## Notes of Romain Tessera’s Rennes lecture nr 2

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

1. A step towards a positive answer

Theorem 1 Let ${(G_n,S_n)}$ be a sequence of finite Cayley graphs, ${|S_n|=k}$. The following are equivalent.

1. ${Cay(G_n,S_n)}$ does not coarsely embed in Hilbert space.
2. There are probability measures ${\mu_n}$ on ${G_n\times G_n}$ such that
1. ${\mu_n\{(g,g')\,;\,d(g,g')\geq r_n\}}$ is bounded away from 0.
2. Poincaré inequality holds,

$\displaystyle \begin{array}{rcl} \sum_{x,\,y\in G_n}|f(x)-f(y)|^2\mu_n(x,y)\leq C\frac{1}{|G_n|}\sum_{x\sim y\in G_n}|f(x)-f(y)|^2. \end{array}$

The proof is by Hahn-Banach.

Question (James Lee). Can we take for ${\mu_n}$ the normalized counting measure on ${A_n\times A_n}$ for some subset ${A_n\subset G_n}$ ?

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

2. Relative property (T)

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

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

Proposition 3 Suppose that the pair ${(G,H)}$ has relative property (T). Let ${G_n}$ be a sequence of finite quotients of ${G}$. Then there is ${C>0}$ such that

$\displaystyle \begin{array}{rcl} \sum_{x,\,y\in G_n}|f(x)-f(y)|^2\mu_n(x,y)\leq C\frac{1}{|G_n|}\sum_{x\sim y\in G_n}|f(x)-f(y)|^2. \end{array}$

with ${\mu_n}$ the uniform measure on the set of pairs ${(x,xh)}$, ${h\in H}$.

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 ${G={\mathbb Z}^2 \times Sl(2,{\mathbb Z})}$, ${H={\mathbb Z}^2}$. This is known to have relative property (T) (Kazhdan). Take ${G_n=({\mathbb Z}/n{\mathbb Z})^2 \times Sl(2,{\mathbb Z}/n{\mathbb Z})}$ ? Unfortunately, ${Cay(Sl(2,{\mathbb Z}/n{\mathbb Z}),S)}$ is an expander.

3.2. Successful attempt

Replace ${Sl(2,{\mathbb Z}/n{\mathbb Z})}$ with a sequence ${T_n}$ of finite groups mapping onto them, but which coarsely embed into Hilbert space.

Let ${n=2^k}$. Let ${J_k\subset Sl(2,{\mathbb Z}/n{\mathbb Z})}$ the index 3 subgroup generated by matrices

$\displaystyle \begin{array}{rcl} \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix},\quad \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix},\quad \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. \end{array}$

It has order ${2^{3k-1}}$.

For an arbitrary group ${G}$, consider the following characteristic subgroups ${\Gamma_0(G)=G}$, ${\Gamma_{k+1}(G)=}$ subgroup of ${\Gamma_k(G)}$ generated by squares.

We define ${H_n=F_3/\Gamma_n(F_3)}$. One easily checks that ${H_n}$ is a finite 2-group.

Lemma 4 Let ${G}$ be a finite 3-generated group. If ${|G|=2^n}$, ${\Gamma_n(G)=1}$, therefore there is an epimorphism ${H_n\rightarrow G}$.

Corollary 5 There is an epimorphism ${H_{3k-1}\rightarrow J_k}$, mapping the standard generators to the 3 matrices above.

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

The interest of these examples is that the embedding does not arise from Yu’s property (A). Locally, ${H_n}$ looks more and more like a tree. The embedding arises from a wall-space structure. The first one ${H_1=({\mathbb Z}/2{\mathbb Z})^3}$ is a cube, and this continues.

3.3. Absence of weakly embedded expanders

Remember that ${G_n}$ the semi-direct product ${({\mathbb Z}/2^k{\mathbb Z})\times H_{3k-1}}$, where each factor coarsely embeds in Hilbert space. This suffices to prove that ${G_n}$ does not weakly contain an expander. Indeed, if there was one, compose with projection to ${H_{3k-1}}$. A subset ${A_n}$ representing a positive proportion of points would be mapped to a point in ${H_{3k-1}}$. Thus this subset is mapped to ${{\mathbb Z}/2^k{\mathbb Z}}$, and therefore to Hilbert space again.

Lemma 7 Let ${X_n}$ be an expander, let ${A_n\subset X_n}$, ${|A_n|\geq c|X_n|}$. Then, for all 1-Lipschitz maps of ${A_n}$ to Hilbert space,

$\displaystyle \begin{array}{rcl} \frac{1}{|A_n|^2}\sum_{x,\,y\in A_n\times A_n}|f(x)-f(y)|^2 \end{array}$

stays bounded.

Indeed, use the Poincaré inequality in ${X_n}$. Given a function ${f:A_n\rightarrow\mathcal{H}}$, define ${\tilde{f}:X_n\rightarrow\mathcal{H}}$ by ${\tilde{f}=f}$ on ${A_n}$ and ${\tilde{f}(x)=f(z_x)}$ where ${z_x}$ is the point of ${A_n}$ closest to ${x}$. Then

$\displaystyle \begin{array}{rcl} \frac{1}{|A_n|^2}\sum_{x,\,y\in A_n\times A_n}|\tilde{f}(x)-\tilde{f}(y)|^2 &\leq&\frac{1}{|X_n|^2}\sum_{x,\,y\in X_n\times X_n}|\tilde{f}(x)-\tilde{f}(y)|^2\\ &\leq&C\frac{1}{|X_n|}\sum_{x\sim y\in X_n}|\tilde{f}(x)-\tilde{f}(y)|^2\\ &\leq&C\frac{1}{|X_n|}(\sum_{x\sim y\in A_n}|\tilde{f}(x)-\tilde{f}(y)|^2\\ &&+\sum_{x\sim y\in B_1}|\tilde{f}(x)-\tilde{f}(y)|^2+\sum_{x\sim y\in B_2}|\tilde{f}(x)-\tilde{f}(y)|^2\cdots), \end{array}$

where ${B_i}$ is the set of points at distance ${i}$ from ${A_n}$. Using the isoperimetric inequality in ${X_n}$, one shows that ${|B_i|}$ decays exponentially. This leads to the required upper bound.