## Notes of Lior Silberman’s lecture nr 3

In the first two lectures, I showed that property (T) was related to exponential convergence of averages in Hilbert space, a fact that can alternatively be formulated as a Poincaré inequality. And I explained how this extends to certain non linear metric spaces, like ${CAT(0)}$ spaces.

Today, I will describe random groups, a machine that produces groups which have fixed point properties.

1. From groups actions to random walks

1.1. Invariant random walks

Let a group ${\Gamma}$ act isometrically on metric spaces ${X}$ and ${Y}$. Assume that the action on ${X}$ is free and properly discontinuous.

Let ${\mu:x\mapsto \mu_x}$ be a ${\Gamma}$-invariant Markov chain on ${X}$, i.e. a ${\Gamma}$-equivariant map from ${X}$ to the set of probability measures on ${X}$. Assume ${\mu}$ is reversible with respect to ${\nu}$, i.e. for every ${\Gamma}$-invariant function ${f:X\times X\rightarrow {\mathbb R}}$,

$\displaystyle \begin{array}{rcl} \int_{\Gamma\setminus X}d\nu(x)\int d\mu(x\rightarrow y)f(x,y)=\int_{\Gamma\setminus X}d\nu(y)\int d\mu(y\rightarrow x)f(x,y). \end{array}$

Idea: Study ${B=B^{\Gamma}(X,Y)=\{f:X\rightarrow Y\,;\,\Gamma-\textrm{equivariant, finite energy}\}}$. Find a constant function in ${B}$.

Assume that ${\Gamma\setminus X}$ is a finite polyhedron (this guarantees that ${B}$ is non empty).

Definition 1 If ${f\in B(X,Y)}$ is ${\Gamma}$-equivariant, set

$\displaystyle \begin{array}{rcl} E_{\mu}(f)=\frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu(x\rightarrow y)d_Y (f(x),f(y))^2 , \end{array}$

The connection with our previous discussion of isometric group actions on ${Y}$ is provided by the following

Example 1 ${X=Cay(\Gamma,S)}$ is the Cayley graph, ${\mu}$ is the nearest neighbour random walk.

Then mapping ${f\mapsto f(e)}$ identifies ${B(X,Y)}$ with ${Y}$ and energy

$\displaystyle \begin{array}{rcl} E_{\mu}(f)=\frac{1}{2|S|}\int_{s\in S}d_Y (y,sy)^2 =E_S (y),\quad y=f(e). \end{array}$

1.2. Assumptions on ${Y}$

Definition 2 Say a metric space ${Y}$ is ${CAT(0)}$ if it is geodesic and the following inequality holds for all ${x}$, ${y}$, ${z\in Y}$.

$\displaystyle \begin{array}{rcl} d(x,m(y,z))^2 \leq \frac{1}{2}d(x,y)^2 +\frac{1}{2}d(x,z)^2 -\frac{1}{4}d(y,z)^2 . \end{array}$

1.3. Center of mass

In a ${CAT(0)}$ space, given a probability measure ${\sigma}$, there is a unique point ${c(\sigma)}$, called the center of mass of ${\sigma}$, where the function

$\displaystyle \begin{array}{rcl} y\mapsto d^{2}(y,\sigma):=\int_{Y}d(y,y')^{2}\,d\sigma(y')=\mathop{\mathbb E}_{\sigma}(d_y^2). \end{array}$

attains its minimum.

Proof: Let ${m=\inf d^2 (\cdot,\sigma)}$. Let ${C_{\epsilon}=\{y\,;\,d^{2}(y,\sigma)\leq m+\epsilon\}}$. Let ${y_0}$, ${y_1 \in C_{\epsilon}}$. Integrating the ${CAT(0)}$ inequality yields

$\displaystyle \begin{array}{rcl} m\leq d([y_0 ,y_1]_t ,\sigma)\leq (1-t)d^2 (y_0 ,\sigma) +t d^2 (y_1 ,\sigma)-t(1-t)d(y_0 ,y_1)^2 . \end{array}$

Take ${t=\frac{1}{\epsilon}}$, find

$\displaystyle \begin{array}{rcl} m\leq m+\epsilon -\frac{1}{4}d(y_0 ,y_1)^2 \end{array}$

