Uniform Tits Alternative III: the proof of the Uniform Height Gap

Today we will finish what we started and prove the height gap theorem. We will first reduce to the case of a 2-element set, for which we recall some terminology. Let $\mathbb{G}$ be a connected semisimple algebraic group over $\overline{\mathbb{Q}}.$ A semisimple group element $a\in \mathbb{G}(\overline{\mathbb{Q}})$ is said to be regular if $\ker(Ad(a)-1)$ has the minimal possible dimension (namely equal to the absolute rank of $\mathbb{G}$). For $A_{1}\in \mathbb{N},$ we will say that $a\in \mathbb{G}(\overline{\mathbb{Q}})$ is $A_{1}$-regular if $\ker (Ad(a)-\omega )$ has minimal possible dimension for every root of unity $\omega $ of order at most $A_{1}$ (namely dimension $0$ if $\omega \neq 1$ and the absolute rank if $\omega =1$). It is clear that the subset of $A_{1}$-regular elements of $\mathbb{G}$ is a non-empty Zariski open subset of $\mathbb{G}$ consisting of semisimple elements.

If $Z$ is a proper Zariski closed subset of $\mathbb{G}$ invariant under conjugation by a maximal torus $T,$ then we let $\widehat{Z}$ be the Zariski-closure of $\{(gag^{-1},gbg^{-1})\in \mathbb{G}^{2}$ with $g\in \mathbb{G}$, $a\in T$ and $b\in Z,$ or $a\in Z$ and $b\in T\}.$ It is a proper algebraic subset of $\mathbb{G\times G}$ of dimension at most $2\dim\mathbb{G}-1.$

Proposition 17: Let $\mathbb{G}$ be a connected semisimple algebraic subgroup of $GL_{d}(\overline{\mathbb{Q}})$ with maximal torus $T$. Let $Z$ be a proper Zariski closed subset of $\mathbb{G}$ invariant under conjugation by $T $. Then there is an integer $c=c(\mathbb{G},Z)>0$ such that if $F$ is a finite subset of $\mathbb{G}(\overline{\mathbb{Q}})$ generating a Zariski-dense subgroup in $\mathbb{G}$, then $(F\cup
\{1\})^{c(d)}$ contains two elements $a$ and $b$ which are regular semisimple, generate a Zariski dense subgroup of $\mathbb{G}$, and satisfy $(a,b)\notin \widehat{Z}.$ For any given integer $A_{1}\in \mathbb{N}$, by allowing $c$ to depend also on $A_{1},$ i.e. $c=c(\mathbb{G},Z,A_{1})>0,$ we may further assume that $a$ and $b$ are $A_{1}$-regular.

This is an immediate consequence of the escape lemma and Proposition 19


The key ingredient is the following lemma. For an algebraic variety $X$ we will denote by $m(X)$ the sum of the degree and the dimension of each of its irreducible components.

Eskin-Mozes-Oh escape lemma: Given an integer $m\geq 1$ there is $N=N(m)$ such that for any field $K$, any integer $d\geq 1$, any $K$--algebraic subvariety $X$ in $GL_{d}(K)$ with $m(X)\leq m$ and any subset $F\subset GL_{d}(K)$ which contains the identity and generates a subgroup which is not contained in $X(K)$, we have $F^{N}\nsubseteq X(K)$.

This result is a consequence of a generalized version of Bezout's theorem about the intersection of finitely many algebraic subvarieties:

Generalized Bezout Theorem: Let $K$ be a field, and let $Y_{1},\ldots ,Y_{p}$ be pure dimensional algebraic subvarieties of $K^{n}$. Denote by $W_{1},\ldots ,W_{q} $ the irreducible components of $Y_{1}\cap \ldots \cap Y_{p}$. Then $\sum_{i=1}^{q}$deg$(W_{i})\leq \prod_{j=1}^{p}$deg$(Y_{j}).$

To prove lemma 19 below we need
Lemma 18: The set $U$ of regular semisimple elements of $\mathbb{G}$ is a non-empty Zariski-open subset of $\mathbb{G}$.
Proof: The set $U$ coincides with the set of $g\in \mathbb{G}$ such that $\ker (Ad(g)-1)$ is of minimal dimension. This is clearly a Zariski-open condition.

