Posted by: matheuscmss | August 30, 2018

## Benoist’s minicourse “Arithmeticity of discrete groups”. Lecture I: A survey on arithmetic groups

Last week, Jon ChaikaJing Tao and I co-organized the Summer School on Teichmüller Theory and its Connections to Geometry, Topology and Dynamics at Fields Institute.

This activity was part of the Thematic Program on Teichmüller Theory and its Connections to Geometry, Topology and Dynamics, and it consisted of four excellent minicourses by Yves BenoistHee OhGiulio Tiozzo and Alex Wright.

These minicourses were fully recorded and the corresponding videos will be available at Fields Institute video archive in the near future.

Meanwhile, I decided to transcript my notes of Benoist’s minicourse in a series of four posts (corresponding to the four lectures delivered by him).

Today, we shall begin this series by discussing the statement of the main result of Benoist’s minicourse, namely:

Theorem 1 (Oh, Benoist–Miquel) Let ${G}$ be a semisimple algebraic Lie group of real rank ${\textrm{rank}_{\mathbb{R}}(G)\geq 2}$. Suppose that ${U}$ is a horospherical subgroup of ${G}$, and assume that ${\Gamma}$ is a Zariski dense and irreducible subgroup of ${G}$ such that ${U\cap \Gamma}$ is cocompact. Then, there exists an arithmetic subgroup ${G_{\mathbb{Z}}}$ such that ${\Gamma}$ and ${G_{\mathbb{Z}}}$ are commensurable.

The basic reference for the proof of this theorem (conjectured by Margulis) is the original article by Benoist and Miquel. This theorem completes the discussion in Hee Oh’s thesis where she dealt with many families of examples of semisimple Lie groups ${G}$ (as Hee Oh kindly pointed out to me, the reader can find more details about her contributions to Theorem 1 in these articles here).

Remark 1 I came across Benoist–Miquel theorem during my attempts to understand a question by Sarnak about the nature of Kontsevich–Zorich monodromies. In particular, I’m thankful to Yves Benoist for explaining in his minicourse the proof of a result that Pascal Hubert and I used as a black box in our recent preprint here.

Below the fold, the reader will find my notes of the first lecture of Benoist’s minicourse (whose goal was simply to discuss several keywords in the statement of Theorem 1).

1. Examples

Let ${G}$ be a Lie group and consider ${\Gamma\subset G}$ a discrete subgroup.

Definition 2

• ${\Gamma}$ is a lattice when ${\textrm{vol}(G/\Gamma)<\infty}$, i.e., there exists ${\mathcal{D}\subset G}$ such that ${G=\mathcal{D} \Gamma}$ and ${\lambda(\mathcal{D})<\infty}$ where ${\lambda}$ is a right-invariant Haar measure on ${G}$.
• ${\Gamma}$ is cocompact if ${G/\Gamma}$ is compact (i.e., the subset ${\mathcal{D}}$ can be chosen compact).

Example 1 ${\mathbb{Z}^d}$ is (discrete and) cocompact in ${\mathbb{R}^d}$: indeed, ${\mathbb{R}^d = \mathcal{D}\mathbb{Z}^d}$ for ${\mathcal{D}:=[0,1]^d}$.

In general, ${\Gamma}$ is cocompact ${\implies}$ ${\Gamma}$ is lattice. However, the converse is not true:

Example 2 ${\Gamma=SL(d,\mathbb{Z})}$ is a lattice in ${G=SL(d,\mathbb{R})}$ which is not cocompact. In fact, the compact subsets of ${X = G/\Gamma = \{\Lambda \textrm{ lattice in } \mathbb{R}^d \textrm{ of covolume } 1\}}$ are described by the so-called Mahler’s compactness criterion asserting that ${Y\subset X}$ is relatively compact if and only if ${\inf\limits_{\Lambda\in Y} \inf\limits_{v\in \Lambda\setminus\{0\}} \|v\| > 0}$.

Example 3 (Siegel) Let ${Q = \sum\limits_{i,j\leq d} Q_{ij} x_i x_j}$ be a non-degenerate quadratic form in ${\mathbb{R}^d}$, ${d\geq 3}$, with ${Q_{ij}\in\mathbb{Z}}$ for all ${i,j\leq d}$. In this setting, ${\Gamma = SO(Q,\mathbb{Z})}$ is a lattice in ${G=SO(Q,\mathbb{R})}$.

