Uniform Tits Alternative IV: Proof of the Uniform Tits Alternative

Today we finally discuss the Uniform Tits Alternative!

Uniform Tits alternative: For every $d \in \mathbb{N}$ there is $N(d) \in \mathbb{N}$ such that for any field $K$ and $F$ a finite symmetric subset of $\mathrm{GL}_d(K)$ containing 1, either $F^{N(d)}$ contains two elements which freely generate a non-abelian free gorup or the group generated by $F$ is virtually soluble.

As previously discussed, this part involves meticulously redoing Tits' proof to ensure uniform control over the constants. This will be involved and we will say relatively little about the details. First we record some well-known facts about Arakelov heights. Let $K$ be a global field. The Arakelov height on the projective space $\mathbb{P}(K^{d})$ is defined as follows for $x=(x_{1}:...:x_{d})$,

\[h_{Ar}(x)=\frac{1}{[K:K_{0}]}\sum_{v\in V_{K}}n_{v}\log ||x||_{v}\]
where $||x||_{v}$ is the standard norm on $K_{v}^{d}$ as defined above. It is well defined thanks to the product formula for norms over completions over a global field and always non-negative. This allows us to define the height of a projective linear subspace of $\mathbb{P}(K^{d}).$ Indeed if $W\leq \mathbb{P}(K^{d})$ is such then we set

\[h_{Ar}(W)=h_{Ar}(\Lambda ^{\dim W}W)\]
where $\Lambda ^{\dim W}W$ is the wedge product of $W$ viewed as a projective point in the projective space $\mathbb{P}(\Lambda ^{\dim W}K^{d})$. By convention we set $h_{Ar}(\{0\})=0$. Recall that the following holds for two projective linear subspaces $V$ and $W$

\[h_{Ar}(V)+h_{Ar}(W)\geq h_{Ar}(V+W)+h_{Ar}(V\cap W)\]
Moreover for every linear form $f$, seen as a point in the dual space $(K^{d})^{*}$, $h_{Ar}(f)$ makes sense as

\[h_{Ar}(\ker f)=h_{Ar}(f)\]
and more generally, $h_{Ar}(W)=h_{Ar}(W^{\bot })$, where $W^{\bot }$ is the orthogonal of $W$ in $(K^{d})^{*}.$

Also note that if $g\in SL_{d}(K)$ and $W$ is a subspace of $K^{d},$ then

\[h_{Ar}(gW)\leq d\cdot h(g)+h_{Ar}(W)\]
where $h(g)=h(\{g\})$ as defined in the last paragraph. Note also that $h(g^{-1})\leq (d-1)h(g).$

Definition: Given $A\in SL_{d}(K),$ we will say that a vector subspace $W$ (or its projectivization) is $A$\textbf{-admissible} if it is a sum of generalized eigenspaces of $A.$ We also denote by $W^{c}$ its complementary subspace, i.e. the sum of the remaining generalized eigenspaces.
Lemma 26: Let $W$ be an $A$-admissible subspace. Then

\[h_{Ar}(W)\leq d^{2}\cdot (2h(A)+\varepsilon _{\Omega }\log 2)\]
where $\varepsilon _{\Omega }=0$ if $car(K)>1$ and $1$ if $car(K)=0.$

Note that if $A,B\in M_{d}(K),$ then $h(A+B)\leq h(A)+h(B)+\varepsilon_{\Omega }\log 2.$ Moreover, if $\alpha $ is an eigenvalue of $A$, then $h(\alpha )\leq h(A).$ Also $h_{Ar}(\ker A)\leq (rk(A))\cdot h(A).$ Indeed $h_{Ar}(\ker A)=h_{Ar}((\ker A)^{\bot })=h_{Ar}(\mathrm{Im}A^{t}).$ But for $B\in M_{d}(K),$ $h_{Ar}(\mathrm{Im}B)\leq (rk(B))\cdot h(B),$ since we may choose a subset of the canonical basis, say $e_{1},...,e_{k}$ such that $Be_{1},...,Be_{k}$ generates $\mathrm{Im}B,$ and $||Be_{1}\wedge ...\wedge Be_{k}||_{v}\leq ||B||_{v}^{k}$ where $k=rk(B).$ With these observations in hand we can prove Lemma 26.