Which implies that ${diameter(C_{\epsilon})\leq 2\sqrt{\epsilon}}$. By completeness, the intersection of ${C_{\epsilon}}$‘s is non empty, whence the minimum is attained. $\Box$

The proof shows that

$\displaystyle \begin{array}{rcl} d^2 (y,\sigma)\geq d(y,c(\sigma))^2 +d^2 (c(\sigma),\sigma). \end{array}$

1.4. Averaging

Back to our setting: ${\Gamma}$ acts on ${X}$ and ${Y}$, ${f:X\rightarrow Y}$ ${\Gamma}$-equivariant. When ${Y}$ is ${CAT(0)}$, one can define

Definition 3

$\displaystyle \begin{array}{rcl} (A_{\mu}f)=c(f_{*}\mu_x), \end{array}$

We need a lazy variant of this.

Definition 4 Given ${x\in X}$, define ${g_x : X\rightarrow Y}$ by

$\displaystyle \begin{array}{rcl} g_x (x')=[f(x),f(x')]_{1/2}. \end{array}$

Then set

$\displaystyle \begin{array}{rcl} (\tilde{A}_{\mu}f)=c((g_{x})_{*}\mu_x), \end{array}$

Example 2 ${X=Cay(\Gamma,S)}$. Let ${\tilde{X}}$ be the barycentric subdivision of ${X}$. On ${\tilde{X}}$, alternate two operations,

1. Set ${f(\textrm{midpoint})=}$ midpoint of ends of edge.
2. Do nearest neighbour averaging on ${Y}$.

Operation (1) is an embedding ${B(X,Y)\rightarrow B(\tilde{X},Y)}$. Thanks to reversibility, it reduces energy on

Remark 1 Again, by an ultrafilter argument, fixed point property in ${Y}$ is equivalent to decreasing energy property for this averaging procedure.

In our non linear setting, this is not quite the same as ${A_{\mu\star\mu}}$.

1.5. Powers

Note that, due to non linearity, ${A_{\mu\star\mu}\not=A_{\mu}\circ A_{\mu}}$. Nevertheless, it turns out that the adequate substitute for powers of ${A_{\mu}}$ is ${A_{\mu^{\ell}}}$.

For every ${x\in X}$, ${y\in Y}$,

$\displaystyle \begin{array}{rcl} d((A_{\mu^{\ell}}f)(x),y)^2 \leq d^2 (y,f_{*}\mu_x)-d^2 ((A_{\mu^{\ell}}f)(x),f_{*}\mu_x)^2 , \end{array}$

so

$\displaystyle \begin{array}{rcl} E_{\mu^{\ell}}(A_{\mu^{\ell}}f) &\leq& \frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu(x\rightarrow x')\left(d^2 (y,f_{*}\mu_x)-d^2 ((A_{\mu^{\ell}}f)(x),f_{*}\mu_x)^2 \right)\\ &=& \frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu(x\rightarrow x') \int_{X}d\mu^{\ell}(x\rightarrow x'')\,d((A_{\mu^{\ell}}f)(x'),f(x''))^2 \\ &&-\frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu(x\rightarrow x') \int_{X}d\mu^{\ell}(x\rightarrow x'')\,d((A_{\mu^{\ell}}f)(x),f(x''))^2 \\ &=& \frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu^{\ell+1}(x\rightarrow x')\,d((A_{\mu^{\ell}}f)(x),f(x'))^2\\ &&-\frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}d\mu^{\ell}(x\rightarrow x')\,d((A_{\mu^{\ell}}f)(x),f(x'))^2 . \end{array}$

by reversibility. We are facing a difficulty due to parity. We solve it here by using ${E_{\mu^2}}$.

$\displaystyle \begin{array}{rcl} E_{\mu^2}(A_{\mu^{2\ell}}f)\leq\frac{1}{2}\int_{\Gamma\setminus X}d\nu(x)\int_{X}\left(d\mu^{2\ell+2}(x\rightarrow x')-d\mu^{2\ell}(x\rightarrow x')\right)\,d((A_{\mu^{\ell}}f)(x),f(x'))^2 \end{array}$

From now on, assume that ${X}$ is a tree, i.e. the Cayley graph of the free group ${F_S}$ (this is a covering space of ${Cay(\Gamma,S)}$). A direct calculation shows that, on a tree,

$\displaystyle \begin{array}{rcl} d\mu^{2\ell+2}(x\rightarrow x')-d\mu^{2\ell}(x\rightarrow x') \leq g(\ell)d\mu^{2\ell}(x\rightarrow x') +h(\ell), \end{array}$

where ${g}$, ${h}$ are numerical functions,

1. ${g(t)\sim \sqrt{\frac{\log\log t}{\log t}}}$,
2. ${h}$ is exponentially small.

