BS-topology and invariant random subgroups
Joint work with Abert, Bergeron, Biringer, Nikolov, Raimbault, Sanet.
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 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 1 is an interval. is homeomorphic to the 4-sphere. is complicated.
1.2. Invariant random subgroups
acts on by conjugation.
Definition 2 An IRS is a -invariant probability measure on .
The set of IRS in topology is compact.
Example 2 Let 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 3 IRS which are Dirac measures correspond to normal subgroups.
So IRS are a generalization of normal subgroups.
Example 4 Let act on a probability space , preservig measure. Stabilizers of points yield an IRS.
Every IRS arises in this way.
Example 5 Let 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 3 is 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 4 Let , 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 5 Suppose 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 6 Let 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 6 A 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 7 Let 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 8 Let 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
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 -manifolds is uniformly discrete.
This implies Lehmer’s conjecture on Mahler measures of polynomial.
Theorem 10 Let be a sequence of uniformly discrete -manifolds which converges to . Then normalized Betti numbers converge, i.e.
Theorem 11 Assume 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.
is the Betti number of ,
The ordinary Betti number of is
So the point is to estimate the difference
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 , 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.