Proposition 19: Let $\mathbb{G}$ be a connected semisimple algebraic group over $\mathbb{C}$. There is a proper algebraic subvariety $X$ of $\mathbb{G\times G}$ such that any pair $(x,y)\notin X$ is made of regular semisimple elements which generate a Zariski-dense subgroup of $\mathbb{G}$.


Proof: We will make use of Jordan's theorem on finite subgroups of $GL_{d}(\mathbb{C})$. See here for Breuillard's discussion of this and the Tits alternative. Recall that according to this theorem, there is a constant $C=C(d)\in \mathbb{N},$ such that if $\Gamma $ is a finite subgroup of $GL_{d}(\mathbb{C}),$ then $\Gamma $ contains a abelian subgroup $A$ with $[\Gamma :A]\leq C(d).$ As the kernel of the adjoint representation coincides with the center of $\mathbb{G}$, it follows that the same bound apply for all finite subgroups of $\mathbb{G}(\mathbb{C})$ as long as $\dim (\mathbb{G})\leq d.$ Let $V(\mathbb{G})$ be the proper Zariski-closed subset of $\mathbb{G\times G}$ consisting of all couples $(x,y)$ such that $[x^{C!},y^{C!}]=1.$ By Jordan's theorem, if $(x,y)\notin V,$ then the subgroup generated by $x$ and $y$ infinite.

Let $(\mathbb{G}_{i})_{1\leq i\leq k}$ be the $\mathbb{C}$-simple factors of $\mathbb{G}$, together with their factor maps $\pi _{i}:\mathbb{G\rightarrow G}_{i}.$ For convenience, let us denote $\mathbb{G}_{0}=\mathbb{G}$. Let $X_{i}$, for $0\leq i\leq k$, be the subset of $\mathbb{G}\times \mathbb{G}$ consisting of couples $(x,y)$ such that the $\mathbb{C}$-subalgebra of $End(\mathfrak{g}_{i})$ generated by $Ad(\pi _{i}(x))$ and $Ad(\pi _{i}(y))$ is of strictly smaller dimension than the subalgebra generated by the full of $Ad(\mathbb{G}_{i}),$ where $\mathfrak{g}_{i}$ is the Lie algebra of $\mathbb{G}_{i}.$ This is a Zariski-closed subset of $\mathbb{G}\times \mathbb{G}.$
It is known/ is an exercise in Bourbaki that each $\mathfrak{g}_{i}$ is generated by two elements. If follows that $X_{i}$ is a proper closed subvariety. Also let $V_{i}$ be the set of couples $(x,y)\in \mathbb{G\times G}$ such that $(\pi _{i}(x),\pi _{i}(y))\in V(\mathbb{G}_{i}),$ where $V(\mathbb{G}_{i})$ is the proper closed subset defined above.