Proof: We have $W=\bigoplus E_{\alpha }$ for some eigenvalues $\alpha $ of $A,$ where $E_{\alpha }$ is the corresponding generalized eigenspace. Hence $h_{Ar}(W)\leq \sum h_{Ar}(E_{\alpha }).$ If $n_{\alpha }=\dim E_{\alpha },$ then $E_{\alpha }=\ker (A-\alpha )^{n_{\alpha }}.$ Hence $h_{Ar}(E_{\alpha})\leq d\cdot h((A-\alpha )^{n_{\alpha }})=dn_{\alpha }\cdot
(2h(A)+\varepsilon _{\Omega }\log 2). \blacksquare$

Proximality

In this paragraph we recall the notion of a proximal element in $SL_{d}(k),$ where $k$ is a local field, and we show some precise estimates
as to how such elements act on the projective space $\mathbb{P}(k^{d}).$

An element $a\in SL_{d}(k)$ is said to be \textbf{proximal} if there is a unique (multiplicity one) eigenvalue of $a$ with maximum modulus $\Lambda_{k}(a).$ We will also need to consider almost proximal elements where the eigenvalues which are larger than, say, some $\omega $ are much larger than all other eigenvalues.

The Fubini-Study metric on $\mathbb{P}(k^{d})$

Let $k$ be a local field and $\overline{k}$ an algebraic closure of $k$. Endow the projective space $\mathbb{P}(\overline{k}^{d})$ with the standard (Fubini-Study) distance defined by
\[d([u],[v])=\frac{||u\wedge v||}{||u||\cdot ||v||} \]

for any $u,v\in \overline{k}^{d}\backslash \{0\}$ and $||\cdot ||$ is the standard norm on $\overline{k}^{d}$ (i.e. Euclidean norm if $k$ is
archimedean and sup norm if $k$ is non archimedean). To avoid heavy notation, we will denote by the same letter a non zero vector, or subspace of $\overline{k}^{d}$ and its projectivization in $\mathbb{P}(\overline{k}^{d}). $

We denote by $\mathbb{K}_{k}$ the maximal compact subgroup of $SL_{d}(k)$ equal to $SO(d,\mathbb{R})$ if $k=\mathbb{R}$, $SU(d,\mathbb{R})$ if $k=\mathbb{C}$, and $SL_{d}(\mathcal{O}_{k})$ is $k$ is ultrametric. Its action on $\mathbb{P}(k^{d}) $ preserves $d$ (in fact this characterizes $d$ up to composition by some positive function).

Lemma 27: Let $h\in SL_{d}(k).$ Then $Lip(h)\leq (||h||\cdot ||h^{-1}||)^{2}\leq ||h||^{2d},$ where $Lip(h)$ is the smallest constant $L\geq 0$ such that $d(hx,hy)\leq L\cdot d(x,y)$ for all $x,y\in \mathbb{P}(k^{d}).$

Proof: Writing $h$ in Cartan's $\mathbb{K}_{k}A\mathbb{K}_{k}$ decomposition of $SL_{d}(k),$ one sees that we can assume that $h$ is diagonal and we are thus reduced to a straightforward verification. $\blacksquare$

Recall that if $H$ is a hyperplane in $k^{d},$ and $f$ a non zero linear form on $k^{d}$ with kernel $H,$ then if $u\in k^{d}\backslash \{0\},$ its distance to $H$ is

\[d(u,H)=\frac{|f(u)|}{||f||\cdot ||u||} \]
where $||f||=\sup \{|f(x)|,||x||\leq 1,x\in k^{d}\}.$ More generally, if $V$ and $W$ are two $k$-subspaces in direct sum, i.e. $V\oplus W=k^{d},$ then

\[d(V,W)=\frac{||\underline{v}\wedge \underline{w}||}{||\underline{v}||\cdot ||\underline{w}||} \]
where $\underline{v}=v_{1}\wedge ...\wedge v_{l}$ and $\underline{w}%
=w_{1}\wedge ...\wedge w_{d-l}$ for any basis $(v_{1},...,v_{l})$ of $V$ and
$(w_{1},...,w_{d-l})$ of $W.$ In particular, when $k$ is archimedean, two subspaces are orthogonal if and only if they are at distance $1$. Let $(e_{1},...,e_{d})$ be the canonical basis in $k^{d}.$

Lemma 28: Let $f$ be a non-zero linear form on $k^{d}$ and $H=\ker f$. Let $V$ a $k$-subspace in $k^{d}$ and $V^{*}$ the orthogonal of $V$ in the dual of $k^{d}.$ Then for every $v\in V, d(v,H)=d(v,V\cap H)\cdot d(f,V^{*})$

