These are my live-TeXed notes (reorganized according to my tastes) for the course Discrete subgroups of Lie groups and discrete transformation groups by Professor Lizhen Ji, July 19 — August 9, at Math Science Center of Tsinghua University.

TopThe plan of this course

  1. Introduction and basic notions
  2. General questions on discrete subgroups and discrete transformation groups.
  3. Fuchsian groups, namely discrete subgroups of $SL(2,\mathbb{R})$ ($PSL(2,\mathbb{R})$). In particular, the modular group $SL(2,\mathbb{Z})$ and its congruence subgroups. Kleinian groups, namely discrete subgroups of $SL(2,\mathbb{C})$. They provide special, important examples of lattices of semisimple Lie groups and also nonlattices.
  4. Crystallographic groups (Bieberbach theorems)
  5. Nilpotent Lie groups and their lattices
  6. Solvable Lie groups and their lattices (Mostow rigidity)
  7. Semisimple Lie groups and their lattices

The main purpose of this course is to study lattices of semisimple Lie groups. But we need to understand lattices of non-semisimple Lie groups too. (Unfortunately, only the first three topics were covered in this course).


The topic of this course is a central subject of modern mathematics. It is closely related to many areas, e.g., group theory, geometry (differential geometry, algebraic geometry, arithmetic geometry), analysis, number theory (automorphic forms) and algebra (algebraic $K$-theory).

Major figures in this subject includes Poincare, Klein, Fricke, and recently C. L. Siegel, A. Weil, A. Selberg, A. Borel, Harish-Chandra, Piatetski-Shapiro, G. Margulis, D. Kazhdan, Maltsev, Auslander, Hsien-Chung Wang, R. Langlands. Two important books are Raghunathan's and Zimmer's.

Now let us explain the "what, why and how" of the words in the course title.

Definition 1 A discrete group is a group with the discrete topology.

This definition is not very interesting at first sight since the discrete topology may not appear "naturally".

Example 1
  • $\mathbb{Z}$ with the natural topology.
  • $\mathbb{R}$ with the discrete topology. But the usual topology is more natural since it is a connected Lie group. The inclusion of $\mathbb{Z}$ into $\mathbb{R}$ induces the usual discrete topology on $\mathbb{Z}$.
  • $\mathbb{Q}$ with the discrete topology. The induced topology of $\mathbb{Q}$ from $\mathbb{R}$ is not discrete. Nevertheless, The discrete topology on $\mathbb{Q}$ is natural and important since it can be realized as a discrete subgroup of an important locally compact topological space, i.e., the ring of adeles.
Definition 2 Let $G$ be a topological group (usually we assume $G$ is locally compact). A subgroup $\Gamma\subseteq G$ is called a discrete subgroup if the induced topological on $\Gamma$ is discrete.
Example 2 An important subclass is given by taking $G$ to be a Lie group and $\Gamma$ be a discrete subgroup.

Here is a less trivial example.

Example 3 Let $G=SL(2,\mathbb{R})$. Then $SL(2,\mathbb{R})$ is a Lie group since it is the inverse image of the regular value 1 under $\det$ (or, it is a closed subgroup of a Lie group). The modular group $SL(2,\mathbb{Z})$ is a discrete subgroup of $SL(2,\mathbb{R})$ since $\mathbb{Z}$ is discrete in $\mathbb{R}$. Also, $SL(2, \mathbb{Z})$ is a discrete subgroup of $GL^+(2,\mathbb{R})$ and $GL(2, \mathbb{R})$. The first one of the three embeddings is the most natural one. Firstly, $SL(2,\mathbb{Z})$ is a lattice, namely $\Vol(SL(2,\mathbb{Z})\backslash SL(2,\mathbb{R}))$ is finite, but $\Vol( SL(2, \mathbb{Z})\backslash GL^+(2, \mathbb{R}))$ is infinite. Secondly, from the point of view of algebraic groups, $SL(2,\mathbb{Z})$ is an arithmetic subgroup of the semisimple linear algebraic group $SL(2)$ and $SL(2,\mathbb{R})$ is the $\mathbb{R}$-locus of $SL(2)$.
Example 4 Let $G=\mathbb{R}^2$, for any nonzero vector $v\in \mathbb{R}^2$, we have an embedding $$\mathbb{Z}\hookrightarrow \mathbb{R}^2,\quad n\mapsto nv.$$ Also, given two linear independent vectors $v_1, v_2\in \mathbb{R}^2$, we have an embedding $$\mathbb{Z}^2\hookrightarrow \mathbb{R}^2,\quad (n_1, n_2)\mapsto n_1v_1+n_2v_2,$$ which is more natural since $\Vol(\mathbb{Z}^2\backslash \mathbb{R}^2)$ is finite. In other words, we are interested in discrete subgroups $\Gamma$ of $G$ that are not "too small" in $G$.

Now let us come to the second key word in the course title — transformation groups, which first appeared in Lie's book Theorie der transformations gruppe I, II, II.

Definition 3 Let $\Gamma$ be a topological group and $X$ be a topological space. A topological action of $\Gamma$ on $X$ is a continuous map $\Gamma\times X\rightarrow X$ satisfying
  1. $e.x=x$ for any $x\in X$,
  2. for any $\gamma_1, \gamma_2\in \Gamma$ and any $x\in X$, $\gamma_1(\gamma_2(x))=(\gamma_1\gamma_2)(x)$.

