## Notes of Tsachik Gelander’s lecture

BS-topology and invariant random subgroups

Joint work with Abert, Bergeron, Biringer, Nikolov, Raimbault, Sanet.

1. Definition

1.1. Benjamini-Schramm topology

The space of all rooted metric spaces has a topology, the Gromov-Hausdorff topology. The space of probability measures on that space (“random metric spaces”) also has a topology. I. Benjamini and O. Schramm used it for graphs. Here is a similar idea, in the category of groups.

Definition 1 Let ${G}$ be a locally compact group. Consider the space ${Sub_G}$ of closed subgroups of ${G}$. It carries the Chabauty topology: ${H_n}$ converges to ${H}$ if

1. ${\forall x\in H}$, there exists ${x_n\in H_n}$ converging to ${H}$.
2. If ${x_{n_k}\in H_{n_k}}$ and ${x_{n_k}}$ converge to ${x}$, then ${x\in H}$.

Example 1 ${Sub_{\mathbb R}}$ is an interval. ${Sub_{{\mathbb R}^2}}$ is homeomorphic to the 4-sphere. ${Sub_{Sl(2,{\mathbb R})}}$ is complicated.

1.2. Invariant random subgroups

${G}$ acts on ${Sub_G}$ by conjugation.

Definition 2 An IRS is a ${G}$-invariant probability measure on ${Sub_G}$.

The set of IRS in ${w^*}$ topology is compact.

Example 2 Let ${\Sigma_g}$ be a surface of genus ${g}$, equipped with a hyperbolic metric. Choose randomly a point and unit tangent vector. This gives an embedding of ${\Gamma=\pi_1(\Sigma_g)}$ into ${PSL(2,{\mathbb R})}$. This gives an IRS.

Example 3 IRS which are Dirac measures correspond to normal subgroups.

So IRS are a generalization of normal subgroups.

Example 4 Let ${G}$ act on a probability space ${X}$, preservig measure. Stabilizers of points yield an IRS.

Every IRS arises in this way.

Example 5 Let ${\Gamma\subset G}$ be a lattice, i.e. a closed subgroup, and ${G}$ preserves a probability measure on ${G/\Gamma}$. Then this action is an IRS, supported on the conjugacy orbit of ${\Gamma}$.

So IRS are a generalization of lattices.

1.3. Connection with BS topology

Fix a left-invariant metric on ${G}$. The map ${Sub_G \rightarrow M}$, ${H\mapsto H\setminus G}$ induces a continuous map ${IRS(G)\rightarrow Prob(M)}$. In general, it need not be a homeomorphism onto its image, but is is so when ${G}$ is semi-simple.

Therefore, from now on, we shall assume that ${G}$ is a simple Lie group.

The Borel density theorem extends to IRS.

Proposition 3 ${G}$ is isolated in ${Sub_G}$. Every non-atomic IRS is supported on discrete and Zariski-dense subgroups of ${G}$.

Therefore, we concentrate on discrete IRS.

Let ${X=G/K}$ be the symmetric space of ${G}$.

Lemma 4 Let ${\mu_n}$, ${\mu_{\infty}}$ be discrete IRS. The following are equivalent.

1. ${\mu_n}$ converge to ${\mu_{\infty}}$.
2. ${\forall k}$, ${\epsilon}$ and every ${R}$-ball ${B}$ in a locally ${X}$-manifold, the ${\mu_n}$-probability that ${B}$ embeds ${\Gamma\setminus X}$ with distorsion ${\epsilon}$ converges to the ${\mu_{\infty}}$-probability of the same event.

Corollary 5 Suppose ${\Gamma_n \subset G}$ are lattices, corresponding to measures ${\mu_n}$. Then ${\mu_n}$ converge to the Dirac measure on the trivial subgroup ${\{1\}}$, if and only if the random metric spaces ${\Gamma_n \setminus X}$ converge to ${X}$. I.e., a random ${R}$-ball in ${M_n}$ is contractible with probability tending to ${1}$.

2. Higher rank phenomena

2.1. Key theorem

The following fails in real rank ${1}$.