Proof: Observe that as $\mathbb{K}_{k}$ permutes transitively the $k$-subspaces of given dimension and preserves $d,$ we may assume that $V=\left\langle e_{1},...,e_{p}\right\rangle $ for some $p\in [0,d].$ Then we may write $f$ in the dual canonical basis $f=\sum f_{i}e_{i}^{*}=f^{<}+f^{>}$ where $f^{<}$ is the part of the sum involving indices $i\leq k$ and $f^{>}$ the other part.
Let $\underline{e}^{>}=e_{p+1}^{*}\wedge ...\wedge e_{d}^{*}$. Then $||f||\cdot d(f,V^{*})=||f\wedge \underline{e}||=||f^{<}\wedge \underline{e}||=||f^{<}||.$ On the other hand note that $f^{<}$ coincides with $f$ on $V.$
Hence for $v\in V$, $d(v,V\cap H)=\frac{|f(v)|}{||f^{<}||\cdot ||v||}.$ As $d(v,H)=\frac{|f(v)|}{||f||\cdot ||v||}$, combining these relations we are done. $\blacksquare$

Lemma 29: Let $V\oplus W=k^{d}$ and $H$ a hyperplane in $k^{d}$ with $V\nsubseteq H.$ Let $\pi $ be the linear projection onto $V$ with kernel $W$. Then for every $u\in \mathbb{P(}k^{d})\backslash W$ we have
\[d(\pi u,V\cap H)\geq d(u,W+V\cap H)\cdot d(V,W) \]


Proof: Write $u=\pi u+\pi u^{\bot }\in V\oplus W$. If $v_{1},...,v_{k-1}$ is a basis of $V\cap H$ and $w_{1},...,w_{d-k}$ a basis of $W$ we set $\underline{v}=v_{1}\wedge ...\wedge v_{k-1}$ and $\underline{w}=w_{1}\wedge ...\wedge w_{d-k}$. We have $d(\pi u,V\cap H)\geq d(\pi u,W+V\cap H)=\frac{||\pi u\wedge \underline{v}\wedge \underline{w}||}{||\pi u||\cdot ||\underline{v}\wedge \underline{w}||}=d(u,W+V\cap H)\cdot \frac{||u||}{||\pi u||}.$ We may assume $u\notin V$, then on the other hand $d(V,W)\leq d(\pi u,\pi u^{\bot})=\frac{||u\wedge \pi u^{\bot }||}{||\pi u||\cdot ||\pi u^{\bot }||}\leq \frac{||u||}{||\pi u||}.$ We are done. $\blacksquare$

Contraction properties of proximal and almost proximal elements

For $a\in SL_{d}(k)$ we set $E_{\lambda }$ its generalized eigenspace with eigenvalue $\lambda $. In this paragraph, we will assume that eigenvalues of $a$ belong to $k.$ We let $\Lambda _{k}(a)=\max \{|\mu |_{k},\mu $eigenvalue of $a\}$ and $\lambda _{k}(a)$ the modulus of the second heighest eigenvalue of $a$. An element $a\in SL_{d}(k)$ is said to be \textbf{proximal} if $\lambda _{k}(a)<\Lambda _{k}(a).$

To deal with non proximal elements we introduce some positive real number $\omega >0$, such that $\Lambda _{k}(a^{-1})^{-1}<\omega \leq \Lambda _{k}(a). $ We set $\Lambda _{k}^{\omega }(a)=\min \{|\mu |_{k},\mu $ eigenvalue of $a, $ $|\mu |_{k}\geq \omega \}$ and $\lambda _{k}^{\omega}(a)=\max \{|\mu |_{k},\mu $ eigenvalue of $a,|\mu |_{k}<\omega \}.$

Lemma 30: Suppose $a\in SL_{d}(k)$ and let $A=\Lambda _{k}(a)\Lambda_{k}(a^{-1})\geq 1.$ For every $\varepsilon >0$ there is $\eta =\eta (\varepsilon ,d)>0$ and $\omega $ such that \[\Lambda _{k}(a^{-1})^{-1}<\omega \leq \Lambda _{k}(a)\] and

\[A^{\eta }\cdot \left( \frac{\Lambda _{k}(a)}{\Lambda _{k}^{\omega }(a)}\right) ^{\frac{1}{\varepsilon }}\leq \frac{\Lambda _{k}^{\omega }(a)}{\lambda _{k}^{\omega }(a)} \]


