## Notes of Lior Silberman’s lecture nr 2

Last time, we talked about fixed point properties, how they interact with algebra (${F\mathcal{A}\Rightarrow}$ finitely generated), Kazhdan’s property (T) and its spectral characterization. Today, we shall connect property (T) with fixed point properties. For that, we need a technical tool: ultrafilter arguments.

1. Ultrafilter arguments

1.1. Ultrafilters

Let ${a_n}$ be a sequence of real numbers. Imagine every ${n\in{\mathbb N}}$ is invited to vote yes or no to the question “is ${a_n}$ ${\epsilon}$-close to ${A}$ ?”. If all but finitely many votes are yes, then ${a_n}$ tends to ${A}$. Thus the obvious “voting scheme”

$\displaystyle \begin{array}{rcl} \mathcal{F}_{cf}=\{I\subset {\mathbb N}\,;\,{\mathbb N}\setminus I \textrm{ is finite}\} \end{array}$

plays a key role in the notion of convergence. We intend to define convergence according to more general voting schemes. The requirements are

• ${\mathcal{F}}$ is non empty.
• ${\emptyset\notin\mathcal{F}}$.
• If ${I\subset\mathcal{F}}$, and ${I\subset J}$, then ${J\in\mathcal{F}}$.
• If ${I}$, ${J\in\mathcal{F}}$, then ${I\cap J\in\mathcal{F}}$.

The last axiom is there to guarantee uniqueness of limit. These axiom define filters.

There is a difficulty with of obvious filter. For ${a_n =(-1)^n}$, half of the votes are yes, half are no, and no decision can be taken. So one should make such situations impossible.

Definition 1 Say a filter ${\mathcal{F}}$ is an ultrafilter if for all ${I\subset {\mathbb N}}$, exactly one of ${I}$ and ${{\mathbb N}\setminus I}$ belongs to ${\mathcal{F}}$.

Lemma 2 For every filter on ${{\mathbb N}}$, there exist ultrafilters than contain it.

Proof: Zorn’s lemma. Indeed, ultrafilters are maximal filters. $\Box$

Also, one should avoid “dictatorships”, i.e. ultrafilters that contain a singleton. These are called principal ultrafilters.

Theorem 3 (Bolzano-Weierstrass) If ${X}$ is a compact metric space and ${\omega}$ a non-principal ultrafilter, then for any sequence ${(a_n)\subset X}$, ${\lim_{\omega}a_n}$ exists.

Indeed, what the ultrafilter does is extract converging subsequences. It is particularly convenient when arranging sequences ${(a_n ,b_n)}$ in a product space to converge. This would required extracting a converging subsequence in ${a_n}$, and then re-extracting for ${b_n}$ to converge. The ultrafilter makes the choice once and for all.

Here is a limitation in the use of ultrafilters: One is not allowed to permute elements in a sequence. This will be harmless in our use of ultrafilters.

Usual rules of logic hold and constructions go through. For instance, ultralimits of groups are groups.

1.2. Ultralimits of pointwise metric spaces

A pointed metric space is a metric space with a base point. Given a sequence ${(X_n ,p_n)}$ of such, define

$\displaystyle \begin{array}{rcl} \hat{X}=\{(x_n)_{n\in{\mathbb N}}\,;\forall n,\,\,x_n \in X_n \textrm{ and }\sup_{n}d(x_n , p_n) <\infty\}. \end{array}$

If ${\hat{x}=(x_n)_{n\in{\mathbb N}}}$ and ${\hat{y}=(y_n)_{n\in{\mathbb N}}\in\hat{X}}$, define

$\displaystyle \begin{array}{rcl} d_{\omega}(\hat{x},\hat{y})=\lim_{\omega}d_{X_n}(x_n ,y_n), \end{array}$

which exists, by Bolzano-Weierstrass.

Definition 4 Let ${\omega}$ be a non principal ultrafilter on ${{\mathbb N}}$. The ${\omega}$-ultralimit of ${(X_n ,p_n)}$ is

$\displaystyle \begin{array}{rcl} X_{\omega}:=(\hat{X},d_{\omega})/\sim \end{array}$

where two sequences are identified iff their ${d_{\omega}}$ vanishes.