Theorem 6 Let ${G}$ be a simple Lie group of real rank ${\geq 2}$. Then

1. Ergodic IRS of ${G}$ are ${\delta_{\{1\}}}$, ${\delta_G}$ and lattices.
2. The only accumulation point of the IRS associated to a lattice is ${\delta_{\{1\}}}$

1. follows from Stuck and Zimmer, building upon Nevo and Zimmer.

2. follows from Kazhdan’s property (T). I explain it now. Suppose ${\mu_n =\mu_{\Gamma_n}}$ converge to some measure ${\mu}$. Then ${\mu}$ is ergodic (this already follows from propoerty (T)).

Example 6 A sequence of Riemann surfaces with one geodesic pinched converges to a non ergodic IRS.

From claim 1, ${\mu}$ is either ${\delta_{\{1\}}}$ or ${\mu_\Gamma}$. Assume it is ${\mu_\Gamma}$. Then ${\Gamma\setminus X}$ contains large balls filling nearly all the volume. In ${\Gamma_n}$, there is a similar ball with small boundary. Propoerty (T) implies large Cheeger constant, so all the volume is in the ball, conradicting the fact that volume tends to infinity.

Corollary 7 Let ${X}$ be a higher rank irreducible symmetric space. Let ${M_n=\Gamma_n\setminus X}$ be ${X}$-orbifolds of finite volume tending to infinity. Then for all ${R>0}$, the probability that an ${R}$-ball in ${M_n}$ is contractible tends to 1.

Again, this fails in rank one (take abelian covering of a Riemann surface).

Theorem 8 Let ${X}$ be a higher rank irreducible symmetric space. Let ${M_n=\Gamma_n\setminus X}$ be ${X}$-orbifolds of finite volume tending to infinity. Then ${M_n}$ BS-converges to ${X}$.

3. Convergence of Betti numbers

3.1. Statement

Definition 9 A sequence of Riemannian manifolds is uniformly discrete if the minimal injectivity radius is bounded below.

Conjecture (Margulis). The family of all compact arithmetic ${X}$-manifolds is uniformly discrete.

This implies Lehmer’s conjecture on Mahler measures of polynomial.

Theorem 10 Let ${M_n}$ be a sequence of uniformly discrete ${X}$-manifolds which converges to ${X}$. Then normalized Betti numbers converge, i.e.

$\displaystyle \begin{array}{rcl} \lim_{n\rightarrow\infty}\frac{b_k(M_n)}{vol(M_n)}=\beta_k(X). \end{array}$

Theorem 11 Assume ${X}$ has rank one. Then convergence of Betti numbers holds without the uniform discreteness assumption.

In other words, in higher rank, ${M_n}$ converge to ${X}$ in a very strong sense.

3.2. Proof

${\beta_k}$ is the ${L^2}$ Betti number of ${X}$,

$\displaystyle \begin{array}{rcl} \beta_k=\lim_{t\rightarrow\infty}\mathrm{Trace}(e^{-t\Delta_k^{(2)}}(x,x)). \end{array}$

The ordinary Betti number of ${M}$ is

$\displaystyle \begin{array}{rcl} b_k(M)=\lim_{t\rightarrow\infty}\int_{M}\mathrm{Trace}(e^{-t\Delta_k}(x,x))\,dx. \end{array}$

So the point is to estimate the difference

Lemma 12

$\displaystyle \begin{array}{rcl} \|e^{-t\Delta_k}(x,x)-e^{-t\Delta_k^{(2)}}(x,x)\|\leq C\,\mathrm{InjRad}(x)^{-d}. \end{array}$

In rank one, when injectivity radius is small, the exponent can be improved to be the absolute rank, i.e. the rank of the complexification.

This amounts to estimating the number of orbit points in a ball. Near a closed geodesic, this is controlled by the absolute rank and not by the rank.

Example 7 In hyperbolic space ${H^3}$, a screw motion can generate a cyclic subgroup whose orbits are ${\epsilon}$-dense in a cylinder, i.e. 2-dimensional. Here, the real rank is 1 but the absolute rank is 2.