Finally, let $X$ be the proper closed subvariety $X=U^{c}\cup \bigcup_{i}X_{i}\cup \bigcup_{i}V_{i}.$ Let us verify that $X$ satisfies the
conclusion of the proposition. Suppose $(x,y)\notin X.$ Then $(x,y)\in U$ and $x$,$y$ are regular semisimple. Let $\mathbb{H}$ be the Zariski closure of the group generated by $x$ and $y.$ Let $\mathfrak{h}_{i}$ be the Lie algebra of $\pi _{i}(\mathbb{H}),$ which is a Lie subalgebra of $\mathfrak{g}_{i}.$ As $\mathfrak{h}_{i}$ is invariant under $Ad(\pi _{i}(x))$ and $Ad(\pi _{i}(y)),$ it must be invariant $Ad(\mathbb{G}_{i})$, by the assumption that $(x,y)\notin X_{i}.$
Therefore $\mathfrak{h}_{i}$ is an ideal of $\mathfrak{g}_{i}.$ As $\mathfrak{g}_{i}$ is a simple Lie algebra, either $\mathfrak{h}_{i}=\{0\}$ or $\mathfrak{h}_{i}=\mathfrak{g}_{i}.$ In the former case, this means that $\pi _{i}(\mathbb{H})$ is finite. However,
by assumption $(\pi _{i}(x),\pi _{i}(y))\notin V(\mathbb{G}_{i}),$ this means that the group generated by $\pi _{i}(x)$ and $\pi _{i}(y)$ is
infinite. So $\pi _{i}(\mathbb{H})$ is not finite, $\mathfrak{h}_{i}=\mathfrak{g}_{i}$ and $\pi _{i}(\mathbb{H})=\mathbb{G}_{i}.$

On the other hand, since $(x,y)\notin X_{0}$, the same argument shows that the Lie algebra of $\mathbb{H}$ itself is an ideal in $\mathfrak{g.}$ Hence $\mathbb{H}^{\circ }$ is a normal subgroup of $\mathbb{G}$, hence is the product of the simple factors of $\mathbb{G}$ contained in it. The fact that $\pi _{i}(\mathbb{H})=\mathbb{G}_{i}$ for each $i$ forces $\mathbb{H=G}$. $\blacksquare$

Reduction to semisimple $\mathbb{G}$

Proposition 20: In order to prove the Height Gap Theorem, it is enough to prove the
following assertion. There is $\varepsilon =\varepsilon (d)>0$ such that :if $\mathbb{G}\subseteq SL_{d}$ is a semisimple algebraic group over $\overline{\mathbb{Q}}$ acting irreductibly on $\overline{\mathbb{Q}}^{d}$, and $F=\{Id,a,b\}$ is a subset of $\mathbb{G}$ generating a Zariski-dense subgroup, then $e(F)>\varepsilon (d).$

The proof of this will rest mainly on the following proposition:

Proposition 21: There are constants $C=C(d)>0$ and $m=m(d)\in \mathbb{N}$
such that if $F$ is a finite subset of $GL_{d}(\overline{\mathbb{Q}})$
containing $1$ and generating a non virtually soluble subgroup, there
exists a subset $F_{1}\subset F^{m}$, a connected semisimple algebraic group
$\mathbb{H}$ together with a faithful irreducible representation $(\rho
_{0},V_{0})$ of $\mathbb{H}$ with $\dim V_{0}\leq d$ and a homomorphism $\pi
:\Gamma _{0}\rightarrow \mathbb{H}(\overline{\mathbb{Q}})$, where $\Gamma
_{0}$ contains $F_{1}$ and has index at most $m$ in $\Gamma =\left\langle
F\right\rangle $, such that $\pi (\Gamma _{0})$ is Zariski dense in $\mathbb{%
H}$ and

\[e(\rho _{0}\circ \pi (F_{1}))\leq C(d)\cdot e(F). \]

The point of the proof of proposition 20 is that the gap can be proved for $e(F)$, and then the adjoint matrix $Ad(F_1)$ has determinant one and generates a nonvirtually soluble subgroup. $\hat{h}$ is bounded away from 0 if and only if $e$ is, but $e(Ad(F_1)) \geq e(\rho((Ad(F_1)) ,$ and $\rho(Ad(F_1))$ generates a Zariski dense subgroup of a semisimple algebraic group. Now apply Proposition 17. 

At long last, the proof

Before beginning the proof of the Height Gap Theorem, we state a couple of results we will need, notably Zhang's theorem on small points of algebraic tori. Let $\mathbb{G}_{m}$ be the multiplicative group and $n\in \mathbb{N}$. On the $\overline{\mathbb{Q}}$-points of the torus $\mathbb{G}_{m}^{n}$ we define a notion of height in the following natural way. If $\mathbf{x}=(x_{1},...,x_{n})\in \mathbb{G}_{m}^{n}$ then $h(\mathbf{x}):=h(x_{1})+...+h(x_{n})$ where $h(x_{i})$ is the standard logarithmic Weil height we have been using so far.