Proof: Let $\lambda _{1},...,\lambda _{d}$ be the eigenvalues of $a$ ordered as $|\lambda _{1}|_{k}\geq ...\geq |\lambda _{d}|_{k}.$ Let $\ell _{i}=\log  \frac{|\lambda _{i}|}{|\lambda _{i+1}|}\geq 0.$ Fix $\varepsilon >0$ and take some $\eta >0.$ We claim that for $\eta $ small enough, there exists $i_{0}\in [0,d-2]$ such that $\ell _{i_{0}+1}-\eta \log A\geq \frac{1}{\varepsilon }\cdot (\ell _{1}+...+\ell _{i_{0}}).$ Indeed, otherwise we would have $\ell _{1}<\eta \log A$, $\ell _{2}<\eta \log A+\frac{1}{\varepsilon }\ell _{1}$, etc, until we get $\log A=\ell _{1}+...+\ell_{d-1}\leq C(\varepsilon ,d)\eta \log A$ for some computable constant $C(\varepsilon ,d),$ a contradiction if $\eta $ is smaller than say $\frac{1}{2C(\varepsilon ,d)}$. Let $\omega =|\lambda _{i_{0}+1}|_{k}. \blacksquare$

When $\varepsilon $ is fixed and $\omega $ so given by Lemma 30, we will refer to $a$ as being \textit{almost proximal for }$\omega $.

We will let $H_{a}^{\omega }$ be the vector subspace equal to the sum of the  $E_{\lambda }$'s for which $|\lambda |_{k}\leq \lambda _{k}^{\omega }(a).$
Similarly, we denote its complementary subspace by $V_{a}^{\omega}=\bigoplus E_{\lambda }$, the sum being over those $\lambda $'s such that $|\lambda |_{k}\geq \Lambda _{k}^{\omega }(a)$. We let $\pi _{a}^{\omega }$ be the linear projection onto $V_{a}^{\omega }$ with kernel $H_{a}^{\omega}. $ We also set $l_{\omega }=\dim V_{a}^{\omega }.$ If $a$ is proximal, we will drop the superscript $\omega $ (and set it to be $\Lambda _{k}(a)$) andsimply denote by $V_{a}$, $H_{a}$, and $\pi _{a}$ the corresponding quantities.


Remark: Note that if $a\in GL_{d}(k),$ then its eigenvalues belong to the extension of $k$ generated by all algebraic extensions of $k$ in a given algebraic closure $\overline{k}$ of degree at most $d$ (there are finitely many such). So this extension is also a local field. Hence up to passing to this finite extension one may always assume that the eigenvalues of $a$ belong to $k.$

The following Tits Converse Lemma is useful when one needs to build an element $x$ such that both $x$ and $x^{-1}$ are proximal. It was originally used by Tits in the proof of his alternative which gives a sufficient condition for $a\in SL_{d}(k)$ to be proximal : it is as soon as $a$ stabilizes some open subset where it contracts distances.

Tits Converse Lemma: Let $a\in SL_{d}(k)$. Assume that there exists a point $v\in \mathbb{P}(k^{d})$ and an open neighborhood $U$ of $v$ such that $\overline{aU}\subset U$ and such that $Lip(a_{|U})<1$, where $Lip(a_{|U})$ is the smallest constant $L>0$ such that
$d(ax,ay)\leq L\cdot d(x,y)$ for every $x,y\in U.$ Then $a$ is proximal, $V_{a}\in U$ and $\frac{\lambda _{k}(a)}{\Lambda _{k}(a)}\leq Lip(a_{|U}).$

Proof: The compact subset $\overline{aU}$ is stable under $a$ and on it $a$ contracts distances. It follows immediately that all orbits $(a^{n}u)_{n\geq 0}$ converge to the unique fixed point $v_{a}$ of $a$ in $\overline{aU}.$
Let $\alpha $ be the eigenvalue of $a$ with eigenvector $v_{a}.$ Let $\beta $ be another eigenvalue of $a$ (if $\alpha $ has multiplicity higher than $1,$ we may take $\beta =\alpha $). There is a non zero vector $w$ such that $aw=\beta w+\kappa v_{a}$ for some $\kappa \in k.$ Let $\varepsilon \in k\backslash \{0\}$ with $|\varepsilon |_{k}$ arbitrarily small. Then one computes from the definition of the Fubini-Study distance $\lim_{|\varepsilon |\rightarrow 0}\frac{d(a(v_{a}+\varepsilon w),v_{a})}{d(v_{a}+\varepsilon w,v_{a})}=\frac{|\beta |}{|\alpha |}.$ If $|\varepsilon |_{k}$ is small enough, $v_{a}+\varepsilon w\in U$ and thus $\frac{|\beta |}{|\alpha |}\leq Lip(a_{|U})<1. \blacksquare$