Theorem 5 Any statement about finite sets of points in metric spaces including universal quantifiers and bounded existential quantification (e.g. ${\exists x}$ such that ${d(x,p)\leq C}$) holds in the ultralimit ${\lim_{\omega}(X_n ,p_n)}$ if it holds for ${\omega}$-almost every ${(X_n ,p_n)}$.

Example 1 Existence of midpoints passes to ultralimits.

Indeed, the midpoint stays a bounded distance away to the base point. To conclude from this that the ultralimit of geodesic metric spaces is geodesic requires some extra convexity assumption.

Example 2 ${CAT(0)}$ passes to ultralimits.

Indeed, ${CAT(0)}$ amounts to the following inequality:

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

It is an equality for Hilbert space, and characterizes Hilbert space.

Example 3 An ultralimit of Hilbert spaces is a Hilbert space.

Example 4 An ultralimit of geodesic ${CAT(0)}$ spaces is geodesic and ${CAT(0)}$.

Example 5 Every sequence of isometries ${\rho_n :X_n \rightarrow X_n}$ such that ${d_{X_n}(p_n ,\rho_n (p_n))}$ stays bounded defines an isometry ${\rho}$ of the ultralimit. Every group homomorphism ${\Gamma\rightarrow Isom(X_n)}$ defines a group homomorphism ${\Gamma\rightarrow Isom(\lim_{\omega}X_n)}$

In fact, it is sufficient to assume that ${\rho_n}$ is close to being an isometry (e.g. ${\|\rho_n \|_{Lip}}$ tends to ${1}$). Also, it is sufficient to have a sequence of near homomorphisms (${\rho_n (\gamma\gamma')}$ close to ${\rho_n (\gamma)\rho_n (\gamma')}$).

1.3. Property ${F\mathcal{H}}$

Proposition 6 Let ${\Gamma}$ be a group finitely generated by ${S}$. The following are equivalent.

1. ${\Gamma}$ has property ${F\mathcal{H}}$.
2. There exists ${r<1}$ such that for every isometric action of ${\Gamma}$ on a Hilbert space ${Y}$ and every ${y\in Y}$,$\displaystyle \begin{array}{rcl} E_S (\frac{I+A_S}{2}y)\leq r\,E_S (y). \end{array}$

Proof: 2. ${\Rightarrow}$ 1. Pick arbitrary ${y_0 \in Y}$, define ${y_{n+1}=\frac{I+A_S}{2}y_n}$. By assumption, ${E_S (y_n)\leq r^n E_S (y_0)}$. One checks that for all ${y}$,

$\displaystyle \begin{array}{rcl} d(\frac{I+A_S}{2}y,y)^2 \leq 2 E_S (y). \end{array}$

This shows that ${d(y_{n+1},y_n)\leq 2\,r^n}$, ${y_n}$ is a Cauchy sequence, the limit is a fixed point.

1. ${\Rightarrow}$ 2. By contradiction. Assume that there exist pointed Hilbert spaces ${(Y_n ,p_n)}$ with ${\Gamma}$-actions such that ${E_S (p_n)>0}$ and

$\displaystyle \begin{array}{rcl} E_S (\frac{I+A_S}{2}(p_n))\geq (1-\frac{1}{n})E_S (p_n). \end{array}$

Rescale the metric so that ${E_S (p_n)=1}$. Form the ultralimit ${Y_{\infty}}$, based at ${p}$, of ${Y_n}$ based at ${p_n}$.

Since ${E_S (p_n)=1}$, each generator translates ${p_n}$ a bounded amount, so we get a limiting action of ${\Gamma}$ on ${Y_{\infty}}$ such that ${E_S (p)=1}$. Also

$\displaystyle \begin{array}{rcl} E_S (\frac{I+A_S}{2}(p))\geq E_S (p). \end{array}$

so

$\displaystyle \begin{array}{rcl} E_S (A_S (p))= E_S (p)=1. \end{array}$

Assume that ${\Gamma}$ has a fixed point on ${Y_{\infty}}$, which take to be the origin. Then ${y}$ and all the ${sy}$ sit on the same sphere. Recall that ${A_S (y)}$ is the point ${z}$ that minimizes

$\displaystyle \begin{array}{rcl} \frac{1}{|S|}\sum_{s\in S}|z-sy|^2 . \end{array}$

