5 Tits Alternative

5.1 Statements

Theorem 5.1 (Tits Alternative) Let $G$ be a finitely generated linear group then $G$ contains a free subgroup or $G$ is virtually solvable.

This alternative implies that a linear group is amenable for a simple reason (it is virtually solvable) or it is not amenable for a simple reason (it contains a free subgroup).

Let us recall that a linear group is a subgroup $G$ of some $\mathrm{GL}_n(k)$ where $n$ is some integer and $k$ any field. In some case, like $k=\mathbb{C}$, the assumption that $G$ is finitely can be removed. The proof depends on the field and to avoid number theoretical difficulties, we will prove the following. Up to replacing $\mathbf{R}$ by another locally compact field, the proof of the Tits alternative reduces (if the group is not virtually solvable) to prove the following.

Theorem 5.2 Let $G$ be a unbounded subgroup of $\mathrm{SL}_n(\mathbf{R})$ whose Zariski closure acts irreductibly on $\mathbf{R}^n$ then $G$ contains a free groups on two generators.

Let us recall that a subgroup $G\leq\mathrm{SL}_n(\mathbf{R})$ acts irreductibly on $\mathbf{R}^n$ if there is non-trivial $G$-invariant subspace of $\mathbf{R}^n$.

A subgroup $H$ of $\mathrm{SL}_n(\mathbf{R})$ is algebraic (or Zariski closed) if there is a familly of polynomial maps $P_i\colon \mathrm{M}_n(\mathbf{R})\to \mathbf{R}$ such that $g\in H\iff P_i(g)=0\ \forall i$. The Zariski closure of subgroup $G$ is the intersection of all algebraic subgroups that contains $G$.

Remark. For a linear space $V$ of dimension $n$, one can think the notion of algebraic group depends of a choice of an identification $V\simeq\mathbf{R}^n$, i.e. the choice of a base. Since change of bases corresponds to conjugation for matrices and conjugation are linear maps, this notion of algebraic groups do not depend of the choice of a base.

Example 5.1 The following groups are algebraic:

• $\mathrm{SO}(p,q)$
• $\mathrm{SO}(n)$
• $\mathrm{SL}_n(\mathbf{R})$
• stabilizers of subspaces of $\mathbf{R}^n$.

An algebraic group $H$ is connected if it can’t be written as the union of two proper Zariski closed subspaces. This strange definition is equivalent to connectedness for the Zariski topology.

5.2 Cartan decomposition and contracting projective transformations

Proposition 5.1 (Cartan decomposition) Let $g\in\mathrm{SL}_n(\mathbf{R})$. There exist $k_1,k_2\in\mathrm{SO}_n(\mathbf{R})$ and $a\in A^+$ such that

$$g=k_1ak_2$$

where $A^+=\{a=\mathrm{diag}(a_1,\dots,a_n),\ a_1\geq a_2\geq \dots\geq a_n>0\}$.

Moreover, for two such decompositions, the $a_i$ are the same and we denote them by $a_i(g)$.

Proof. The matrix $^tgg$ is symmetric definite positive, so there is $k\in\mathrm{SO}(n)$ such that $^tk^tggk$ is $\mathrm{diag}(b_1,b_2,\dots,b_n)$ where the $b_i's$ are positive and in decreasing order. Let $a_i=\sqrt{b_i}$ and $a=\mathrm{diag}(a_1,a_2,\dots,a_n)$ then $^t(a^{-1})^tk^tggka^{-1}$ is the identity, i.e. $gka^{-1}\in\mathrm{SO}(n)$, so, there is $k_2\in\mathrm{SO}(n)$ such that $g=k_2ak^{-1}$. By setting $k_1=k^{-1}$, we have the desired decomposition.

One should observe that the $a_i$’s are the square roots of the eigenvalues of $^tgg$ and thus unique.

Observe that $a_1(g)=||g||$ (the operator norm associated to the Euclidean norm on $\mathbf{R}^n$) and thus if $||g_m||\to+\infty$ then $a_1(g_m)\to\infty$.

Let $d\leq n$ and $g\in\mathrm{SL}_n(\mathbf{R})$. The linear map $g$ induces a linear map $\overline{g}\in\mathrm{SL}(\wedge^d \mathbf{R}^n)$ given by $\overline{g}(e_{i_1}\wedge\dots\wedge e_{i_d})=ge_{i_1}\wedge\dots\wedge ge_{i_d}$ for $i_1,\dots ,i_d$ distincts in $\{1,\dots,n\}$.