Remark 2 It is possible to prove that ${\Gamma}$ is cocompact if and only if ${Q}$ doesn’t represent zero (i.e., ${Q^{-1}(\{0\})\cap\mathbb{Z}^d=\{0\}}$).Nevertheless, this information is not very useful to produce cocompact lattices because it is possible (to use Hasse’s principle) to show that if ${d\geq 5}$ and ${Q}$ is not definite, then ${Q}$ represents zero.

Example 4

• ${SL(d,\mathbb{Z}[i])}$ is a lattice in ${SL(d,\mathbb{C})}$.
• ${SL(d,\mathbb{Z}[\sqrt{2}])}$ can be viewed as a lattice in ${SL(d,\mathbb{R})\times SL(d,\mathbb{R})}$ via the map ${g\mapsto (g, g^{\sigma})}$ where ${\sigma(a+b\sqrt{2}) = a-b\sqrt{2}}$ is Galois conjugation.

Historically, the basic idea of the previous example was adapted to produce the first examples of discrete cocompact subgroups of ${SO(n,1)}$:

Example 5 Let ${Q_0 = x_1^2+\dots+x_n^2-\sqrt{2} x_{n+1}^2}$. Then, ${SO(Q_0,\mathbb{Z}[\sqrt{2}])}$ can be viewed as a lattice in ${SO(Q_0,\mathbb{R})\times SO(Q_0^{\sigma},\mathbb{R})}$ via the map ${g\mapsto (g, g^{\sigma})}$.Note that ${Q_0^{\sigma} = x_1^2+\dots+x_n^2+\sqrt{2} x_{n+1}^2}$ is definite and, hence, it doesn’t represent zero. Therefore, ${SO(Q_0,\mathbb{Z}[\sqrt{2}])}$ is a discrete cocompact subgroup of ${SO(Q_0,\mathbb{R})\simeq SO(n,1)}$ (thanks to Mahler’s compactness criterion).

In the sequel, we will put all examples above in a single framework.

2. Arithmetic groups

Let ${G\subset SL(D,\mathbb{R})}$ be an algebraic subgroup (i.e., a subgroup described by the zeros of polynomial functions of the matrix entries of elements of ${SL(D,\mathbb{R})}$).

Definition 3

• ${G}$ is simple if its Lie algebra ${\mathfrak{g}=Lie(G)}$ is simple in the sense that its ideals are trivial.
• ${G}$ is semi-simple if ${\mathfrak{g} = \bigoplus\limits_{i} \mathfrak{g}_i}$ where ${\mathfrak{g}_i}$ are simple ideals.

If ${G}$ is semi-simple, the adjoint map ${\textrm{Ad}:G\rightarrow \textrm{Aut}(\mathfrak{g})}$ has finite kernel and finite index image.

In other terms, if ${G}$ is semi-simple, then ${G}$ equals to the group of matrices ${\textrm{Aut}(\mathfrak{g})}$ up to finite index. In particular, the adjoint map of a semi-simple group allows to replace the “extrinsic” algebraic structure ${G\subset SL(D,\mathbb{R})}$ by the “intrinsic” algebraic structure ${G = \textrm{Aut}(\mathfrak{g})}$ modulo finite index.

Definition 4 A ${\mathbb{Q}}$-form of ${\mathfrak{g}}$ is the choice of ${\mathbb{Q}}$-vector subspace ${\mathfrak{g}_{\mathbb{Q}}}$ of ${\mathfrak{g}}$ such that

• ${\mathfrak{g}_{\mathbb{Q}}}$ is a Lie subalgebra;
• ${\mathfrak{g} = \mathbb{R}\otimes_{\mathbb{Q}}\mathfrak{g}_{\mathbb{Q}}}$.

In other words, a ${\mathbb{Q}}$-form of ${\mathfrak{g}}$ is a choice of basis where the Lie bracket ${[.,.]}$ is described by a matrix with rational coefficients.

Definition 5 An arithmetic subgroup of ${G}$ is ${G_{\mathbb{Z}}:=\textrm{Ad}^{-1}(\textrm{Aut}(\mathfrak{g}_{\mathbb{Z}}))}$ for some choice of ${\mathfrak{g}_{\mathbb{Z}}}$.

It is possible to check that Examples 2345 above describe arithmetic subgroups ${\Gamma}$ of semisimple Lie groups ${G}$. In particular, the fact that these examples provide lattices ${\Gamma}$ in ${G}$ can be viewed as concrete applications of Borel–Harish-Chandra theorem:

Theorem 6 (Borel–Harish-Chandra) An arithmetic subgroup ${G_{\mathbb{Z}}}$ of an algebraic semisimple group ${G}$ is a lattice.