For any $\gamma\in\Gamma$, the $\gamma$-action is a homeomorphism. Let $\Homeo(X)$ be the group of homeomorphisms of $X$, then we get a homomorphism $\Gamma\rightarrow \Homeo(X)$. The following two concepts are closely related.

Definition 4 The space $X$ with a $\Gamma$-action is called a $\Gamma$-space.
Definition 5 The group $\Gamma$ with an action on $X$ is called a transformation group.

We can even put more structures on the spaces $X$ and the groups $\Gamma$:

  1. $X$ is a manifold, $\Gamma$ is either a discrete group or a Lie group and the action is by diffeomorphisms. This is the subject of the classical transformation group theory.
  2. $X$ is a complex manifold (or complex space), $\Gamma$ is either discrete or a complex Lie group and the action is by holomorphisms.
  3. $X$ is an algebraic variety, $\Gamma$ is an algebraic group and the action is by morphisms.

We are interested in the case that $\Gamma$ is an infinite discrete group and $X$ is a topological space or manifold, which is important for, e.g., geometric group theory.

Definition 6 Let $X$ be a Hausdorff space with $\Gamma$-action on $X$. We say the action of $\Gamma$ is discontinuous, if for any $x\in X$, the orbit $\Gamma\cdot x$ of $x$ is a discrete subset of $X$. We say the action of $\Gamma$ is properly discontinuous, if for any compact subset $C\subseteq X$, the set $\{\gamma\in \Gamma: \gamma C\cap C\ne0\}$ is finite.
Proposition 1 If the $\Gamma$-action on $X$ is properly discontinuous, then the quotient $\Gamma\backslash X$ with the quotient topology is Hausdorff.

Here are a few reasons why we study discrete subgroups.

  1. It occurs naturally (e.g., $\mathbb{Z}\hookrightarrow \mathbb{R}$).
  2. It provides important examples in geometric group theory and combinatorial group theory.
  3. Given a manifold $M$ with nontrivial fundamental group, then $\pi_1(M)$ acts on $\tilde M$ as Deck transformations.
  4. Many natural spaces arises from quotients $\Gamma\backslash X$ (e.g., locally symmetric space).

Groups are interesting in themselves, to understand their properties, the "only" effective way is to use their actions on suitable spaces. So groups that admit good actions on nice spaces are interesting and special. Besides discrete subgroups of Lie groups, two other very important discrete transformation groups are:

  1. Mapping class groups of surfaces with the actions on the Teichmuller spaces.
  2. Let $F_n$ be the free group on $n$ generators, the outer automorphism group $\Out(F_n)=\Aut(F_n)/\Inn(F_n)$ is the most important group in combinatorial group theory. In 1980, Culler-Vogtmann discovered the Outer space $X_n$, a contractible space (which is a simplicial complex) where $\Out(F_n)$ acts simplically and properly. $X_n$ is the space of marked metrics, which is related to the hot topic of tropical geometry.
Definition 7 A discrete subgroup of $\Gamma$ of a topological group $G$ is called a uniform lattice if $\Gamma\backslash G$ is compact.
Proposition 2
  1. Every discrete subgroup $\Gamma$ of $\mathbb{R}^n$ is of the form $$\Gamma=\mathbb{Z}v_1+\cdots \mathbb{Z} v_k.$$ where $v_i$'s are linearly independent.
  2. $\Gamma$ is a uniform lattice if and only if $\Vol(\Gamma\backslash \mathbb{R}^n)$ is finite.

We can define the more general notion of volume on the quotient $\Gamma\backslash G$ using a fundamental set of the $\Gamma$-action on $G$.

Definition 8 A subset $F $ of $G$ is called a fundamental set for the $\Gamma$-action on $G$ if $F $ meets every $F $-orbit exactly once, namely $\bigcup_{\gamma\in\Gamma}\gamma F=G$ and $\gamma_1F\cap\gamma_2F=\varnothing$ for any $\gamma_1\ne\gamma_2$ in $\Gamma$.
Remark 1 We always have a fundamental set $F $ by the axiom of choice, i.e., by picking one element from each orbit. However, it may not be a Borel subset.
Proposition 3 Let $G$ be a secondly countable locally compact topological group and $v$ be a left invariant Haar measure on $G$, then for any discrete subgroup $\Gamma$ of $G$, there exists an induced measure on $\Gamma\backslash G$.
Proof The proof goes back to [1]. First let us construct a Borel fundamental set $F $. Since $\Gamma$ is a discrete subgroup of $G$, there exists an open subset $V\subseteq G$ containing the identity such that $V\cap\Gamma=\{e\}$. Since $G$ is a topological group , there exists an open neighbourhood $U$ containing $e$ such that $UU^{-1}\subseteq V$. Then the $\Gamma$-translates of $U$ are disjoint. Since $G$ is secondly countable, there exists a sequence $g_1, g_2,\ldots$ of $G$ such that $G=\bigcup_{i=1}^\infty Ug_i$. Now let $$F=\bigcup_{n=1}^\infty \left(Ug_n- \bigcup_{i<n} \Gamma Ug_i\right),$$ one can check that $F $ is a Borel fundamental set.