Zhang's theorem: Let $V$ be a proper closed algebraic subvariety of $\mathbb{G}_{m}^{n}$ defined over $\overline{\mathbb{Q}}.$
Then there is $\varepsilon >0$ such that the Zariski closure $V_{\varepsilon}$ of the set $\{\mathbf{x}\in V$, $h(\mathbf{x})<\varepsilon \}$ consists of a finite union of torsion coset tori, i.e. subsets of the forms $\mathbf{\zeta }H$, where $\mathbf{\zeta }=(\zeta _{1},...,\zeta _{n})$ is a torsion point and $H$ is a subtorus of $\mathbb{G}_{m}^{n}.$


Let $\mathbb{G}$ be a semisimple algebraic group over an algebraically closed field, $T$ a maximal torus together with a choice of simple roots $\Pi $, and $f$ is the regular function defined for proposition 11 in the last post:

Lemma 22: For every $k\in \mathbb{N}$, the regular functions $f_{1},...,f_{k}$ defined on $\mathbb{G}$ by $f_{i}(g)=f(g^{i})$ are multiplicatively independent. Namely, if for each $i,$ $n_{i}$ and $m_{i}$ are non-negative integers and the $f_{i}$'s satisfy an equation of the form $f_{1}^{n_{1}}\cdot ...\cdot f_{k}^{n_{k}}=f_{1}^{m_{1}}\cdot ...\cdot f_{k}^{m_{k}}$ then $n_{i}=m_{i}$ for each $i$.


Proof: To prove this lemma it is enough to show that for each $i$ one can find a group element $g\in \mathbb{G}$ such that $f_{i}(g)=0$ while all other $f_{j}(g)$'s are non zero. Let $H$ be the copy of $PGL_{2}$ corresponding to the roots $\alpha =\alpha _{d}$ and $-\alpha =\alpha _{1}$ with Lie algebra $\mathfrak{h}$ generated by $X_{\alpha },$ $X_{-\alpha }$ and $H_{\alpha }.$
Clearly it is enough to prove the lemma for the restriction of $f$ to $H$. Therefore without loss of generality we may assume that $\mathbb{G}=PGL_{2}$, hence $f(g)=a^{2}$ if $g=\left(
\begin{array}{ll}
a & b \\
c & d%
\end{array}%
\right) \in PGL_{2}.$ Let for instance $D_{\lambda }=\left(
\begin{array}{ll}
\lambda & 0 \\
0 & \lambda ^{-1}%
\end{array}%
\right) \in PGL_{2}$ and $P=\left(
\begin{array}{ll}
1 & 1 \\
1 & 2%
\end{array}%
\right) .$ Set $g_{\lambda }=PD_{\lambda }P^{-1}.$ Then compute $%
f(g_{\lambda })=2\lambda -\lambda ^{-1}$ and $f_{i}(g_{\lambda
})=f(g_{\lambda ^{i}}).$ Hence $f_{i}(g_{\lambda })=0$ if and only if $%
2\lambda ^{2i}=1.$ These conditions are mutually exclusive for distinct
values of $i$. So we are done. $\blacksquare$

At long last, we are at the end.