Remark 3 One can prove that ${G_{\mathbb{Z}}}$ is cocompact if and only if ${\mathfrak{g}_{\mathbb{Z}}\setminus \{0\}}$ doesn’t contain nilpotent elements, i.e., there is no ${0\neq x\in\mathfrak{g}}$ such that the matrix ${\mathfrak{g}\ni v\mapsto [x,v]\in\mathfrak{g}}$ is nilpotent.

Remark 4 This theorem naturally leads to the question of the existence of non-arithmetic lattices. As it turns out, this question is answered by the so-called Margulis arithmeticity theorem.

Let us now pursue the discussion of the statement of Theorem 1 by introducing the notion of irreducible subgroups.

Definition 7 Let ${G}$ be an algebraic semisimple group with Lie algebra ${\mathfrak{g} = \bigoplus\limits_{i} \mathfrak{g}_i}$, where ${\mathfrak{g}_i}$ are simple ideals. A discrete subgroup ${\Gamma\subset G}$ is irreducible if

$\displaystyle \Gamma\cap \textrm{ker}(\textrm{Ad}_{i}:G\rightarrow \textrm{Aut}(\mathfrak{g}_i))$

is finite for all ${i}$.

Remark 5 Any ${\Gamma\subset G}$ is irreducible when ${G}$ is simple.

Finally, let us close this section by noticing that Theorem 1 is a sort of “converse” to the so-called Borel density theorem:

Theorem 8 (Borel) Let ${G}$ be a connected semisimple algebraic group with Lie algebra ${\mathfrak{g} = \bigoplus\limits_{i} \mathfrak{g}_i}$, where ${\mathfrak{g}_i}$ are simple ideals. Assume that none of the factors ${G_i}$ of ${G}$ (associated to ${\mathfrak{g}_i}$‘s) is compact. Then, any lattice ${\Gamma\subset G}$ is Zariski dense (i.e., ${\Gamma}$ is not included in a proper algebraic subgroup of ${G}$).

3. Horospherical group

The last bit of information needed to understand the statement of Theorem 1 is the concept of horospherical group. In the literature, this notion is usually phrased in terms of unipotent radicals of parabolic subgroups. For the sake of exposition, we will give an alternative elementary definition of this notion.

Definition 9 Let ${G}$ be a semisimple group with neutral element ${e\in G}$.

• an element ${u\in G}$ is unipotent whenever ${\exists\, g\in G}$ such that ${\lim\limits_{n\rightarrow\infty} g^{-n} u g^n = e}$;
• a non-trivial subgroup ${U\subset G}$ is horospherical when ${\exists\, g\in G}$ such that

$\displaystyle U=U_g:=\{u\in G: \lim\limits_{n\rightarrow\infty} g^{-n} u g^n = e \}$

Example 6 Let ${G=SL(2p,\mathbb{R})}$, ${p\geq 2}$, and ${g=\left(\begin{array}{cc} \lambda I & 0 \\ 0 & \lambda^{-1} I \end{array}\right)\in G}$ where ${\lambda>1}$ and ${I}$ is the ${p\times p}$ identity matrix.Note that if ${u = \left(\begin{array}{cc} A & B \\ C & D \end{array}\right)\in G}$, then ${g^{-n} u g^n = \left(\begin{array}{cc} A & \lambda^{-2n} B \\ \lambda^{2n} C & D \end{array}\right)}$. Hence,

$\displaystyle U = \left\{\left(\begin{array}{cc} I & B \\ 0 & I \end{array}\right): B \textrm{ is a } p\times p \textrm{ matrix}\right\}$

is a horospherical subgroup.

As it was shown by Kazhdan and Margulis, the cocompactness of lattices is detected by horospherical groups:

Theorem 10 (Kazhdan–Margulis) Let ${\Gamma}$ be a lattice in a semisimple Lie group ${G}$. Then, ${\Gamma}$ is not cocompact ${\iff}$ ${\Gamma\setminus\{e\}}$ contains unipotent elements ${\iff}$ ${\exists\, U\subset G}$ horospherical such that ${\Gamma\cap U}$ is cocompact in ${U}$.

This completes our discussion of the statement of Theorem 1. Next time, we will start the proof of Theorem 1 in the particular case of Example 6, i.e., ${G=SL(2p,\mathbb{R})}$, ${p\geq 2}$, and ${U = \left\{\left(\begin{array}{cc} I & B \\ 0 & I \end{array}\right): B \textrm{ is a } p\times p \textrm{ matrix}\right\}}$.

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