Let $\pi:G\rightarrow\Gamma\backslash G$ be the projection. Then $\pi|_F: F\rightarrow \Gamma\backslash G$ is one-to-one. Define an induced measure $v$ on $\Gamma\backslash G$ by $$v(B)=v(F\cap \pi^{-1}(B)),$$ where $B$ is a Borel subset of $\Gamma\backslash G$. We need to show that this induced meausre is independent on the choice of $F $. Suppose $F, F'$ are two Borel fundamental sets. Then it suffices to show that $v(F\cap A)=v(F'\cap A)$ for any subset $A\subseteq G$ which is invariant under $\Gamma$. By the invariance of $A$ and the definition of $F, F'$, we have
v(F\cap A)&=\sum_{\gamma\in\Gamma}v(\gamma F'\cap F\cap A) \\
&=\sum_{\gamma\in\Gamma}v(F'\cap \gamma^{-1}F\cap \gamma^{-1}A) \\
&=\sum_{\gamma\in\Gamma}v(\gamma^{-1}F\cap F'\cap A) \\
&=v(F'\cap A).
This completes the proof.

Definition 9 A discrete subgroup $\Gamma$ of a locally compact topological group $G$ is called a lattice (or lattice subgroup) if $\Vol(\Gamma\backslash G)$ is finite. In other words, let $\mu$ be a left invariant measure on $G$, a discrete subgroup $\Gamma$ is called a lattice if $\mu(\Gamma\backslash G)$ is finite, where $\mu$ is the induced meausre on $\Gamma\backslash G$.
Remark 2 In general, lattices are not necessarily uniform (e.g., $SL(2,\mathbb{Z})\subseteq SL(2,\mathbb{R})$).

Let $G$ be a locally compact topological group. Then $G$ admits a left invariant Haar measure unique up to multiple. Let $v$ be a left invariant Haar measure and let $R_g$ be the right action, then $R_g$ induces a measure $R_g^*(v)(A)=v(Ag^{-1})$ and $R_g^*v$ is also left invariant. By the uniqueness of left invariant Haar measure, there exists a constant $\Delta(g)$ such that $R_g^*v=\Delta(g)v$.

Definition 10 The function $\Delta(g)$ is called the modular function of $G$. If $\Delta(g)\equiv1$, the $G$ is called unimodular (hence any left invariant Haar measure is also right invariant).
Example 5 Let $G$ be a real semisimple Lie group, then $G$ is unimodular, since the modular function $\Delta: G\rightarrow \mathbb{R}_+$ is homomorphism, $\mathbb{R}_+$ is abelian and $G=[G,G]$.
Proposition 4 If $G$ admits a lattice, then $G$ is unimodular.
Proof There exists a Borel fundamental set $F $ such that $v(F)=v(\Gamma\backslash G)$. One can check that for any $g\in G$, $F g^{-1}$ is also a fundamental set. So $v(F)=v(F g^{-1})$ (finite) and $G$ is unimodular.
Example 6 The affine group $\mathrm{Aff}(\mathbb{R})=\left\{
  a & b \\
  0 & 1
\end{bmatrix}: a\ne0, b\in \mathbb{R}\right\}$ is not unimodular, as the measure $dg=da db/a$ is left invariant but not right invariant. Hence it doesn't admit lattices. Similarly the $\left\{
  a & b \\
  0 & c
\end{bmatrix}: a\ne0, b\in \mathbb{R}, c\ne0\right\}$ is not unimodular, hence it doesn't admit lattices either.
Question Does every unimodular (e.g., abelian) group admit lattices?
Answer No, $\mathbb{Q}_p$ is abelian but doesn't have any notrivial discrete subgroups. If $\Gamma\subseteq \mathbb{Q}_p$ is a discrete subgroup containing $a\ne0$, then $p^na\in\Gamma$ but $p^na\rightarrow0$, contradicting the fact that $\Gamma$ is discrete.
Remark 3 Any locally compact field that contains $\mathbb{Q}$ as a dense subfield is either $\mathbb{R}$ or $\mathbb{Q}_p$ for some prime $p$.

It is still an open problem to decide when a unimodular Lie group group admits lattices. Nevertheless, the anwser is affirmative for semisimple Lie groups.

Theorem 1 (Borel) If $G$ is a real semisimple Lie group with finitely many connected components, then $G$ admits lattices (both uniform and nonuniform).
Remark 4 For a $p$-adic semisimple Lie group, any lattice is uniform.

TopQuestions and problems about discrete subgroups

We start by listing several problems concerning a discrete subgroup $\Gamma$ to motivate our discussion.

Finite generation Namely, find finitely many elements $\gamma_1, \ldots\gamma_n$ such that every element of $\Gamma$ can be written as $\gamma_{i_1}\cdots,\gamma_{i_k}$ where $\gamma_{i_j}\in \{\gamma_1^\pm,\ldots\gamma_n^\pm\}$. The existence of the generation is relatively easy (e.g., Kazhdan property (T)). However, finding the explicit generators are generally much harder, e.g., how about the generators of $SL(n,\mathbb{Z})$ for $n\ge3$? The finite generation is useful for geometric group theory (word metric, Cayely graph, etc.)