On a tree,

$\displaystyle \begin{array}{rcl} E_{\mu^2}(f)\sim\max_{\gamma\in S^2} d(f(x),f(x\gamma))^2 . \end{array}$

By triangle inequality,

$\displaystyle \begin{array}{rcl} d((A_{\mu^{\ell}}f)(x),f(x'))^2 \leq \ell^{C} E_{\mu^2}(f). \end{array}$

This yields

$\displaystyle \begin{array}{rcl} E_{\mu^2}(A_{\mu^{2\ell}}f)\leq g(\ell)\int_{\Gamma\setminus X\times X}d\nu(x)d\mu^{2\ell}(x\rightarrow x')d((A_{\mu^{\ell}}f)(x),f(x'))^2 +h(\ell) E_{\mu^2}(f). \end{array}$

By definition of average,

$\displaystyle \begin{array}{rcl} \frac{1}{2}\int_{\Gamma\setminus X\times X}d\nu(x)d\mu^{\ell}(x\rightarrow x')d((A_{\mu^{\ell}}f)(x),f(x'))^2 &\leq& \frac{1}{2}\int_{\Gamma\setminus X\times X}d\nu(x)d\mu^{\ell}(x\rightarrow x')d(f(x),f(x'))^2 \\ &=& E_{\mu^{\ell}}(f). \end{array}$

So

$\displaystyle \begin{array}{rcl} E_{\mu^2}(A_{\mu^{2\ell}}f)\leq g(\ell)E_{\mu^{2\ell}}(f)+h(\ell)E_{\mu^{\ell}}(f). \end{array}$

Assume that we also have a Poincaré inequality

$\displaystyle \begin{array}{rcl} E_{\mu^{2\ell}}(f)\leq C\,E_{\mu^2}(f), \end{array}$

with ${C}$ independant of ${\ell}$. Then, for ${\ell}$ large enough,

$\displaystyle \begin{array}{rcl} E_{\mu^2}(A_{\mu^{2\ell}}f)\leq r E_{\mu^{2}}(f), \end{array}$

with ${r<1}$. Therefore ${A_{\mu^{2\ell}}f}$ converges to a constant map as ${\ell}$ tends to infinity, providing a fixed point.

Conclusion: To show that ${\Gamma\in F\mathcal{C}}$ where ${\mathcal{C}}$ is a family of ${CAT(0)}$ spaces, it is enough to establish the follwing Poincaré inequality. There exist ${r<1}$, ${C>0}$ such that for all ${Y\in\mathcal{C}}$, for all ${f\in B^{\Gamma}(X,Y)}$, there exists ${\ell}$ such that

$\displaystyle \begin{array}{rcl} E_{\mu^{2\ell}}(f)\leq C\,E_{\mu^2}(f) \quad \textrm{and}\quad Cg(\ell)+h(\ell)\leq r. \end{array}$

2. Random groups in the graph model

Due to Gromov, 2003.

2.1. Idea

Let ${G=(V,E)}$ be a graph, let ${\alpha:\vec{E}\rightarrow S}$ be a symmetric labelling (i.e. switching an oriented edge gives inverse label). ${\alpha}$ extends to a multiplicative map on paths, with values in the free group ${F_S}$. Words read along cycles of ${G}$ form a normal subgroup ${R_{\alpha}}$ of ${F_S}$. We are interested in the group ${\Gamma_{\alpha}=\langle S\,|\,R_{\alpha} \rangle}$ and its Cayley graph ${X_{\alpha}}$.

There is a projection from ${X_{\alpha}}$ to ${G}$. Take base points ${v_0 \in V}$, ${x_0 \in X_{\alpha}}$. Set

$\displaystyle \begin{array}{rcl} \pi_{v_0 \rightarrow x_0}(v)=x_0 \alpha(\vec{p}), \end{array}$

where ${\vec{p}}$ is a path from ${v_0}$ in ${v}$. Using this projection, one can try to simulate random walk on ${X_{\alpha}}$ by random walk on ${G}$.

