Our main concern: let be a bounded degree graph which does not coarsely embed into Hilbert space. Does this imply that weakly contains an expander ? I will show that the answer is no.

**1. A step towards a positive answer **

Theorem 1Let be a sequence of finite Cayley graphs, . The following are equivalent.

- does not coarsely embed in Hilbert space.
- There are probability measures on such that

- is bounded away from 0.
- Poincaré inequality holds,

The proof is by Hahn-Banach.

**Question** (James Lee). Can we take for the normalized counting measure on for some subset ?

This would put us very close to getting an expander in . I will show that the answer is again no. The counterexample relies on relative property (T).

**2. Relative property (T) **

Definition 2Let be a finitely generated group. Let be an infinite subgroup of . We say that the pair has relative property (T) if, for every affine isometric action of on a Hilbert space, the orbits of are bounded.

There is again a formulation in terms of 1-cocycles: such cocycles should be bounded along .

Proposition 3Suppose that the pair has relative property (T). Let be a sequence of finite quotients of . Then there is such that

with the uniform measure on the set of pairs , .

Clearly, this implies non embeddability into Hilbert space, but for a substantially different reason than expansion.

**3. A counterexample **

From now on, joint work with Goulnara Arzhantseva.

** 3.1. First attempt **

Use , . This is known to have relative property (T) (Kazhdan). Take ? Unfortunately, is an expander.

** 3.2. Successful attempt **

Replace with a sequence of finite groups mapping onto them, but which coarsely embed into Hilbert space.

Let . Let the index 3 subgroup generated by matrices

It has order .

For an arbitrary group , consider the following characteristic subgroups , subgroup of generated by squares.

We define . One easily checks that is a finite 2-group.

Lemma 4Let be a finite 3-generated group. If , , therefore there is an epimorphism .

Corollary 5There is an epimorphism , mapping the standard generators to the 3 matrices above.

Theorem 6 (Arzhantseva-Guentner-Spakula 2012)The sequence embeds uniformly into Hilbert space.

The interest of these examples is that the embedding does not arise from Yu’s property (A). Locally, looks more and more like a tree. The embedding arises from a wall-space structure. The first one is a cube, and this continues.

** 3.3. Absence of weakly embedded expanders **

Remember that the semi-direct product , where each factor coarsely embeds in Hilbert space. This suffices to prove that does not weakly contain an expander. Indeed, if there was one, compose with projection to . A subset representing a positive proportion of points would be mapped to a point in . Thus this subset is mapped to , and therefore to Hilbert space again.

Lemma 7Let be an expander, let , . Then, for all 1-Lipschitz maps of to Hilbert space,

stays bounded.

Indeed, use the Poincaré inequality in . Given a function , define by on and where is the point of closest to . Then

where is the set of points at distance from . Using the isoperimetric inequality in , one shows that decays exponentially. This leads to the required upper bound.