Proof of the height gap theorem: According to the reductions made in we may assume that $F\subset \mathbb{G}(\overline{\mathbb{Q}})$ where $\mathbb{G}$ is a connected absolutely almost simple algebraic group $\mathbb{G}$ of adjoint
type (viewed as embedded in $GL(\mathfrak{g})$ via the adjoint representation), and that the group $\left\langle F\right\rangle $ is
Zariski dense in $\mathbb{G}$. Let $T$ be a maximal torus in $\mathbb{G}$ and $\Phi $ be the corresponding set of roots with set of simple roots $\Pi $and let $\alpha _{1}=-\alpha _{d}$ be the highest root. The function $f\in\overline{\mathbb{Q}}[\mathbb{G}]$ was defined at the before Proposition 11 by $f(g)=g_{dd}$ where $\{g_{ij}\}_{1\leq i,j\leq d}$ is thematrix of $Ad(g)$ in the Chevalley basis $(Y_{1},...,Y_{d})$. Let $ f_{i}(g)=f(g^{i})$ and let $\Omega $ be the Zariski open subset of $\mathbb{G}$ defined by $\{g,~f_{i}(g)\neq 0$ for each $i\leq d+1\}$. Let $\mathbf{f}$ be the regular map $\mathbf{f}(g):=(f_{1}(g),...,f_{d+1}(g))$ $:$ $\Omega \rightarrow \mathbb{G}_{m}^{d+1}$. Since $d=\dim \mathbb{G}$, $Im(\mathbf{f})$ is not Zariski dense in $\mathbb{G}_{m}^{d+1}.$ Let $V$ be its Zariski closure. According to the above theorem of Zhang, there is $\mu >0$ such that the Zariski closure $V_{\mu }$ of $\{\mathbf{x} =(x_{1},...,x_{d+1})\in V \textnormal{ such that } h(\mathbf{x})<\mu \}$ is a finite union of torsion coset tori. On the other hand, Lemma 22 and the Zariski connectedness of $\mathbb{G}$ shows that $V$ cannot be equal to a finite union of torsion coset tori.
Hence $V_{\mu }$ is a proper Zariski closed subset of $V$. Let $Z_{\mu }=\Omega ^{c}\cup \mathbf{f}^{-1}\{V_{\mu}\}.$ Then $Z_{\mu }$ is a proper Zariski-closed subset of $\mathbb{G}$. Note that since $f$ is invariant under conjugation by $T,$ $Z_{\mu }$ also is invariant under conjugation by $T.$ Let $\widehat{Z}_{\mu }$ the Zariski closure of the set $\{(gag^{-1},gbg^{-1})\in \mathbb{G}^{2}$ with $g\in\mathbb{G}$, $a\in T$ and $b\in Z_{\mu },$ or $a\in Z_{\mu }$ and $b\in T\}$ in $\mathbb{G\times G}.$ It is a proper Zariski closed subset, since $\dim \widehat{Z}\leq 2\dim \mathbb{G}-1.$ Take $n=d+1$ and $\alpha =\mu /n$ in Proposition 11, which gives us an $A_{1}>0$ and an $\eta >0.$
According to Proposition 17 there is a number $c=c(\mathbb{G},Z_{\mu },A_{1})>0$ such that $F^{c}$ contains two elements $a$ and $b$ which are $A_{1}$-regular semisimple elements, generate a Zariski-dense subgroup of $\mathbb{G}$ and satisfy $(a,b)\notin \widehat{Z}_{\mu }.$ Now
let $\varepsilon =\eta /c$ and assume that $e(F)<\varepsilon .$ Then $e(\{a,b\})<\eta $. For some $g\in \mathbb{G}(\overline{\mathbb{Q}})$, $gag^{-1}\in T$, and since $e(\cdot )$ is invariant under conjugation by elements from $\mathbb{G}(\overline{\mathbb{Q}}),$ we have $%
e(\{gag^{-1},gbg^{-1}\})<\eta .$ We can now apply Proposition 11 to see that we must have $h(\mathbf{f}(gbg^{-1}))<\mu $,
therefore $gbg^{-1}\in Z_{\mu }$ and hence $(gag^{-1},gbg^{-1})\in \widehat{Z}_{\mu }$. which gives the desired contradiction. Hence $e(F)>\varepsilon $ and we are done. $\blacksquare$

We finish today's post with a discussion of some corollaries. First, we show the following, which is needed at some point in the proofs of the lemmas we omitted:

Lemma 23: Let $F$ be a finite subset of a group $\Gamma $containing $1$. Assume that the elements of $F$ (together with their inverses) generate $\Gamma .$ Let $\Gamma _{0}$ be a subgroup of index $k$ in $\Gamma .$ Then $F^{2k-1}$ contains a generating set of $\Gamma _{0}.$


Proof: It is clear that $F^{k-1}$ contains a set of representatives for each left coset in $\Gamma /\Gamma _{0},$ say $\{s_{1},...,s_{k}\}.$ Similarly, $(F^{-1})^{k-1}$ contains a set of representatives of the left cosets, say $\{u_{1},...,u_{k}\}.$ Consider all elements of $\Gamma _{0}$ of the form $s_{i}fu_{j}^{-1}$ for $i,j\in \lbrack 1,k]$ and $f\in F.$ They all belong to $F^{2k-1}.$ It is straightforward to verify that, together with their inverses, they generate $\Gamma _{0}$.

Corollary 24: There are constants $\varepsilon =\varepsilon (d)$, $\kappa =\kappa (d)\in \mathbb{N}$ and $C=C(d)\in \mathbb{N}$ such that if $F$ is any finite subset of $GL_{d}(\overline{\mathbb{Q}})$ containing $1$, there
is some $a\in F^{\kappa }$ and some eigenvalue $\lambda $ of $a$ such that
\[h(\lambda )\geq \frac{1}{|F|^{C}}\cdot \widehat{h}(F). \]

Proof:  First we assume that $F$ generates a non virtually soluble group. From the spectral radius formula, we have for any set $F$ containing $1$, $\sum_{a\in F^{d^{2}}}e(\{a\})\geq e(F)-|\log c|$. In particular

\[\max \{e(\{a\}),a\in F^{nd^{2}}\}\geq \frac{1}{|F|^{nd^{2}}}(n\widehat{h}(F)-|\log c|)  \]
and for every $n\in \mathbb{N}$. Now by Height Gap theorem, we have $\widehat{h}(F)>\varepsilon =\varepsilon (d)>0$. Hence for some $%
n_{0}=n_{0}(d)\in \mathbb{N}$,
\[\max \{e(\{a\}),a\in F^{n_{0}}\}\geq \frac{d}{|F|^{n_{0}}}\cdot \widehat{h}(F). \]

On the other hand, we clearly have $e(\{a\})\leq \sum h(\lambda )$ where the sum is over the $d$ eigenvalues of $a.$ so we are done in this case

Now assume that $F$ generates a virtually soluble subgroup. It is well known that there is an integer $n_{0}=n_{0}(d)\in \mathbb{N}$ such that any virtually soluble subgroup of $GL_{d}(\mathbb{C})$ contains a subgroup of index at most $n_{0}$ which can be conjugated inside the upper-triangular matrices. Applying Lemma 23 (and its proof), we may find $F_{1}\subset F^{2n_{0}-1}$ such that $F^{n}\cap B\subset (F_{1}\cup F_{1}^{-1})^{2n}$ for all $n,$ where $B=T_{d}(\mathbb{C})$ is the subgroup of upper-triangular matrices. But $F^{n}=\cup (F^{n}\cap f_{i}^{-1}B)$ for at most $n_{0}$ elements $f_{i}$ in $F^{n_{0}}.$ Hence $F^{n}\subset \cup f_{i}^{-1}(F^{n+n_{0}}\cap B)$ and $R_{v}(F)\leq \lim \inf ||F^{n}\cap B||^{1/n}\leq R_{v}(F_{1}\cup F_{1}^{-1})^{2}.$
However, since $F_{1}\subset B,$ it is straightforward to observe that $R_{v}(F_{1}\cup F_{1}^{-1})=\Lambda _{v}(F_{1}\cup F_{1}^{-1}). $ Summing over all places, we obtain $\widehat{h}(F)\leq 2\sum_{a\in F_{1}}e(\{a\})+e(\{a^{-1}\})\leq 2|F|^{2n_{0}}\max \{\sum h(\lambda )+h(\lambda ^{-1}),\lambda $ eigenvalue of $a\in F_{1}\}.$ Since $h(\lambda )=h(\lambda ^{-1}),$ we get the desired result. $\blacksquare$

