## Notes of Emmanuel Breuillard’s fourth Cambridge lecture 11-04-2017

Introduction to approximate groups, IV

Today, I will enter deeper into the proof of the structure theorem.

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 ${|A\cap gH|\geq \frac{1}{C(K)}|A|}$. Furthermore, ${H}$ has a normal nilpotent finite index subgroup ${N}$ with number of generators and nilpotency class ${\leq 6\log K}$.

This is still a weak version. In the strong version, ${H=FN}$ where ${F}$ is contained in the product of bounded many copies of ${A}$.

It is enough to prove the theorem when ${A}$ is a ${K}$-approximate subgroup, i.e. ${1\in A}$, ${A=A^{-1}}$ and ${AA\subset XA}$, ${|X|\leq K}$. Indeed, we have seen in lecture 1 that, with some combinatorics, this is equivalent, up to passing to a large subset.

If ${G}$ has finite exponent (i.e. satisfies the law ${x^e=1}$ for some integer ${e}$), then ${H}$ must be finite. Hence ${A}$ is contained in boundedly many cosets of a finite subgroup of ${G}$.

0.1. Strategy of proof

Inspired from Gromov, expanded in model theory language by Hrushovski.

1. Ultraproducts

This is the model theory of the poor.

Let ${G_n}$ be a sequence of groups and ${A_n\subset G_n}$ be finite subsets of ${G_n}$ which are ${K}$-approximate subgroups. Fix a nonprincipal ultrafilter ${\mathcal{U}}$ on ${{\mathbb N}}$. Form the ultraproduct

$\displaystyle \begin{array}{rcl} \mathbb{G}=\prod_{\mathcal{U}}G_n \end{array}$

Los’ theorem asserts that any first order sentence ${\phi}$ in the language of groups holds for ${\mathbb{G}}$ iff it holds for ${\mathcal{U}}$-almost every ${G_n}$. Usually, it is easy to reprove it in simple situations.

Example 1 Let ${X_n\subset G_n}$ be a finite subset. Let ${\mathbb{X}=\prod_{\mathcal{U}}X_n}$ (such subsets are called internal subsets of ${\mathbb{G}}$). Given ${k\in{\mathbb N}}$, ${|\mathbb{X}|=k}$ iff ${|X_n|}$ a.e..

${\mathbb{G}}$ is agroup, ${\mathbb{X}}$ is a subgroup iff ${X_n}$ is a subgroup a.e., ${\mathbb{X}}$ is a ${K}$-approximate subgroup iff ${X_n}$ is a ${K}$-approximate subgroup a.e. However, ultraproducts of finite groups need not be finite, they are by definition pseudofinite.

The structure theorem is equivalent to the following.

Theorem 2 If ${\mathbb{A}}$ is a pseudofinite ${K}$-approximate subgroup of ${\mathbb{G}}$, then there exist internal subgroups ${\mathbb{M}<\mathbb{H}<\mathbb{G}}$ such that ${\mathbb{M}}$ is normal in ${\mathbb{H}}$, ${\mathbb{M}}$ is nilpotent, ${\mathbb{H}_{|\mathbb{M}}}$ is pseudofinite, and ${\mathbb{A}\subset\mathbb{H}X}$, with ${|X|<\infty}$.

Note that this does not give a bound on ${|X|}$, i.e. on the number of cosets required.

2. Hrushovski’s Lie model theorem

Theorem 3 Let ${\mathbb{A}\subset\mathbb{G}}$ be a pseudofinite ${K}$-approximate subgroup. There is a locally compact group ${G}$ and a surjective group homomorphism

$\displaystyle \begin{array}{rcl} \pi:\langle \mathbb{A}\rangle \rightarrow G \end{array}$

such that

1. ${\pi(\mathbb{A}^2)}$ is a compact neighborhood of ${1\in G}$,
2. For every open set ${U\subset G}$ and compact set ${K\subset U}$, there exists ${k\in{\mathbb N}}$ and an internal subset ${\mathbb{X}\subset\mathbb{A}^k}$ such that ${\pi^{-1}(K)\subset\mathbb{X}\subset\pi^{-1}(U)}$.