Lemma 5.1 Let $g_m$ be an unbounded sequence of elements of $\mathrm{SL}_n(\mathbf{R})$ then there some $l$ such the action if $\overline g_m$ is the image of $g_m$ for the action on $\wedge^l \mathbf{R}^n$ then up to extraction $\alpha_1(\overline g_m)/\alpha_2(\overline g_m)\to \infty$.

On the projective space $P(\mathbf{R}^n)$, for two lines $l_1,l_2$, we define $d(l_1,l_2)=\sin(\alpha)$ where $\alpha$ is the geometric angle between the two lines. For a metric space $(X,d)$ and a subspace $Y\subset X$, we denote by $Y_\varepsilon$, the $\varepsilon$-neighborhood of $Y$.

Definition 5.1 Let $g$ be a projective transformation of $P(\mathbf{R}^n)$ and $\varepsilon>0$. This transformation is said to be $\epsilon$-contracting if there a point $v\in P(\mathbf{R}^n)$ and a hyperplane $H\subset P(\mathbf{R}^n)$ such that $g(P(\mathbf{R}^n)\setminus H_\varepsilon)\subset v_\varepsilon$.

Lemma 5.2 If $g$ is $\varepsilon$-contracting then for any $k_1,k_2\in\mathrm{SO}(n)$ $k_1gk_2$ is $\varepsilon$-contracting as well.

Proposition 5.2 Let $g\in \mathrm{SL}(V)$. For any $\varepsilon>0$, there is $M>0$ such that if $a_1(g)/a_2(g)>M$ then the projective transformation associated to $g$ is $\varepsilon$-contracting and $\varepsilon$-Lipschitz outside $H_\varepsilon$ .

Lemma 5.3 Let $g\in \mathrm{SL}(V)$ and $\delta>0$ then there is an open subset $O\subset P(V)$ such that $g|_O$ is $(1+\delta)$-Lipschitz.

Proposition 5.3 For every $\varepsilon>0$ there is $\delta>0$ such that if $g\in \mathrm{SL}(V)$ is $\delta$-Lipschitz on some open subspace of $P(V)$ then the projective transformation associated to $g$ is $\varepsilon$-contracting.

5.3 Separating points and hyperplanes

Definition 5.2 A subset $F\subset \mathrm{SL}(V)$ is $(k,r)$-separating (where $k\in\mathbf{N}$, $r>0$) if for any hyperplanes $H_1,\dots,H_k$ in $P(V)$ and $v_1,\dots,v_k\in P(V)$ there is $f\in F$ such that

$$d(fv_i,H_j)>r$$ and $$d(f^{-1}v_i,H_j)>r$$

for all $i,j\in\{1,\dots,k\}$.

Proposition 5.4 Let $G$ be a subgroup of $SL(V)$ with a connected Zariski closure that acts irreductibly on $V$ then for any $k$, there is $r>0$ such that $G$ contains a finite $(k,r)$-separating subset.

Proof. Let $H$ be the Zariski closure of $G$ and let $H_1,\dots, H_k$ be hyperplanes in $P(V)$ and $v_1,\dots, v_k$ points in $P(V)$. We define

$$V_{i,j}^+=\{h\in H,\ hv_i\in H_j\}$$

and

$$V_{i,j}^-=\{h\in H,\ h^{-1}v_i\in H_j\}$$ .

These subsets are algebraic subsets. Define $V_{i,j}=V_{i,j}^+\cup V_{i,j}^-$. Since $H$ is irreducible $\cup_{i,j}V_{i,j}\subsetneq H$. Since $G$ is Zariski dense, there is $\gamma\in H\setminus \cup_{i,j}V_{i,j}$.

Now, define for $\gamma\in G$,

$$X_\gamma=\{(H_1,\dots,H_k,v_1,\dots,v_k),\ \gamma v_i, \gamma^{-1} v_i\notin H_j,\ \forall i,j\}.$$

Let $X=\mathrm{Grass}_{n-1}(V)^k\times P(V)^k$. We proved that $X=\cup_{\gamma\in G}X_\gamma$. Since each $X_\gamma$ is open, by compactness of $X$, there is $F\subset G$ finite such that $X=\cup_{\gamma\in F}X_\gamma$. Now let

$r=\max_{\gamma\in F}\min_{i,j}\{d(\gamma v_i, H_j),d(\gamma^{-1}v_i, H_j)\}$. Now, $F$ is the desired $(k,r)$-subset of $G$.

5.4 Completion of the proof

In this subsection we prove Theorem 5.2 and we will need the following definition.

Definition 5.3 Let $\varepsilon,r>0$. A projective map $g\colon P(V)\to P(V)$ is $(\varepsilon,r)$-proximal if it $\varepsilon$-contracting with respect to a pair $(v,H)$ with $d(v,H)>r$.

