Notes of Emmanuel Breuillard’s third Cambridge lecture 14-03-2017

Introduction to approximate groups, III

1. Almost flat manifolds

Recall our structure result, in simplified form.

Theorem 1 (Breuillard-Green-Tao) Let {k\geq 1}. Let {A\subset G} be a finite group such that {|AA|\leq K|A|}. Then there exists a virtually nilpotent subgroup {H} of {G} and {g\in G} such that {|A\cap gH|\geq \frac{1}{C(K)}|A|}.

Corollary 2 Given an integer {d}, there exists {n(d)} such that if {G} is a group generated by a finite symmetric set {S}, if there exists {n\geq n(d)} such that {|S^n|\leq n^d |S|}, then {G} is virtually nilpotent.

This improves on Gromov’s theorem, since only one scale {n} is required. Note that Gromov had observed that a finitary version of his theorem did hold: {\forall d} and {C}, {\exists n_0(d,C)} such that if {|S^n|\leq C\,n^d} for all {n\leq n_0(d,C)}, then {G} is virtually nilpotent. Our {n(d)} does not depend on the multiplicative constant {C} in the polynomial growth assumption. We shall see that this implies Gromov’s almost flat manifolds theorem.

1.1. Proof of Corollary 2

The assumption implies that some large ball has small doubling, {|S^{2n_1}|\leq 3^d |S^{n_1}|}. The structure theorem implies that {S^{n_1}\subset XH} where {H} is virtually nilpotent, {|X|\leq C(3^d)}. Taking {n_1>|X|} implies that {G=\langle S\rangle\subset XH}, i.e. {H} has finite index in {G}.

Note however that our argument does ot allow to bound the index of {H}.

1.2. Gromov’s almost flat manifold theorem

Here is a weakening of Gromov’s theorem.

Theorem 3 Given an integer {d}, there exists {\epsilon(d)} such that every compact Riemannian {d}-manifold of diameter 1 and absolute value of sectional curvature {\leq \epsilon(d)}, then the fundamental group is virtually nilpotent.

This is weaker since Gromov determines the diffeomorphism type, not the mere fundamental group.

Here is how this weak form follows from Corollary 2. Assume curvature is {\geq -\epsilon} (in fact, Ricci {\geq-(d-1)\epsilon} suffices). Bishop-Gromov’s volume comparison implies that, in the universal cover {M},

\displaystyle \begin{array}{rcl} \frac{vol(B^M(x,r))}{vol(B^M(x,1))}\leq \frac{vol(B^{\mathbb{H}^d}(x,r))}{vol(B^{\mathbb{H}^d}(x,1))}\rightarrow r^d \quad \textrm{as}\quad \epsilon\rightarrow 0, \end{array}

where {\mathbb{H}^d} denotes hyperbolic space of constant curvature {\epsilon}.

Let {S} be the set of homotopy classes of based loops of length {\leq 3}. This is a finite generating set of the fundamental group. Using the Dirichlet fundamental domain {D} (piece of Voronoi tesselation defined by the orbit of {x}), which satisfies

\displaystyle \begin{array}{rcl} B(x,n)\subset S^n D\subset B(x,3n+1), \end{array}

one gets

\displaystyle \begin{array}{rcl} \frac{|S^n|}{|S|}\leq \frac{vol(B^M(x,3n+1))}{vol(B^M(x,1))}\leq 2(3n+1)^d \end{array}

provided {\epsilon} is small enough. So Corollary 2 applies.

V. Kapovich and Wilking get a sharper result: they estimate the index.

Question. Can one derive such an index bound from the structure theorem? The point is to exploit the fact manifolds have dimension {d} to show that there cannot be large finite subgroups nearly fixing a point. This is known (Jordan) for finite subgroups fixing a point.

2. Special classes of groups

The structure theorem can be improved for certain classes.

2.1. Free groups

Theorem 4 (Safin) Let {G} be a non-abelian free group. Let {A\subset G} be a subset that does not generate an abelian subgroup. Then {|AAA|\geq c|A|^2}, {c=\frac{1}{100}}. More generally, for all {d},

\displaystyle \begin{array}{rcl} |A^{2d+1}|\geq c^d|A|^{d+1}. \end{array}

 

There is a more complicated statement for small doubling. Note that the union of a point and an arithmetic progression has small doubling.

