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 1 Let 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 2 Let 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 3 Suppose 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 4 Let be a finite 3-generated group. If , , therefore there is an epimorphism .
Corollary 5 There 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 7 Let be an expander, let , . Then, for all 1-Lipschitz maps of to Hilbert space,
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.