Let’s start the proof. So, let $G$ be an unbounded subgroup of $SL_n(\mathbf{R})$ whose Zariski closure is connected and acts irreducibly on $\mathbf{R}^n$. We know by Proposition 5.4 that there is some finite set $F\subset G$ that is $(2,r)$-separating.

Let $C$ be an upper bounds for Lipschitz constants of elements of $F^{\pm 1}$ and $\delta\in(0,r)$. We know there is $\gamma$ that is $\delta$-contracting outside $H_\delta$ for some hyperplane $H$. We proved that $\gamma^{-1}$ is 2-Lipschitz on some open subset $O\subset P(\mathbf{R}^n)$. Let $v\in O$ and $f\in F$ such that $f(\gamma^{-1}v)\notin H_r$ and $f^{-1}(\gamma^{-1}v)\notin H_r$. Now let $\gamma_0=\gamma f\gamma^{-1}$. It is $2C\delta$-Lipschitz and its inverse as well.

Let $\varepsilon>0$. By choosing $\delta$ small enough, this implies that $\gamma_0^{\pm 1}$ are $\varepsilon$-contracting with respect to some $(v_0^+,H_0^+)$ and $(v_0^-,H_0^-)$.

By separation, there is $f_0$ such that

$$d(f_0v_0^+,H_0^+)>r$$

and $$d(f_0^{-1}v_0^-,H_0^-)>r.$$ Since $f_0^{-1}$ is $C$-Lipschitz, $d(v_0^-,f_0H_0^-)>r/C$.

For $x\notin (H_0^+)_\varepsilon$, since $f$ is $C$-Lipschitz

$$d(f_0\gamma_0x,f_0v_0^+)\leq Cd(\gamma_0 x,v_0^+)\leq C\varepsilon.$$

So if we set $\gamma_1=f_0\gamma_0$, it is $(C\varepsilon, r)$-proximal with respect to $(f_0v_0^+, H_0^+)$.

Its inverse $\gamma^{-1}=\gamma_0^{-1}f_0^{-1}$ is $(C\varepsilon, r/C)$-proximal with respect to $(v_0^-,f_0H_0^-)$. Actually, if $x\notin (f_OH_0^-)_{C\varepsilon}$ then

$$d(\gamma_0^{-1}f_0^{-1}x,v_0^-)<\varepsilon$$ since $\gamma_0^{-1}$ is $\varepsilon$-contracting with respect to $(v_0-, H_0^-)$ and we have proved that $d(v_0^-,f_OH_0^-)>r/C$.

Set $v_1^+=f_0v_0^+$, $H_1^+=H_0^+$, $v_1^-=v_0^-$ and $H_1^-=f_OH_0^-$. So $\gamma_1$ is $(C\varepsilon, r)$-proximal with respect to $(v_1^+, H_1^+)$ and $\gamma^{-1}$ is $(C\varepsilon, r/C)$-proximal with respect to $(v_1^-,H_1^-)$.

By separation, find $f_1\in F$ such that

$$d(f_1v_1^+,H_1^+\cup H_1^-)>r$$ and

$$d(f_1v_1^-,H_1^+\cup H_1^-)>r$$ .

Set $\gamma_2=f_1\gamma_1f_1^{-1}$. By similar computations, one can prove that it is $(C^2\varepsilon,R/C)$-proximal with respect to $(f_1v_1^+,f_1H_1^+)$. Similarly, $\gamma_2^{-1}$ is $(C^2\varepsilon,r/C)$-proximal with respect to $(f_1v_1^-,f_1H_1^-)$.

Set $v_2^\pm$ and $H_2^\pm$ and we have

$$d(v_2^\pm,H_1^+\cup H_1^-)>r$$

and

$$d(v_1^\pm, H_2^+\cup H_2^-)>r/C.$$

So for $\varepsilon$ small enough, one can play ping-pong.

5.5 Exercises

Exercise 5.1 Prove that the Zariski closure of a group is also a group.

Exercise 5.2 Prove that any algebraic group is a Lie group.

Exercise 5.3 Prove that a Zariski closed set has finitely many irreducible components. Prove that if $G$ is an algebraic subgroup then the identity is contained in a unique component and this component is open. Prove that this component is a normal finite index subgroup $G^0$ that coincides with the connected component of the identity for the Zariski topology. This component is called the identity component of $G$.

What is the identity component (in the algebraic sense) of $\mathrm{GL}_n(\mathbf{R})$ ? Compare it with the identity component for the usual topology.

Prove that any projective transformation is Lipschitz.