By strict convexity, unless all ${sp}$‘s are equal, ${E_S (A_S (p)), contradiction. So all ${sp=tp}$, ${s}$, ${t\in S}$. So ${p}$ is a fixed point for ${\Gamma^2}$, which has index at most ${2}$. The midpoint of ${p}$ and ${sp}$ is a fixed point. $\Box$

1.4. Property ${F\mathcal{H}}$ implies property (T)

Proposition 7 (Guichardet) Property ${F\mathcal{H}}$ implies property (T)

Proof: Let ${\Gamma}$ be a group with finite generating set ${S}$. Let ${\pi:\Gamma\rightarrow U(Y)}$ be a unitary representation which almost has invariant unit vectors, but no invariant vectors.

By assumption, for all ${\epsilon>0}$, there exists a unit ${y\in Y}$ such that ${E_S (y)<\epsilon}$. In fact, for every ${R>0}$, ${\epsilon>0}$, ${\delta>0}$, there is a unit ${y\in Y}$ such that every ${y'\in B(y,R\sqrt{E_S (y)})}$ has

$\displaystyle \begin{array}{rcl} E_S (y')\geq (1-\delta)E_S (y). \end{array}$

Indeed, otherwise, for all unit ${y}$ with ${E_S (y)\leq\epsilon}$, there ecists ${y'}$ such that

$\displaystyle \begin{array}{rcl} d(y,y')\leq R\sqrt{E_S (y)} \quad \textrm{ and }E_S (y')\leq (1-\delta)E_S (y). \end{array}$

Iterate. This produces a sequence ${y_n}$, ${R_n \rightarrow \infty}$, ${\delta_n \rightarrow 1}$, such that ${E_S (y_n)\rightarrow 0}$. Rescale so that ${E(y_n)=1}$. Take an ultralimit. There, energy is ${\geq 1}$ everywhere, so no fixed points. This contradicts property ${F\mathcal{H}}$. $\Box$

Remark 1 Although we started with a linear action, that had a fixed point, this has disappeared from the resulting isometric action on the ultralimit.

Indeed, ${|y_n|=1}$ makes that in the rescaled space, ${d(0,y_n)}$ tends to infinity.

This is not Guichardet’s initial proof. I took this proof from Gromov’s random walk in random groups paper. We could probably have used Proposition 6 instead (relating energy fro affine actions to energy for unitary representations), but I prefered to show an ultrafilter argument. Also, this argument has a wider scope, it does not require an averaging procedure.

2. Averaging in ${CAT(0)}$ spaces

Now we pass to the nonlinear setting.

2.1. ${CAT(0)}$ spaces

In a geodesic space, given points ${y}$, ${z}$, let ${[y,z]}$ denote a geodesic segment, and ${[y,z]_t}$ the point on that geodesic at relative distance ${t}$ from ${y}$.

Definition 8 A geodesic metric space ${Y}$ is ${CAT(0)}$ is for all ${x}$, ${y}$, ${z\in Y}$ and ${t\in[0,1]}$,

$\displaystyle \begin{array}{rcl} d(x,[y,z]_t)^2 \leq (1-t)d(x,y)^2 +td(x,z)^2 -t(1-t)d(y,z)^2 . \end{array}$

This is a uniform convexity statement about the function ${d(x,\cdot)^2}$. This makes midpoints unique, geodesic segments between points unique.

If ${C\subset Y}$ is a closed convex set, the function ${x\mapsto d(x,C)=\inf_{y\in C}d(x,y)}$ achieves a unique minimum. Indeed,

$\displaystyle \begin{array}{rcl} C_{\epsilon}=\{Y\in C\,;\,d(x,y)^2 \leq d(x,C)^2 +\epsilon\} \end{array}$

has diameter ${\leq O(\sqrt{\epsilon})}$, so ${\bigcap_{\epsilon>0}C_{\epsilon}}$ is non empty and a single point, denoted by ${\pi(x)}$. ${\pi :Y\rightarrow C}$ is ${1}$-Lipschitz. Furthermore, any geodesic in ${C}$ starting at ${\pi(x)}$ makes there an obtuse angle with ${[x\pi(x)]}$ (angles make sense in ${CAT(0)}$ spaces).