Definition 11 Let $S$ be a set of generators of $\Gamma$. Define $|\gamma|_S=\inf\{k:\gamma=\gamma_{i_1}\cdots\gamma_{i_k}: \gamma_{i_j}\in S\cup S^{-1}\}$. Define the word metric $d_S(\gamma, \gamma')=|\gamma^{-1}\gamma'|_S$ for any $\gamma,\gamma'\in\Gamma$. This metric is left invariant under $\Gamma$, namely $d_S(g\gamma, g\gamma')=d_S(\gamma,\gamma')$.
Remark 5 $(\Gamma, d_S)$ is a discrete metric space, hence not connected.
Definition 12 The Cayely graph $\Ca(\Gamma)$ consists of the elements of $\Gamma$ as vertices. An edge between $\gamma,\gamma'$ exists if and only if $\gamma^{-1}\gamma'\in S\cup S'$. Then $\Ca(\Gamma)$ is connected.
Remark 6 There are many generating sets, but the word metrics defined by them are quasi-isometric in the following sense.
Definition 13 Let $S$, $S'$ be two generating sets of $\Gamma$. Then there exists constants $c_1>1, c_2>0$ such that for any $\gamma, \gamma'\in\Gamma$, $$c_1^{-1}d_S(\gamma,\gamma')-c_2\le d_{S'}(\gamma,\gamma')\le c_1d_S(\gamma,\gamma')+c_2.$$
Remark 7 The Calay graph $\Ca(\Gamma)$ is a metric space by declaring that every edge has length 1. Then the metric space $\Ca_S(\Gamma)$ is quasi-isometric to $(\Gamma, d_S)$. $\Ca_S(\Gamma)$ is connected but is not simply connected in general.
Example 7 $\Gamma=\mathbb{Z}^2$, $S=\{(1,0),(-1,0),(0,1),(0,-1)\}$. Then $\Ca_S(\Gamma)$ is the infinite grid on $\mathbb{R}^2$ and is not simply connected.
Example 8 $\Gamma=F_2=\langle a, b\rangle$, $S=\langle a,b\rangle$. Then $\Ca_S(F_2)$ is a infinite tree, hence simply connected.

Finite presentation Let $\Gamma$ be a finitely generated group, we may ask whether there are only finitely many relations between the generators. Finding the existence and explicit relations are very important (and hard) in combinatorial group theory.

Cohomology $H^i(\Gamma)$ and $H_i(\Gamma)$ can be defined as the cohomology groups and homology groups of the classifying space $B\Gamma$. These are important invariants of $\Gamma$ useful in topology, number theory, representation theory and differential geometry.

We are interested in knowing the conditions under which good models (closed manifolds) exist for the classifying space $B\Gamma$. Furthermore, if $B\Gamma$ can be realized by a closed manifold, the Borel conjecture asks whether it is unique up to homeomorphism.

Large scale geometry The geometric group theory single-handedly established by Gromov asks how "big" $\Gamma$ is, roughly speaking, the growth rate of the volumes of the balls in $\Gamma$. For example, $\mathbb{R}^n$ is "smaller" than the Poincare upper half plane $(\mathbb{H},ds^2)$: the volume $\Vol(B(x_0, r))\sim c r^n$ in $\mathbb{R}^n$, but $\Vol(B(z_0, r))\sim ce^{1/4r}$ in $\mathbb{H}$ grows exponentially.

For $(\Gamma, d_S)$, we consider the asymptotic behaviors of the volume $\Vol(B(\gamma_0, r))=\# B(\gamma_0,r)$.

Example 9 For $\Gamma=\mathbb{Z}^n$, $\#B(\gamma_0, r)$ grows polynomially, but for $\Gamma=F_2$, it grows exponentially.

The following striking result due to Gromov relates the large scale geometry of a group to its own algebraic structure.

Theorem 2 (Gromov) If $\Gamma$ is a lattice of a nilpotent Lie group, then $\Gamma$ has a polynomial growth. Conversely, if $\Gamma$ has a polynomial growth, then $\Gamma$ is virtually nilpotent, namely there exists a finite index subgroup $\Gamma'$ of $\Gamma$ such that $\Gamma'$ is a lattice in a nilpotent Lie group.

Rigidity and action on manifolds Mostow strong rigidity, Margulis super-rigidity, Zimmer program, etc..

Construction and classification of infinite simple groups Margulis normal subgroup theorem and the rough classification up to quasi-isometry.

Theorem 3 (Margulis normal subgroup theorem) If $\Gamma$ is an irreducible lattice in a semisimple Lie group of rank $\ge2$. Then every normal subgroup of $\Gamma$ is finite or of finite index.

Now let us list some problems concerning a discrete transformation group $\Gamma$ acting on $X$.

Properness If $\Gamma$ acts on $X$ properly, the the quotient space $\Gamma\backslash X$ is a Hausdorff sapce. Nevertheless, We are also interested in non-proper actions such as ergodic actions (chaotic actions). They are important in Mostow rigidity, Margulis' work and also number theory.

Definition 14 Let $\mu$ be a measure on $X$ invariant under $\Gamma$, the action of $\Gamma$ is called ergodic if any invariant subset of $X$ is either of measure 0 or of full measure.

