** 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 1Let be a locally compact group. Consider the space of closed subgroups of . It carries the Chabauty topology: converges to if

- , there exists converging to .
- If and converge to , then .

Example 1is an interval. is homeomorphic to the 4-sphere. is complicated.

** 1.2. Invariant random subgroups **

acts on by conjugation.

Definition 2An IRS is a -invariant probability measure on .

The set of IRS in topology is compact.

Example 2Let be a surface of genus , equipped with a hyperbolic metric. Choose randomly a point and unit tangent vector. This gives an embedding of into . This gives an IRS.

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

So IRS are a generalization of normal subgroups.

Example 4Let act on a probability space , preservig measure. Stabilizers of points yield an IRS.

Every IRS arises in this way.

Example 5Let be a lattice, i.e. a closed subgroup, and preserves a probability measure on . Then this action is an IRS, supported on the conjugacy orbit of .

So IRS are a generalization of lattices.

** 1.3. Connection with BS topology **

Fix a left-invariant metric on . The map , induces a continuous map . In general, it need not be a homeomorphism onto its image, but is is so when is semi-simple.

Therefore, from now on, we shall assume that is a simple Lie group.

The Borel density theorem extends to IRS.

Proposition 3is isolated in . Every non-atomic IRS is supported on discrete and Zariski-dense subgroups of .

Therefore, we concentrate on discrete IRS.

Let be the symmetric space of .

Lemma 4Let , be discrete IRS. The following are equivalent.

- converge to .
- , and every -ball in a locally -manifold, the -probability that embeds with distorsion converges to the -probability of the same event.

Corollary 5Suppose are lattices, corresponding to measures . Then converge to the Dirac measure on the trivial subgroup , if and only if the random metric spaces converge to . I.e., a random -ball in is contractible with probability tending to .

**2. Higher rank phenomena **

** 2.1. Key theorem **

The following fails in real rank .

Theorem 6Let be a simple Lie group of real rank . Then

- Ergodic IRS of are , and lattices.
- The only accumulation point of the IRS associated to a lattice is

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

2. follows from Kazhdan’s property (T). I explain it now. Suppose converge to some measure . Then is ergodic (this already follows from propoerty (T)).

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

From claim 1, is either or . Assume it is . Then contains large balls filling nearly all the volume. In , 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 7Let be a higher rank irreducible symmetric space. Let be -orbifolds of finite volume tending to infinity. Then for all , the probability that an -ball in is contractible tends to 1.

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

Theorem 8Let be a higher rank irreducible symmetric space. Let be -orbifolds of finite volume tending to infinity. Then BS-converges to .

**3. Convergence of Betti numbers **

** 3.1. Statement **

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

**Conjecture (Margulis)**. The family of all compact arithmetic -manifolds is uniformly discrete.

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

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

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

In other words, in higher rank, converge to in a very strong sense.

** 3.2. Proof **

is the Betti number of ,

The ordinary Betti number of is

So the point is to estimate the difference

Lemma 12

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 7In hyperbolic space , a screw motion can generate a cyclic subgroup whose orbits are -dense in a cylinder, i.e. 2-dimensional. Here, the real rank is 1 but the absolute rank is 2.