2.2. Basic results

Theorem 5 (Gromov 2003) Assume that the girth ${g(G)}$ is large (say ${\geq 4\ell}$). Choose ${\alpha}$ uniformly at random. With high probability, one can effectively simulate ${\ell}$ steps of random walk on ${X_{\alpha}}$ by random walk on ${G}$.

Theorem 6 (Gromov 2003, Ollivier, Arzhantseva, Delzant) Assume that the girth ${g(G)}$ is large. Choose ${\alpha}$ uniformly at random. With high probability, ${\Gamma_{\alpha}}$ is infinite and Gromov-hyperbolic.

These results reduce the problem of finding groups with ${F\mathcal{C}}$ to finding graphs with large girth having Poincaré inequality for ${Y}$-valued maps. Specifically,

Problem: We want a constant ${C}$ and a family of graphs with increasing girth such that for all graphs in the family, for all ${Y\in\mathcal{C}}$, for all ${f:V\rightarrow Y}$,

$\displaystyle \begin{array}{rcl} E_{\nu_{G}\star \nu_{G}}(f)\leq C\,E_{\mu_G}(f), \end{array}$

where ${\mu_G}$ is random walk on ${G}$ and ${\nu_G}$ a probability measure on ${V}$.

Remark 2 Restriction to maps with values in a geodesic shows that Poincaré inequality must hold for ${{\mathbb R}}$-valued functions, i.e. the family must be a family of expanders.

The resulting groups ${\Gamma_{\alpha}}$ will automatically have property (T).

3. Non linear Poincaré inequalities

Remember that the constant in the scalar Poincaré inequality is the inverse of the spectral gap ${\frac{1}{1-\lambda(G)}}$. For non linear spaces, try to get a constant of the form ${C_Y \frac{1}{1-\lambda(G)}}$.

3.1. The tangent cone

We saw that

$\displaystyle \begin{array}{rcl} \int_{V\times V} \nu_G (x) \nu_G (x')\,d(f(x),f(x') \geq \int_{V}\nu_G (x) d(f(x),c(f_* \nu_G))^2 , \end{array}$

(equality in Hilbert spaces), and by triangle inequality, the reverse inequality holds up to a factor of ${2}$,

$\displaystyle \begin{array}{rcl} \int_{V\times V} \nu_G (x) \nu_G (x')\,d(f(x),f(x') \leq 2\int_{V}\nu_G (x) d(f(x),c(f_* \nu_G))^2 . \end{array}$

Definition 7 Let ${Y}$ be a ${CAT(0)}$ space. Let ${p\in Y}$. Define a new metric ${\bar{d}}$ on ${Y}$ by

$\displaystyle \begin{array}{rcl} \bar{d}(y,y')^2 =d(p,y)^2 +d(p,y')^2 -2d(p,y)d(p,y')\cos(\theta_p (y,y')), \end{array}$

where ${\theta_p (y,y')}$ is the angle at ${p}$ between the geodesics ${[p,y]}$ and ${[p,y']}$. Then ${(Y,\bar{d})}$ is again ${CAT(0)}$. Its completion is the tangent cone ${T_p Y}$ of ${Y}$ at ${p}$.

The ${CAT(0)}$ inequality implies that radial distances are preserved and ${\bar{d}\leq d}$, i.e. the identity map ${(Y,d)\rightarrow (Y,\bar{d})}$ is distance non increasing. Therefore, to get a Poincaré inequality for ${Y}$-valued maps, it is sufficient to prove one for ${T_p Y}$-valued maps (up to a factor of ${2}$). This observation is due to Mu-Tao Wang.

Corollary 8 (Wang 1998) If all tangent cones ${T_p Y}$ embed isometrically in Hilbert spaces, then Poincaré inequality holds with constant ${\frac{2}{1-\lambda(G)}}$.

Example 3 Lattices in higher rank semisimple Lie groups have property (T), so they have a spectral gap. Nevertheless, they act isometrically on a ${CAT(0)}$ space, their symmetric space, without fixed point.