The exponent {d+1} is sharp (same example), as is the exponential nature of the constant. This implies that

\displaystyle \begin{array}{rcl} \lim_{d\rightarrow\infty} \sqrt[d]{|A^d|}\geq\frac{\sqrt{|A|}}{10}. \end{array}

This obviously extends to fully residually free groups. This also extends to certain free products (Button).

2.2. Sketch of Safin’s argument

Before Safin, there were weaker results by Chang and Razborov.

Lemma 5 Let {G} be a non-abelian free group. Let {X,Z\subset G} be finite subsets. Let {y\in G} be longer than any element of {X}. Assume that all products in {XyZ} are reduced. Then {|XyZ|\geq \frac{1}{2}|X||Z|}, unless {y} is a subword of a periodic word and some element of {X} ends with the period of {y}.

Example. If {y=a^s \vec y=\overleftarrow y b^s}, {X=\{a^i\,;\,1\leq i\leq n\}}, {Z=\overleftarrow y X} give a small {|XyZ|\leq 2n}.

Proof of Lemma. Assume that {|XyZ|< \frac{1}{2}|X||Z|}. Then there exist 3 distinct pairs {(x_i,z_i)} such that {x_i y z_i} are equal. Since {y} is longer than the {x_i}‘s and has an overlap of length { \frac{1}{2}|y|} with either the first or third occurrence, it must be periodic.

Proof of Theorem 4.

Step 1. After possibly conjugating {A}, can find {A_0}, {B_0\subset A} with {|A_0|,|B_0|\geq\frac{1}{100}|A|} such that all products in {A_0 B_0} and {B_0 A_0} are reduced, and {|a|\leq |b|}

Step 2. Apply Lemma 5 to products {A_0 bB_0}, {b\in B_0}. If{|A_0 bB_0|\ll |A|^2}, then {b} must be periodic. One verifies that all words in {B_0} are periodic with the same period.

Step 3. Apply Lemma 5 on the right to show that all tails are the same. Therefore {B_0} is an arithmetic progression.

Step 4. Find {x\in A_0} which does not commute with the period, then look at {B_0\times B_0}.

Question. Does Safin’s theorem extend to hyperbolic groups ?

Open problem. Does there exist a uniform lower bound on growth valid for all hyperbolic groups and all generating systems.

2.3. Linear groups

Theorem 6 (Breuillard-Green-Tao) Let {G\subset Gl(d,{\mathbb C})} be a linear group.

  1. Weak version. If {A\subset G} is a finite subset and {|AA|\leq K|A|}, then there exists a nilpotent subgroup{H} and an element {x} such that {|A\cap xH|\geq \frac{1}{C(K)}|A|}, {C(K)} polynomial in {K}.
  2. Strong version. If {A\subset G} is a finite subset, then {|AAA|\geq |A|^{1+\epsilon}} unless {\langle A\rangle} is virtually nilpotent.

 

The is a finitary version for subgroups of {\mathbb{G}(\mathbb{F}_q)}. Then {|AAA|\geq |A|^{1+\epsilon}} unless {AAA=G} or {A} is contained in a proper subgroup.

Open problem. One should be able to tell more when {A} is contained in a proper non-nilpotent subgroup, see work by Pyber-Szabo and Helfgott-Gill.

3. From growth to spectral gap

A spectral gap is stronger than a mere growth lower bound.

1% spectral problem. Is it true that given {\epsilon>0}, there exists {\delta>0} such that if {\mu} is a finitely supported probability measure on {Gl(d,{\mathbb C})}, generating subgroup {\Gamma}, under the condition {\|\lambda_\Gamma(\mu)\|>\epsilon} then {\mu} charges an amenable subgroup with probability {\geq \delta}~? In other words, the structure theorem quantifies the relation betwen polynomial growth and virtual nilpotence. I ask wether one can quantify the relation between spectral gap and amenability.

The analogous 99% spectral structure theorem (replace {\epsilon} with {1-\epsilon} and {\delta} with {1-\delta}) holds, it follows from the strong uniform Tits alternative.

There is no hope for a general structure theorem valid for all groups. Indeed, there exist groups with non-uniform exponential growth (John Wilson), thus even the 99% structure theorem cannot hold in general.

Perhaps a spectral structure theorem holds for free groups.

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About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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