ is Lipschitz if and only if it is a Lipschitz map in the sense of Definition 2.

Lemma 3. If $${\mathcal U}$$ is a filtered uniformity, then for any two generators $$U, V$$ of $${\mathcal U}$$ the metrics $$d_U, d_V$$ are Lipschitz-equivalent.

Lemma 4. Suppose that $$(X,d)$$ is a metric space, $${\mathcal U}={\mathcal U}_d$$ is the corresponding metric uniformity with a generator $$U$$. Then the metrics $$d, d_U$$ are Lipschitz-equivalent in the traditional sense.

So, here you have it: A non-metric definition of bi-Lipschitz equivalence and structure.

Since nobody came up with anything better, let me expand my comment here, at the MSE version of the question.

The substantive part of your question is to give a "metric-free" definition of the Lipschitz equivalence class of metrics on a given topological space. I will deal only with the case of compact spaces since the answer is much cleaner in this setting.

To every compact metric space $$(X,d)$$ one associates its Lipschitz algebra which is the Banach *-algebra of complex-valued Lispchitz functions $$Lip(X)$$ on $$X$$, equipped with the Lipschitz norm $$||f||_{Lip}= ||f|| + \sup_{x\ne y} \frac{||f(x)-f(y)||}{d(x,y)}.$$ Here $$||f||$$ is the supremum-norm of $$f$$.

To the best of my knowledge, these algebra were first analyzed in detail by Donald Scherbert in

Banach algebras of Lipschitz functions, Pacific J. Math. 13 (1963) 1387--1399.

Scherbert proves results for general metric spaces; here are are some basic results that he proved adopted to the compact case:

1. $$Lip(X)$$ is a (unital, which I will assume throughout) *-Banach algebra, i.e. a topological *-algebra (up to an isomorphism in the category of topological *-algebras) which admits a Banach norm. (I am sure somebody who knows analysis better than I do can come up with a reference to a "metrization theorem" for topological *-algebras. Frechet is a necessary condition.)

2. Every *-Banach algebra is isomorphic to $$Lip(X)$$ for some compact metric space $$X$$.

3. Metrics $$d_1, d_2$$ on $$X$$ are (boundedly) equivalent (i.e. the identity map is bi-Lipschitz) if and only if $$Lip(X,d_1)=Lip(X,d_2)$$ as subsets of $$C(X)$$.

4. Every morphism $$Lip(X_1,d_1)\to Lip(X_2,d_2)$$ is induced by a Lipschitz map $$(X_1,d_1)\to (X_2,d_2)$$. In particular, two metric spaces $$(X_1,d_1), (X_2,d_2)$$ are bi-Lipschitz homeomorphic if and only if $$Lip(X_1,d_1)\cong Lip(X_2,d_2)$$.

Thus, Scherbert's results yield a "metric-free" description of the Lipschitz category, i.e. the category of compact metric spaces where morphisms are Lipschitz-continuous. This does not quite answer your question which, in the language of Banach algebras can be formulated as follows:

Given a compact metrizable space $$X$$, describe subalgebras $$A\subset C(X)$$ (the $$C^*$$-algebra of complex-valued continuous functions on $$X$$) such that there exists a metric $$d$$ on $$X$$ for which $$A=Lip(X,d)$$.

One can give several easy necessary conditions:

(1) $$A$$ is "regular" (see Scherbert's paper).

(2) $$A$$ is "large" in the sense that every maximal ideal in $$C(X)$$ is generated by an element of $$A$$.

(3) $$A$$ is "small" in the sense that $$C(X)$$ and $$A$$ admit Banach norms $$||\cdot||$$ and $$||\cdot||_A$$ making them, respectively, a $$C^*$$-algebra and a Banach *-algebra, such that the inclusion map $$(A, ||\cdot||_A)\to (C(X), ||\cdot ||)$$ has unit norm and is compact. (The latter is the Arzela-Ascoli theorem.)

I suspect that a complete list of properties characterizing Lipschitz subalgebras of $$C^*$$-algebras is well-known (to the right group of people).

I would suggest to consider the strong equivalence as a form of preserving of a differentiable structure in metric space case. (I should mention that this answer is not really different from the one given by Piotr Hajlasz.)

To provide some meaning to the first sentence of this answer I suggest to recall the famous Rademacher theorem, its Banach-space analogues (see the book Benyamini, Yoav; Lindenstrauss, Joram Geometric nonlinear functional analysis. Vol. 1. American Mathematical Society, Providence, RI, 2000) and the famous differentiability theory of Cheeger (Differentiability of Lipschitz functions on metric measure spaces. Geom. Funct. Anal. 9 (1999), no. 3, 428–517; This theory is presented in the first reference provided by Piotr Hajlasz.)

However, it seems that for bilipschitz maps no differentiability is preserved in general, analyze, for example, the result of Aharoni (Every separable metric space is Lipschitz equivalent to a subset of $$c^+_0$$. Israel J. Math. 19 (1974), 284–291).