Now we turn to
Corollary 25: There are constants $N_{1}=N_{1}(d)\in \mathbb{N}$, $\varepsilon =\varepsilon (d)>0$ such that if $F$ is any finite subset of $GL_{d}(\overline{\mathbb{Q}})$ containing $1$ and generating a non virtually soluble subgroup, then we may find $a\in F^{N_{1}}$ and an eigenvalue $\lambda $ of $a$ such that $h(\lambda )>\varepsilon ,$ for some fixed $\varepsilon =\varepsilon (d)>0.$
Proof: By the remark above on the bound $n_{0}$ of the index of a triangular subgroup in any virtually soluble subgroup, it is easy to see that the set of pairs $(A,B)$ in $GL_{d}\times GL_{d}$ that generate a virtually soluble subgroup is a closed subvariety. Since every
connected simple algebraic group can be topologically generated (for the Zariski topology) by two elements we can apply the escape from subvarieties lemma and conclude that there is a pair $\{A,B\}$ in $F^{c(d)}$ which generates a non virtually soluble subgroup of $\left\langle F\right\rangle .$ Then apply Corollary 24 to $\{Id,A,B\}.$

As a corollary of this and the height gap theorem we obtain an effective
version of Schur's classical result on torsion linear groups

Corollary: (Effective Schur/no large torsion balls) There is an integer $N_{2}=N_{2}(d)\in \mathbb{N}$ such that if $K$ is a field and if $F$ is a finite subset of $GL_{d}(K)$ containing $1$, then either it generates a finite subgroup, or $(F\cup F^{-1})^{N_{2}(d)}$ contains an element of infinite order. Furthermore if $F$ generates a non virtually nilpotent subgroup, then we can find the element of infinite order already in $F^{N_{2}(d)}$.


The following example gives a situation showing that without the assumption
on $F$ in the last sentence of this corollary, the conclusion may fail.
Consider the subgroup of $GL_{2}(\mathbb{C})$ consisting of affine
transformations of the complex line. Then, for arbitrary $N\in \mathbb{N}$
one may find a finite (non-symmetric!) set $F$ containing the identity such
that the group generated by $F$ is infinite and virtually abelian, while $%
F^{N}$ consists solely of elements of finite order. For instance, take $%
F=\{id,a_{\omega },ta_{\omega }t^{-1}\}$ where $a_{\omega }=\left(
\begin{array}{cc}
\omega & 0 \\
0 & 1%
\end{array}%
\right) $ is multiplication by $\omega $ (a root of $1$ of order $N+1$) and $%
t=\left(
\begin{array}{cc}
1 & 1 \\
0 & 1%
\end{array}%
\right) $ is translation by $1,$ then the commutator $[a_{\omega
},ta_{\omega }t^{-1}]$ is $\neq 1$ if $N\geq 0$ and unipotent so of infinite
order, while $F^{N}$ is made of homotheties of ratio $\omega ^{k}$ with $%
1\leq k\leq N$ (i.e. elements of the form $\left(
\begin{array}{cc}
\omega ^{k} & \ast \\
0 & 1%
\end{array}%
\right) $), which are all torsion elements.