It is the factor ${2}$ that makes the Poincaré inequality too weak to imply fixed point property. This loss has been studied in detail by Kondo, Nayatani, Ozeki. They could

3.2. Building-valued Poincaré inequalities

Theorem 9 (Lang, Schlichenmaier 2004) Buildings of linear groups over local fields have finite Assouad-Nagata dimension.

It follows (Assouad, Lee-Naor) that buildings ${(Y,d)}$ have snowflakes ${(Y,\sqrt{d})}$ which bilipschitz embed in Hilbert space. Now Hilbert space has a linear isometric embedding in ${L^4}$. So to prove a ${Y}$-valued Poincaré inequality, it suffices to prove a scalar ${\ell_4}$-Poincaré inequality, i.e.

$\displaystyle \begin{array}{rcl} \mathop{\mathbb E}_{\nu_G \times \nu_G}|f(x)-f(x')|^4 \leq C\,\mathop{\mathbb E}_{\mu_G}|f(x)-f(x')|^4 . \end{array}$

Proposition 10 (Matousek) If ${\ell_p}$-Poincaré inequality holds for a graph ${G}$, then ${\ell_q}$-Poincaré inequalities holds for ${G}$ up to a constant which depends only on ${p}$ and ${q}$.

This allows to upgrade a spectral gap into a building-valued Poincaré inequality.

Corollary 11 (Naor, Silberman 2010) For every ${n}$, there exist a graph ${G}$ such that, with high probability, random groups modelled on ${G}$ have fixed point preperty on all buildings of rank ${\leq n}$.

3.3. Banach space-valued Poincaré inequalities

Lafforgue, using representation theory of ${PSl(3,\mathbb{Q}_p)}$, and later ??, using zig-zag products, have found graphs which satisfy Poincaré inequalities with values in uniformly convex Banach spaces, but they do not have large girth.

3.4. An application

Let ${M}$ be a smooth compact manifold, equipped with a smooth volume form ${\mu}$. Consider the space ${Y'}$ of all smooth Riemannian metrics on ${M}$ with volume form pointwise equal to ${\mu}$. The ${L_2}$-metric on ${Y'}$ is defined as follows. The tangent space to ${Y'}$ at ${g}$ is the space of fields of trace free quadratic forms. Set

$\displaystyle \begin{array}{rcl} G_g (h)=\int_{M}\mathrm{Trace}((g(x)^{-1}h(x))^2)\,d\mu(x). \end{array}$

View this as a Riemannian metric on the infinite dimensional manifold ${Y'}$, and consider the corresponding distance. Every smooth diffeomorphism of ${M}$ acts isometrically on ${Y'}$.

Here is an alternative definition. Throw away a set of measure zero to make ${M}$ contractible, so that the tangent bundle becomes trivial. Note that ${\mathrm{Trace}((g(x)^{-1}h(x))^2)}$ is the canonical metric on the symmetric space ${S}$ of ${Sl(n,{\mathbb R})}$. So ${Y'}$ embeds isometrically into the space of Borel maps ${M\rightarrow S}$ equipped with the ${L_2}$-metric. Let ${Y}$ be the completion of ${Y'}$ with respect to the ${L_2}$-metric. Then ${Y=L_2 (M,S)}$ is a ${CAT(0)}$ space (an infinite product of ${CAT(0)}$ space).

Corollary 12 Let ${\mathcal{C}}$ be the class of ${CAT(0)}$ spaces with Hilbertian tangent cones. Let ${\Gamma}$ be a group having property ${F\mathcal{C}}$. If ${\Gamma}$ acts smoothly on a compact manifold ${M}$, preserving a smooth volume form. Then ${\Gamma}$ fixes an ${L_2}$ Riemannian metric on ${M}$.

Theorem 13 (Zimmer) If ${M}$ is a smooth compact manifold, and ${g}$ an ${L_2}$ Riemannian metric on ${M}$. Then ${Isom(M,g)}$ embeds in a compact group.

It is easy to make groups which do not embed into any compact group: add to the relations of ${\Gamma}$ an infinite sequence of long random words. So the resulting group (which is a limit of hyperbolic groups but is not hyperbolic itself) cannot act non trivially on any smooth manifold.

Theorem 14 (Fisher, Silberman 2010) There exist torsion free finitely generated groups ${\Gamma}$ such that every homorphism ${\Gamma:Diff_{\mu}(M)}$ is trivial.