2.2. Averaging

Definition 9 Let ${\sigma}$ be a probability measure on ${Y}$. Define its center of mass ${C(\sigma)}$ as the point that minimizes

$\displaystyle \begin{array}{rcl} x\mapsto \int_{Y}d(x,y)^2 \,d\sigma(y). \end{array}$

Existence and uniqueness of ${C(\sigma)}$ follows from the same proof as for ${\pi}$.

Theorem 10 (Sturm) (Jensen’s inequality). Let ${f:Y\rightarrow{\mathbb R}}$ be a convex function. Then ${f(C(\sigma))\leq \int f\,d\sigma}$.

Corollary 11 Let ${\Gamma}$ be a group with finite generating set ${S}$, acting isometrically on a ${CAT(0)}$ space ${Y}$. Set

$\displaystyle \begin{array}{rcl} E_S (y)=\frac{1}{2|S|}\sum_{s\in S}d(sy,y)^2 , \end{array}$

and

$\displaystyle \begin{array}{rcl} A_S (y)=C(\frac{1}{|S|}\sum_{s\in S}\delta_{sy}). \end{array}$

Then

$\displaystyle \begin{array}{rcl} E_S (A_S (y))\leq E_S (y). \end{array}$

2.3. Induction

Proposition 12 Let ${\Gamma\subset G}$ be groups such that ${\Gamma\setminus G}$ is compact and has a finite ${G}$-invariant measure. Then ${G}$ has property ${F\mathcal{H}}$ iff ${\Gamma}$ does.

Proof: Assume that ${\Gamma}$ has property ${F\mathcal{H}}$. Let ${G}$ act on some Hilbert space ${Y}$. Let ${y_0}$ be a point of ${Y}$ fixed by ${\Gamma}$. Use orbit map ${\Gamma\setminus G\rightarrow Y}$ to push forward ${G}$-invariant measure to a measure ${\sigma}$ on ${Y}$. Then the center of mass of ${\sigma}$ is ${G}$-invariant.

Conversely, assume that ${G}$ has property ${F\mathcal{H}}$. Let ${\Gamma}$ act on some Hilbert space ${Y}$. Consider the space

$\displaystyle \begin{array}{rcl} B=\{f:G\rightarrow Y\,;\, f(\gamma g)=\gamma f(g)\}. \end{array}$

It is nonempty (requires partition of unity if ${G}$ is nondiscrete). It has a natural metric. Indeed, if ${f_1}$, ${f_2 \in B}$, the map ${g\mapsto d_Y (f_1 (g),f_2 (g))^2}$ is ${\Gamma}$-invariant, so one can integrate it over the quotient space ${\Gamma\setminus G}$. The resulting metric space is an affine Hilbert space (a direct integral of Hilbert spaces), related to the induced representation. ${G}$ acts on it on the right, by isometries. Let ${f\in B}$ be ${G}$-invariant. This means that ${f}$ is constant, and its value is a fixed point of ${\Gamma}$ on ${Y}$. $\Box$

2.4. From groups actions to random walks

Let ${\Gamma}$ act on ${X}$ with ${\Gamma\setminus X}$ compact. As we just saw, ne can replace the study of points of ${Y}$ under ${\Gamma}$ by the study of ${\Gamma}$-equivariant maps ${f\in B(X,Y)}$. Fixed points correspond to constant functions. Let ${\nu}$ be probability measure on ${\Gamma\setminus X}$. A random walk on ${X}$ is a map ${\mu:X\rightarrow P(X)}$ to the space of compactly supported probability measures on ${X}$. We furthermore assume that

• ${\mu}$ is ${\Gamma}$-equivariant.
• Each ${\mu_x}$ is reversible with respect to ${\nu}$, i.e. for every ${\Gamma}$-invariant function ${f:X\times X\rightarrow {\mathbb R}}$,$\displaystyle \begin{array}{rcl} \int_{\Gamma\setminus X}d\nu(x)\int d\mu(x\rightarrow y)f(x,y)=\int_{\Gamma\setminus X}d\nu(y)\int d\mu(y\rightarrow x)f(x,y). \end{array}$

Here, we use Gromov’s suggestive notation ${d\mu(x\rightarrow y)=d\mu_x (y)}$.

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

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

and

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

and we would like to prove energy decreasing property: ${\exists r<1}$ such that ${\forall f}$,

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