Orbits and quotient space Suppose $\Gamma$ acts properly on $X$, we would like to understand structures (geometry, topology, analsis) of orbits of $\Gamma$ in $X$ and the quotient $\Gamma\backslash X$.

  • Geometry: the geometry of locally symmetic spaces $\Gamma\backslash X$ when $X$ is a symmetic space;
  • Topology: often $\Gamma\backslash X$ provides classification spaces like the classifying spaces.
  • Analysis: spectual theorem of automorphic forms. The Selberg trace formular relates the geometry and analysis.

A crucial role is played by finding good fundamental domains of $\Gamma$.

Definition 15 A domain $\Omega$ in $X$ (somewhere between closed and open) is called a fundamental domain for $\Gamma$ in $X$ if $\bigcup_{\gamma\in\Gamma} \gamma\overline{\Omega}=X$ and the interiors of $\gamma\Omega^{\circ}$'s are disjoint (cf. Definition 8).

Usually we require that $\Omega$ is not too complicated, e.g., the boundary of $\Omega$ is small: $\overline{\Omega^{\circ}}=\overline{\Omega}$. If $X$ has a measure $v$ invariant under $\Gamma$, we usually require that $v(\partial\Omega)=0$. If $X$ is a manifold, we usually require that $\partial\Omega$ consists of submanifolds of smaller dimensions.

Definition 16 A fundamental domain $\Omega$ is called locally finite if for any $x\in X$, there exists a neighbourhood $U$ containing $x$ such that $U$ only meets finitely many $\Gamma$-translates of $\overline{\Omega}$. A fundamental domain $\Omega$ is called globally finite if $\{\gamma\in\Gamma: \gamma\overline{\Omega}\cap\overline{\Omega}\ne\varnothing\}$ is fintie.

Suppose $\Omega$ is a fundamental domain. Let $\pi: X\rightarrow\Gamma\backslash X$ be the projection. Then $\pi(\overline{\Omega})=\Gamma\backslash X$ is obtained by identifying some points on the boundary $\partial\overline{\Omega}$, namely we have a continuous bijection $\theta:\overline{\Omega}/\!\!\sim\rightarrow\Gamma\backslash X$. The significance of the local finiteness and global finiteness can be seen from the following two theorems.

Theorem 4 $\theta$ is a homeomorphism if and only if $\overline{\Omega}$ is locally finite.
Proof We omit the proof here.
Theorem 5 Suppose $X$ is a connected topological space and $\Gamma$ acts properly on $X$. Let $\Omega$ be a fundamental domain. Assume $\Omega$ is open and $\bigcup_{\gamma\in\Gamma}\gamma\Omega=X$. Let $I=\{\gamma\in\Gamma: \gamma\Omega\cap\Omega\ne\varnothing\}$. If $I$ is finite, then $\Gamma$ is finitely generated. Briefly speaking, if $\Gamma$ admits a globally finite fundamental domain then $\Gamma$ is finitely generated.
Proof Let $\Gamma^*$ be the subgroup generated by $I$. We claim that $\Gamma^*=\Gamma$. If not, then by the construction of $\Gamma^*$, $$X=\Big(\bigcup_{\gamma\in\Gamma}\gamma\Omega\Big)\cup\Big(\bigcup_{\gamma\in\Gamma^*-\Gamma}\gamma\Omega\Big)$$ decomposes as two disjoint open subsets, which contradicts the assumption that $X$ is connected.

Finding fundamental domains with global finiteness for lattices in Lie groups is called reduction theory. Legendre and Gauss started this theory while studying number-theoretic problems. The finiteness property is now called Siegel finiteness.

Example 10 Let $(X,d)$ be a proper (i.e., every bounded closed subset is compact) metric space. Let $\Gamma$ acts isometrically. We define the Dirichlet fundamental domain $$D_\Gamma(x_0)=\{x\in X: d(x,x_0)\le d(\gamma x,x_0), \forall\gamma\in\Gamma\}$$ for $x_0\in X$ not fixed by any nontrivial element of $\Gamma$. In other words, we pick the elements which are closest to $x_0$ in each $\Gamma$-orbit to form $D_\Gamma(x_0)$. It is clear that $\bigcup_{\gamma\in\Gamma}\gamma D_\Gamma(x_0)=X$, however, the boundary may be huge. Nevertheless, we know that $D_\Gamma(x_0)$ is a fundamental domain in the following special case.
Proposition 5 Let $X$ be a Riemannian manifold and $d$ be the Riemannian distance. Then $D_\Gamma(x_0)^{\circ}=\{x\in X: d(x,x_0)< d(\gamma x,x_0), \forall\gamma\in\Gamma\}$ and $D_\Gamma(x_0)$ is a fundamental domain.

We can also relax the requirement on the domain in the proof of Theorem 5.

Definition 17 A domain $S\subseteq X$ is called a rough (coarse) fundamental domain for $\Gamma$, if $\bigcup_{\gamma\in\Gamma}\gamma S=X$ and the induces map $\pi: S\rightarrow \Gamma\backslash X$ is finite-to-one, i.e., the size of the fibres are bounded.
Proposition 6 Let $X$ be a connected topological space. If $\Omega$ is a rough fundamental domain (open or closed) and $\{\gamma\in\Gamma: \gamma\Omega\cap\Omega\ne\varnothing\}$ if finite, then $\Gamma$ is finitely generated.
Proposition 7 $\Gamma\backslash X$ is compact if and only if there exists a compact rough fundamental domain.