Proof sketch of the uniform Tits alternative:
The proof is done in three steps. First we reduce to the situation when $F$ generates a Zariski-dense subgroup in $\mathbb{G}(\Omega )$ where $\mathbb{G}$ is a simple Chevalley group of adjoint type to be chosen among a finite list of such.
Second we show that we may assume that $F=\{1,X,X^{-1},Y,Y^{-1}\}$, i.e. $F$ is a symmetric set with $4$ elements plus the identity.
And finally, in the third and most difficult step that we will not discuss, it is checked that there exists a place $v$ of the field $K$ of coefficients for which some very complicated conditions for ping-pong are fulfilled with some explicit choice of constants depending only on $\mathbb{G}$, and thus yield the desired ping-pong pair. One key lemma (4.6 in Breuillard's paper on the strong Tits alternative), whose statement is rather messy, computes the rate of convergence to the attracting point of powers of a given proximal element $a$ in terms of three quantities : its norm $||a||,$ the modulus of its maximal eigenvalue $\Lambda _{k}(a)$ and the modulus of its second to maximal eigenvalue $\lambda _{k}(a).$ A similar estimate is given for an almost proximal element depending on the choice of the cursor $\omega .$ One has to be careful in those estimates to ensure they are uniform over all ultrametric local fields. The multiplicative constants $C_{k,i}$'s that appears in the estimates always disappears when $k$ is ultrametric, which is a crucial part of the proof. Unfortunately this messiness seems unavoidable and I don't see a way of presenting this in an illuminating way, which is why it is omitted. We indicate how the first two steps can be done

Claim (i): We may assume that $F$ generates a Zariski-dense subgroup in $\mathbb{G}(\Omega )$ where $\mathbb{G}$ is a simple Chevalley group of adjoint type.

Proof: Since $F$ generates a non virtually solvable subgroup $\left\langle F\right\rangle $, the connected component $\mathbb{G}^{0}$ of the
Zariski-closure $\mathbb{G}$ of $\left\langle F\right\rangle $ is not solvable. Moding out by the solvable radical of $\mathbb{G}^{0},$ which is a normal subgroup of $\mathbb{G}$, we see that we can assume that $\mathbb{G}^{0}$ is a non-trivial semisimple algebraic group. We let $\mathbb{G}$ act on $\mathbb{G}^{0}$ by conjugation we obtain a homomorphism of $\mathbb{G}$ in $Aut(\mathbb{G}_{ad}^{0}\mathbb{)}$ where $\mathbb{G}_{ad}^{0}$ is the adjoint group of $\mathbb{G}^{0}$ whose image contains $\mathbb{G}_{ad}^{0}.$ However it is known that $Aut(\mathbb{G}_{ad}^{0}\mathbb{)}/\mathbb{G}_{ad}^{0}$ is a subgroup of the automorphisms of the Dynkin diagram of $\mathbb{G}$. In particular it is a finite group whose order is bounded in terms of $\dim \mathbb{G}$ only, hence in terms of $d$ only. By lemma 23 in the third post, we may therefore assume that $\mathbb{G}=\mathbb{G}_{ad}^{0}$ is a semisimple algebraic group of adjoint type. Further projecting to one of the simple factors, we may assume that $\mathbb{G}$ is a simple algebraic group of adjoint type over $\Omega .$ As $\Omega $ is algebraically closed, $\mathbb{G(}\Omega )$ is the group of $\Omega $-points of a Chevalley group $\blacksquare.$

We need the following variant of proposition 9:
Theorem: If $\mathbb{G}$ is a Chevalley group, then there is a constant $C=C(\mathbb{G})>0$ and a Zariski open subset $\mathcal{O}=\mathcal{O}(\mathbb{G})$ of $\mathbb{G\times G}$ such that for any choice of $\Omega $ and for any pair $(a,b)\in \mathcal{O}(\Omega ),$ there is $g\in \mathbb{G}(\Omega )$ such that, setting $F=\{a,b\}$,
\[h(gFg^{-1})\leq C\cdot \widehat{h}(F)  \]


Claim (ii): In the Uniform Tits Alternative, we may assume that $F=\{1,X,X^{-1},Y,Y^{-1}\}$ for some $(X,Y)\in \mathcal{O}(\Omega ).$

This claim was already proven in the special case of characteristic $0$ making key use of Jordan's theorem about finite subgroups of $GL_{n}(\mathbb{C}).$ This argument fails in positive characteristic and a more involved argument, which uses a fair bit of machinery from the theory of algebraic groups, is needed.