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<p<\infty} 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.

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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 http://www.math.ens.fr/metric2011/
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