TopFuchsian groups

We shall study the Fuchsian groups in the general framework of discrete subgroups of semisimple Lie groups.

Recall that $SL(2,\mathbb{R})$ acts on $\mathbb{H}^2$ holomorphically and isometrically by Mobius transformations. This action is transitive, i.e., for any $z_1, z_2\in \mathbb{H}$, there exists $g\in SL(2,\mathbb{R})$ such that $gz_1=z_2$. The stablizer of $i $ is $SO(2)$, hence $\mathbb{H}^2\cong SL(2,\mathbb{R})/SO(2)$. If $\Gamma$ is a discrete subgroup of $SL(2,\mathbb{R})$, then $\Gamma$ acts properly on $\mathbb{H}^2$.

Definition 18 Any discrete subgroup of $SL(2,\mathbb{R})$ acting on $\mathbb{H}^2$ is called a Fuchsian group.

Fuchsian groups divide into two types:

  1. the first kind: if the limit set of $\Gamma$ in the boundary $\mathbb{H}^2(\infty)$ is the whole boundary.
  2. the seconnd kind: if it is not of the first kind.
Proposition 8 If $\Gamma$ is a lattice, then $\Gamma$ is of the first kind.
Example 11 Let $\Gamma=SL(2,\mathbb{Z})$ be the modular group, then $\Gamma$ is a lattice, therefore it is a Fuchsian group of the first kind. Note that $\Vol(\Gamma\backslash SL(2,\mathbb{R}))$ is finite if and only if $\Vol(\Gamma\backslash\mathbb{H}^2)$ is finite. The latter also follows from the well-known explicit description of the fundamental domain of $SL(2,\mathbb{Z})$: $$\Omega=\{|\omega|>1, -1/2\le\Re \omega<1/2\}\cup \{ |\omega|=1, -1/2\le\Re w\le0\},$$ which is also a Dirichlet fundamental domain $D_\Gamma(iy)$ for $y\gg0$. In this case, the strip $$S=\left\{x+iy: -1/2\le x\le 1/2,y\ge\sqrt{3}/2\right\}$$ is a rough fundamental domain. Since the integration of $dxdy/y^2$ is bounded on $S$, we know that $\Gamma$ is a lattice. Also, Proposition 7 implies that $SL(2,\mathbb{Z})$ is not a uniform.
Remark 8 The well-known description of the fundamental domain of $SL(2,\mathbb{Z})$ appeared early in Gauss' Disquisitiones Arithmeticae when he studied the problem of representing integers by quadratic forms. The space of all positive definite quadratic forms corresponds to the space of all positive definite $2\times2$ matrices. Note that $SL(2,\mathbb{R})$ acts on all positive definite matrices of determinant 1 by $g.A=g^\mathrm{t}Ag$ transitively and the stabilizer of $I$ is $SO(2)$, so the space of all positive definite $2\times2$ matrices of determinant 1 corresponds to $SL(2,\mathbb{R})/SO(2)\cong\mathbb{H}^2$. For $\gamma\in SL(2,\mathbb{Z})$, the quadratic forms $Q(x,y)$ and $Q(\gamma(x,y))$ take the same set of values on integers $(x,y)\in \mathbb{Z}^2$. So this problem boils down to finding the simplest quadratic form in each $SL(2,\mathbb{Z})$ orbit. The reduction process goes by starting with any quadratic form and reducing it to the simplest one.

An important part of Minkowskii's geometry of numbers consists of reduction theory of finding fundamental domains of $SL(n,\mathbb{Z})$ on $SL(n,\mathbb{R})/SO(n)$ or $GL(n,\mathbb{Z})$ on $GL(n,\mathbb{R})/O(n)$.

Proposition 9 Assume $X$ is a complete Riemannian manifold, $\Gamma$ acts properly and isometrically on $X$, then the Dirichlet fundamental domain $D_\Gamma(x_0)$ is locally finite.

If $\Gamma$ is torsion-free, then $\Gamma$ acts freely on $\mathbb{H}^2$ and $\mathbb{H}^2\rightarrow \Gamma\backslash\mathbb{H}^2$ is a covering map. However, $SL(2,\mathbb{Z})$ is not torsion-free and some elements of $SL(2,\mathbb{Z})$ fix points in $\mathbb{H}^2$. Nevertheless, $SL(2,\mathbb{Z})$ admits finite index torsion-free subgroups. For such a torsion-free subgroup $\Gamma'$, $\Gamma'\backslash\mathbb{H}^2\rightarrow SL(2,\mathbb{Z})\backslash\mathbb{H}^2$ is a covering map from a smooth manifold to a orbifold. Minkowskii showed that when $N\ge3$, the principal congruence subgroup $\Gamma(N)$ is torsion-free.

The following more general result is due to Selberg.

Lemma 1 (Selberg's Lemma) If $\Gamma$ is a finitely generated subgroup of $GL(n,\mathbb{C})$, then $\Gamma$ admits a finite index torsion-free subgroup.