Proof: Let $k$ be the algebraic closure of $K$ and $\Gamma $ the subgroup generated by $F$. First assume that $\Gamma \leq GL(W)$ acts absolutely irreducibly on $W=k^{d}$. According to Burnside's theorem the $k$-subalgebra generated by the elements of $\Gamma $ is the full algebra $End_{k}(W).$ Since $D=\dim End_{k}(W)=(\dim W)^{2}\leq d^{2}$, there exists a linear basis, say $w_{1},...,w_{D}$ of $End_{k}(W)$ in $F^{d^{2}}$ (start with $w_{1}=1,$ then multiply by the elements of $F$ one after the other). Since $\{x\mapsto tr(zx)\}_{z\in End_{k}(W)}$ account for all linear forms on $End_{k}(W),$ the linear forms $x\mapsto tr(w_{i}x)$ must be linearly independent, and the matrix $\{tr(w_{i}w_{j})\}_{1\leq i,j\leq D}$ is invertible.
Let $L$ be the field generated by the eigenvalues of all elements of $F^{2d^{2}+1}.$ Note that $L$ contains $tr(w_{i}w_{j})$ and $tr(fw_{i}w_{j})$ for $f\in F$ and all $i,j.$ We claim that $\Gamma \leq \bigoplus\limits_{1\leq i\leq D}Lw_{i}\leq End_{k}(W).$ Indeed for each $i,$ and each $f\in F,$ write $fw_{i}=\sum a_{ij}w_{j}$ for some $a_{ij}\in k$. Then as $\{tr(w_{i}w_{j})\}_{1\leq i,j\leq D}$ is invertible, the $a_{ij}$ must belong to $L.$ Since $w_{1}=1,$ we see that positive words in $F$ lie all in $\bigoplus\limits_{1\leq i\leq D}Lw_{i}.$ On the other hand, the Cayley-Hamilton theorem implies that $f^{-1}\in L[f].$ Finally $\Gamma \leq \bigoplus\limits_{1\leq i\leq D}Lw_{i}$ as claimed. The left regular representation of $\Gamma $ on $\bigoplus\limits_{1\leq i\leq D}Lw_{i}$ gives us a faithful representation of $\Gamma $ in $GL_{D}(L).$ If $F^{2d^{2}+1}$ consists only of torsion elements, the field $L,$ is generated over its prime field by finitely many roots of unity. If $char(K)>0$ this already implies that $L$ is finite and thus that $\Gamma $ is finite, a contradiction. If $char(K)=0,$ then $L$ belongs to $\overline{\mathbb{Q}}$ and we are thus reduced to the case when $\Gamma $ lies in $GL_{D}(\overline{\mathbb{Q}})$. Then, by the combination of Corollary 24 with the Height Gap Theorem we are done unless $\Gamma $ is virtually soluble.

If $\Gamma $ does not act irreducibly of $k^{d},$ let $\{0\}\leq V_{1}\leq ...\leq V_{k}=k^{d}$ be a composition series for $\Gamma $ and let $W=V_{i_{0}}/V_{i_{0}+1}$ be an (irreducible) composition factor. If $char(K)>0,$ by the above, the image of $\Gamma $ is $GL(W)$ is finite. It follows that $\Gamma $ is virtually unipotent and hence finite, because finitely generated unipotent subgroups in positive characteristic are finite.

If $char(K)=0,$ then the image of $\Gamma $ on each composition factor is virtually soluble, and hence $\Gamma $ itself is virtually soluble. Recall that there is an integer $n_{0}=n_{0}(d)\in \mathbb{N}$ such that any virtually soluble subgroup of $GL_{d}(\mathbb{C})$ contains a subgroup of index at most $n_{0}$ which can be conjugated inside the upper-triangular matrices. Applying Lemma 23 we may assume without loss of generality that $F$ is made of upper-triangular matrices. Then for every $a,b\in F$, the commutator $[a,b]$ is a unipotent matrix in $SL_{d}(\mathbb{C})$, hence is either trivial or of infinite order. If one of them has infinite order, we are done. Otherwise
this means that the matrices in $F$ commute. But a finitely generated abelian group generated by torsion elements is finite. We are done.

The argument above works verbatim without the need to take inverses until the point in the last paragraph when $F$ is assumed to consist of upper-triangular matrices. Note that if the elements of $F$ are torsion, then their eigenvalues are roots of unity, hence the group generated by $F$ is virtually nilpotent. This completes the proof of the corollary. $\blacksquare$