We see that ${\pi}$ has kernel contained in ${\mathbb{A}^{k_0}}$ for some finite ${k_0}$. Hence ${\pi}$ is almost injective.

Also, ${G}$ must be unimodular. Indeed, ${\langle \mathbb{A}\rangle}$ admits a biinvariant measure, arising as a limit of renormalized counting measures. Its image by ${\pi}$ is a biinvariant Haar measure on ${G}$.

The proof will show that ${G}$ has an open subgroup ${G'=KL}$ such that ${K}$ is normal and compact and ${L}$ is a nilpotent Lie group.

2.1. Special cases

What if ${G}$ is trivial? Then ${\langle \mathbb{A}\rangle\subset \mathbb{A}^{k_0}}$. By (2), ${\mathbb{A}^{k_0}}$ is a subgroup, so a.e. ${A_n^{k_0}}$ is a finite subgroup, countained is finitely many translates of ${A_n}$. This is the structure theorem

What if ${G}$ is compact? The same conclusion holds.

What if ${G}$ has a compact open subgroup ${H}$? By (1), ${\pi(\mathbb{A}^2)}$ is contained in finitely many translates of ${H}$. Since ${\pi^{-1}(H)}$ is internal and contained in ${\langle \mathbb{A}\rangle\subset \mathbb{A}^{k_0}}$, there exists a.e. a finite subgroup ${K_n such that ${A_n}$ is contained in bounded many copies of ${K_n}$.

Clearly, if ${A_n=[-n,n]\subset {\mathbb Z}}$, approximation by finite subgroups does not hold, so ${G}$ does not contain any compact open subgroup.

Corollary 4 (Hrushovski) Given integers ${K}$, ${e}$, if ${G}$ is a group of exponent ${e}$ and ${A}$ a ${K}$-approximate subgroup of ${G}$., then there exists a finite subgroup ${H such that ${|H|\leq C|A|}$ and ${A}$ is countained in a bounded number of translates of ${H}$.

It suffices to prove this for pseudofinite groups of finite exponent. ${G}$ has finite exponent. This implies that ${G}$ is totally disconnected, hence it has a compact open subgroup (Dantzig’s theorem).

3. Proof of the Lie model theorem

No Lie group today, they will arise next time. The point is to find a (possibly non-Hausdorff) locally compact topology on ${\langle\mathbb{A}\rangle}$. Then one divides by the closure of 1.

One would like to say that ${g}$ is close to the identity if ${g\mathbb{A}\delta \mathbb{A}}$ is small, e.g.

$\displaystyle \begin{array}{rcl} |g\mathbb{A}\Delta \mathbb{A}|<\epsilon|A|. \end{array}$

The following combinatorial lemma helps.

Lemma 5 (Hrushovski, Sanders, Croot-Sisak) Given integers ${k,K}$, let ${A\subset G}$ be a finite ${k}$-approximate subgroup. Then there exists a subset ${S\subset A^4}$,

1. ${aS^ka^{-1}\subset A^4}$, for all ${a\in A}$.
2. ${|S|\geq \frac{1}{c(k,K)}|A|}$.

Indeed, applying the Lemma to ${A_n}$ in ${G_n}$, find sets ${S_{n,k}}$. Furthermore, inductively, one can assume that

$\displaystyle \begin{array}{rcl} (S_{n,k+1})^2\subset S_{n,k}. \end{array}$

Let ${\mathbb{S}_k}$ be their ultraproduct. These sets form a basis for a topology. Their intersection is a normal subgroup. One shows that ${ \mathbb{A}}$ mod intersection is compact. This a bit like the bounded covering property characterization of approximate subgroups.

The proofs by Sanders or Croot-Sisak are elementary. Croot-Sisak consider sets

$\displaystyle \begin{array}{rcl} S_\epsilon=\{g\in G\,;\,\|1_A\star 1_A-g1_A\star 1_A\|_2<\epsilon\|1_A\star 1_A\|_2\}. \end{array}$

In model theory, this topology come out naturally. It is called the logic topology, compactness follows from general principles. See Ben Green’s Bourbaki seminar.