Notes of Adriane Kaichouh’s lecture nr 1

Non-embeddability of Urysohn space

Theorem 1 (Pestov 2008) Urysohn space {U} does not embed uniformly in any super-reflexive Banach space.

1. Urysohn space

1.1. Universality

Urysohn space is the universal complete separable metric space. It contains an isometric copy of every complete separable metric space. In particular, it contains {c_0}. So Pestov’s theorem follows from Kalton’s non embeddability result for {c_0} into reflexive Banach spaces. The point of the course is to study the method, different from Kalton’s.

As far as universality, {U} is not so spectacular: Banach and Mazur observed that {C([0,1],{\mathbb R})} is also universal. However, Urysohn space is ultrahomogeneous: any isometry between finite subsets of {U} extends to a global isometry (this generalizes to compact sets). Nevertheless, {U} was long forgotten, until Katetov gave a new construction.

1.2. Katetov’s construction

For any finite subset {A\subset U}, and any isometric embedding of {A} into a finite metric space {B}, {B} also sits in {U} containing {A}. So any time a point is given with distances to points of {A} that satisfy triangle inequality, that point in fact sits in {U}. This is the basic step of Katetov’s construction, “one point extension”.

Given a metric space {X}, let

\displaystyle  E(X)=\{f:X\rightarrow{\mathbb R}\,;\,\forall x'\in X,\,|f(x)-f(x')|\leq d(x,x')\leq f(x)+f(x')\}.

It is equipped with the sup distance. {X} sits there isometrically. Consider {E_2(X)=E(E(X))}, and iterate. Let {U} be the completion of

\displaystyle  \begin{array}{rcl}  \bigcup_{n\in{\mathbb N}}E_n(X). \end{array}

Actually, we take only those functions of {E(X)} which have finite support inthe following sense: {f\in E(X)} if there is a finite subset {A\subset X} such that {f} is the , i.e. {f} is the largest function compatible with its values on {A},

\displaystyle  \begin{array}{rcl}  f(x)=\max_{a\in A}f(a)+d(a,x). \end{array}

This is necessary to get separable spaces.

Naturality: isometries of {X} extend uniquely to isometries of {E(X)}. So they also extend to isometries of {U}.

Theorem 2 (Uspensky) {Isom(U)} is universal for all Polish groups (i.e. separable and admit a complete compatible distance).

Example. {Isom(U)} contains {Homeo^+([0,1])}.

Later, we shall use a result about that group (Megrelishvili 2001): The only continuous isometric linear representation of {Homeo^+([0,1])} on a reflexive Banach space is the trivial representation.

Uspensky relies on the following theorem.

Theorem 3 (Gao-Kechris) Every Polish group is a closed subgroup of some {Isom(X)}, {X} complete separable.

2. The extension property

Let us discuss a strenthening of ultrahomogeneity.

Say a metric space {X} has the extension property if for every finite subset {A\subset X}, there exists a finite subset {B\subset X} containing it such that every partial isometry of {A} (i.e. an isometry between subsets of {A}) extends to a global isometry of {B}.

Example. {{\mathbb N}} with it usual distance has the extesion property. Hrushovski has shown that the random graph (percolation in the complete graph) has the extension property.

Theorem 4 (Solecki) {U} has the extension property.

This a hard theorem.

By iterating the extension property applied to {A\cup B}, and then adding points, one obtains ultrahomogeneity.

Here is one more property that we shall need tomorrow.

Theorem 5 {Isom(U)} contains a dense, locally finite subgroup.

Here, locally finite means that every finitely generated subgroup is finite.

This subgroup is obtained as an increasing union of finite groups, isometry groups of finite subsets, as a consequence of a strenghening of extension property: not only isometries do extend from {A} to {B}, but this is performed by a group homomorphism {Isom(A)\rightarrow Isom(B)}.


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
This entry was posted in Workshop lecture and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s