## Notes of Adriane Kaichouh’s lecture nr 2

1. Super-reflexivity

1.1. Uniform convexity

A Banach space ${E}$ is uniformly convex if the midpoint of two points on the unit sphere which are sufficiently far apart is deep inside the ball.

Example. Hilbert spaces, ${L^p}$ spaces with ${1 are uniformly convex.

Uniform convexity carries to ${\ell^2}$ direct sums ${\ell^2(X,E)}$.

1.2. Super-reflexivity

Uniform convexity is not stable under renorming.

Definition 1 A Banach space is super-reflexive if it admits an equivalent norm which is uniformly convex.

Here is an equivalent definition.

Theorem 2 (Enflo) A Banach space is super-reflexive iff every one of its ultrapowers is reflexive.

In other words, every space which is finitely representable in it is reflexive.

Recall that, given an ultrafilter ${\mathcal{U}}$ on an index set ${I}$, the ultraproduct ${\prod_{\mathcal{U}}E_i}$ of a family of Banach spaces ${(E_i)_{i\in I}}$ is the quotient of ${\ell^{\infty}(I,E_i)}$ by the subspace ${c_0(I,E_i)}$ of sequences that tend to 0 along ${\mathcal{U}}$. This is a Banach space. If all ${E_i=E}$, the ultraproduct is called an ultrapower.

2. Pestov’s theorem

Here is Pestov’s main theorem.

Theorem 3 (Pestov) Let ${X}$ be a metric space. Let ${G}$ be a locally finite group acting on ${X}$ by isometries. Assume that ${X}$ admits a coarse/uniform embedding into a Banach space ${E}$. Then there is a coarse/uniform embedding ${\psi}$ from ${X}$ to some ultrapower of some ${\ell^2(\mathcal{U},E)}$, and the action of ${G}$ on ${\psi(X)}$ extends to an action by affine isometries on the affine span of ${\psi(X)}$.

2.1. Proof

Let ${\phi}$ be the initial coarse embedding. Start with a finite subgroup ${F}$ of ${G}$. Define

$\displaystyle \begin{array}{rcl} \psi_F : X\rightarrow \ell^2(F,E),\quad \psi_F(x)(f)=\frac{1}{|F|}\phi(f^{-1}x). \end{array}$

Then ${\psi_F}$ has the same expansion and compression moduli as ${\phi}$. ${F}$ acts on ${\ell^2(F,E)}$ via left translations, and ${\psi_F}$ is equivariant for this action.

Next we glue all maps ${\psi_F}$ together. Let ${I}$ denote the set of all finite subgroups of ${G}$. Because ${G}$ is locally finite, there exists an ultrafilter ${\mathcal{U}}$ on ${I}$ such that for all ${F\in I}$, the set

$\displaystyle \begin{array}{rcl} \{H\in I\,;\,F\subset H\} \end{array}$

is in ${\mathcal{U}}$. Indeed, the intersection of two such subsets

$\displaystyle \begin{array}{rcl} \{H\in I\,;\,F\subset H\}\cap\{H\in I\,;\,F\subset H\}=\{H\in I\,;\,F\subset H\}=\{H\in I\,;\,F\subset H\} \end{array}$

and the subgroup ${±langle F_1,F_2\rangle}$ generated by two finite subgroups is finite again. So we are dealing with a filter basis.

Fix an origin ${o\in X}$. Let ${V}$ be the ultraproduct over ${\mathcal{U}}$ of spaces ${\ell^2(F,E)}$, based at ${\psi_F(o)}$ instead of 0 (this makes ${\ell^2(F,E)}$ an affine space instead of a vectorspace, and so much for the ultraproduct). Define ${\psi(x)}$ as the sequence ${(\psi_F(x))_{F\in I}}$. It is well defined and has the same expansion and compression moduli as ${\phi}$.

Given ${g\in G}$, observe that ${\mathcal{U}}$-almost surely,

$\displaystyle \begin{array}{rcl} g\psi(x)=\psi(gx) \end{array}$

is well defined, and this provides an isometry of ${\psi(X)}$. Isometries of subsets of Hilbert spaces extend to affine hulls, so the ${G}$ action extends.

As defined, ${V}$ is not quite an ultrapower. For every ${F\in I}$, ${\ell^2(F,E)}$ embeds non canonically into ${\ell^2(G,E)}$, so

2.2. Application to uniform embeddability of Urysohn space

Let ${\phi:U\rightarrow E}$ be a uniform embedding. Let ${G}$ be a dense locally finite subgroup of ${Isom(U)}$. Pestov’s theorem implies that there is a uniform embedding of ${U}$ to a reflexive Banach space ${V}$, equivariant with respect to an affine isometric action of ${G}$. ${\Psi}$ is a homeo onto ${\psi(U)}$. The topologies of pointwise convergence induced on ${G}$ by the two actions of ${\psi(U)}$ coincide, so, by density, the action on the affine span ${S}$ of ${\psi(U)}$ extends to ${Isom(U)}$. I.e. we get a continuous group homomorphism

$\displaystyle \begin{array}{rcl} Isom(U)\rightarrow Aff(S)=O(S)\times S. \end{array}$

Since there is an injective equivariant map, it is injective.

Now we use the fact that ${Isom(U)}$ contains a subgroup ${H}$ isomorphic to ${Homeo^*([0,1])}$. So we get a linear isometric representation of ${H}$ on a reflexive Banach space. According to Megerelishvili’s theorem, this must be the trivial representation. So ${H}$ acts faithfully by translations. Contradiction, since ${H}$ is non abelian.