In particular, every finitely generated Fuchsian group admits a finite index torsion-free subgroup.

Remark 9 A Lie group $G$ is called linear if it can be embedded into some $GL(n,\mathbb{C})$. So a finitely generated subgroup $\Gamma$ of a linear Lie group $G$ admits a finite index torsion-free subgroup $\Gamma'$. However, if $G$ is not linear, this is not necessarily true.
Question When is a discrete subgroup $\Gamma$ of a Lie group $G$ finitely generated?
Theorem 6 (Siegel) If $\Gamma$ is a lattice of $SL(2,\mathbb{R})$, then $\Gamma$ is finitely generated.

More generally, we have:

Theorem 7 If $\Gamma$ is a lattice of a semisimple Lie group $G$, then $\Gamma$ is finitely generally.
Remark 10 $SL(2,\mathbb{R})$ is a semisimple Lie group, but has rather special properties.

We shall discuss the idea of the proof of Theorem 6.

Let $\Gamma\subseteq SL(2,\mathbb{R})$ be a Fuchsian group, then the Dirichlet fundamental domains $D_\Gamma(z_0)$ are bounded by geodesics because the bisector $\{z\in \mathbb{H}^2: d(z,z_0)=d(z,\gamma z_0)\}$ is a geodesic.

Definition 19 A Fuchsian group $\Gamma$ is called geometrically finite if it admits a fundamental domain that is bounded by finitely many geodesics (called a geodesic polygon).
Proposition 10 Let $\Gamma$ be a Fuchsian group. If $\Gamma$ is geometrically finite, then $\Gamma$ is finitely generated.
Proof Suppose $D$ is a fundamental domain of $\Gamma$ bounded by finitely many geodesics in $\mathbb{H}^2$, then there exists a pairing of sides of $D$ by elements of $\Gamma$. In other words, for any geodesic side $F $, there exists another side $F'$ and an element $\gamma\in\Gamma$ such that $\gamma(F')=F$. In fact, by definition, the $\Gamma$-translates of $D$ cover $\mathbb{H}^2$ without overlap in the interior. So there exists a translate $\gamma D$ such that $\gamma D\cap D=F$, which implies that there exists $F'$ such that $\gamma(F')=F$. Note that $\gamma^{-1}(F)=F'$, so this gives a pairing of all sides of $D$.

Let $\gamma_1,\ldots,\gamma_k$ be elements of $\Gamma$ that pairs the sides of $D$. We claim that these elements $\{\gamma_i\}$ generates the whole group $\Gamma$. In fact, we can reach the translate $\gamma D$ by a chain of neighbouring translates of $D$, which lies in $\{\gamma_i\}$.

The Siegel Theorem 6then follows from Proposition 10and the following theorem.

Theorem 8 If $\Gamma\subseteq SL(2,\mathbb{R})$ is a lattice, then every Dirichlet fundamental domain of $\Gamma$ in $\mathbb{H}^2$ has only finitely many geodesic sides, hence $\Gamma$ is geometrically finite.

We have proved that a geometrically finite Fuchsian group is finitely generates. The following converse is also true for Fuchsian groups.

Proposition 11 Let $\Gamma$ be a Fuchsian group. If $\Gamma$ is finitely generated, then it is geometrically finite.

The hyperbolic plane $\mathbb{H}^2$ is a two dimensional simply connected complete Riemannian manifold of constant curvature -1. For each dimension, we have a unique hyperbolic space $\mathbb{H}^n$ with this property, which is given by $$\{\mathbb{H}^n: (x_1, \ldots, x_n): x_1,\ldots, x_{n-1}\in \mathbb{R}, x_n>0\}$$ with metric $$ds^2=\frac{dx_1^2+\cdots dx_n^2}{x_n^2}.$$

Definition 20 If $\Gamma$ is a discrete group acting isometrically and properly discontinuously on $\mathbb{H}^3$, then $\Gamma$ is called a Kleinian group. If $\Gamma$ is torsion-free, then $\Gamma\backslash\mathbb{H}^3$ is a hyperbolic manifold.
Remark 11 It turns out that $SL(2,\mathbb{C})$ acts isometrically and transitively on $\mathbb{H}^3$ and $\Iso^\circ(\mathbb{H}^3)=PSL(2,\mathbb{C})$. So Kleinian groups can be also defined as discrete subgroups of $SL(2,\mathbb{C})$.

We can define similar notions of Kleinian groups like Dirichlet fundamental domains (bounded by geodesic hypersurfaces) and geometrically finiteness. However, unlike the case of Fuchsian groups, a Kleinian group $\Gamma$ is finitely generated implies neither its Dirichlet fundamental domains are bounded by finitely many geodesic sides nor $\Gamma$ is geometrically finite. So Proposition 11 is special for Fuchsian groups.

Question How to construct Fuchsian groups?

The first geometric method due to Poincaré is the reverse procedure of obtaining a fundamental domain for a group $\Gamma$ (see [2] for details).

Theorem 9 (Poincaré polygon theorem) Start with a connected geodesic polygon $D$ in $\mathbb{H}^2$ and a pairing of geodesic sides of $D$ by elements of $SL(2,\mathbb{R})$, then under a suitable conditions, these side-pairing elements generate a Fuchsian group such that $D$ is a fundamental domain.

The second algebraic method depends on the notion of arithmetic subgroups.

Definition 21 A subgroup $\Gamma$ of $SL(2,\mathbb{Q})$ is called an arithmetic subgroup if $\Gamma$ is commensurable with $SL(2,\mathbb{Z})$. For an arithmetic subgroup $\Gamma$, we get a finite common cover $\Gamma\cap SL(2,\mathbb{Z})\backslash\mathbb{H}^2$ for $\Gamma\backslash\mathbb{H}^2$ and $SL(2,\mathbb{Z})\backslash\mathbb{H}^2$. The examples of arithmetic subgroups of $SL(2,\mathbb{Q})$ include congruence subgroup of $SL(2,\mathbb{Z})$ and finite index subgroup of $SL(2,\mathbb{Z})$.
Question Is every arithmetic subgroup of $SL(2,\mathbb{Z})$ a congruence subgroup?
Answer This a major problem for arithmetic groups. Fricke-Klein solved this problem and the anser is NO. There are infinitely many subgroups of finite index in $SL(2,\mathbb{Q})$ that are not congruence subgroups. However, the answer to the same question for $SL(n,\mathbb{Z})$ ($n\ge3$) is YES due to Bass-Milnor-Serre.
Question $SL(2,\mathbb{Z})$ is not uniform. Fuchsian groups commensurable with $SL(2,\mathbb{Z})$ i.e., arithmetic subgroups, are not uniform as they share a common cover with $SL(2,\mathbb{Z})\backslash \mathbb{H}^2$. Is there any algebraic method to construct uniform Fuchsian groups?

So far, we have restricted to subgroups of $SL(2,\mathbb{Q})$. In order to construct unifomr Fuchsian groups algebraically, we need to find discrete subgroups that are not commensurable to $SL(2,\mathbb{Z})$. Let $\Lambda_1, \Lambda_2\subseteq \mathbb{R}^n$ be two lattices, then $\Lambda_1, \Lambda_2$ are commensurable if and only if they define the same $\mathbb{Q}$-structure on $\mathbb{R}^n$, i.e., $\Lambda_1\otimes_\mathbb{Z}\mathbb{Q}=\Lambda_2\otimes_\mathbb{Z}\mathbb{Q}$. So to construct uniform lattices of $SL(2,\mathbb{R})$, we need a different $\mathbb{Q}$-structure on $SL(2,\mathbb{R})$, namely quaternion division algebras over $\mathbb{Q}$. Those Riemann surfaces constructed in this way are called Shimura curves. André Weil proved that these together with the arithmetic subgroups of $SL(2,\mathbb{Q})$ exhaust all the arithmetic subgroups of $SL(2,\mathbb{R})$.

To define the general notion of arithmetic subgroups of $SL(2,\mathbb{R})$ (or a real Lie group), we need to introduce the notion of algebraic groups.

Definition 22 A subgroup $\underline{G}\subseteq GL(n,\mathbb{C})$ is called an algebraic group if $\underline{G}$ is subvariety, i.e., it is a closed subset defined by a set of polynomial equations. $\underline{G}$ is called to be defined over $\mathbb{Q}$ if the coefficients of the defining polynomials are in $\mathbb{Q}$.

Assume $\underline{G}$ is a linear algebraic group defined over $\mathbb{Q}$. Weil showed that $G=\underline{G}(\mathbb{R}):=\underline{G}\cap GL(n,\mathbb{R})$ is a Lie group with finitely many connected components.

Definition 23 A subgroup $\Gamma\subseteq\underline{G}(\mathbb{Q})$ is called an arithmetic subgroup if $\Gamma$ is commensurable with $\underline{G}(\mathbb{Z})$.
Remark 12 A $\mathbb{Q}$-structure on a Lie group $G$ is essentially a linear algebraic group $\underline{G}$ such that $\underline{G}(\mathbb{R})=G$.
Theorem 10 (Borel, Harish-Chandra) If $G$ is a semisimple Lie group, then any arithmetic subgroup $\Gamma$ of $G$ is a lattice.
Proposition 12 Let $\Gamma\subseteq SL(2,\mathbb{R})$ be a lattice, then $\Gamma$ is not uniform if and only if $\Gamma$ contains unipotent elements.
Remark 13 The same thing holds for lattices in semisimple Lie groups. This is a conjeture of Selberg and it is proved by Kazhdan-Margulis.
Question Is every lattice in a semisimple lie group $G$ an arithmetic subgroup with respect to a suitable $\mathbb{Q}$-structure on $G$?

The answer is NO for $G=SL(2,\mathbb{R})$ by the Teichmuller theory. In almost all other cases, the answer is YES due to Margulis, which relies on a long line of his work on the rigidity of discrete subgroups. More precisely, we have:

Theorem 11 (Arithmeticity theorem) Suppose $G$ is a semisimple Lie group without compact factor of rank $\ge2$, then any irreducible lattice of $G$ is an arithmetic subgroup.


[1]Siegel, C.L., Discontinuous groups, Annals of Mathematics 44 (1943), no.4, 674--689.

[2]Alan F. Beardon, The Geometry of Discrete Groups (Graduate Texts in Mathematics) (v. 91), Springer, 1983.