A STRONG TITS ALTERNATIVE 1. Introduction The goal of this ...

A STRONG TITS ALTERNATIVEEMMANUEL BREUILLARDAbstract. We show that for every integer d ∈ N, there is N(d) ∈ Nsuch that if K is any field and F is a finite subset of GL d (K), whichgenerates a non amenable subgroup, then F N(d) contains two elements,which freely generate a non abelian free subgroup. This improves theoriginal statement of the Tits alternative. It also implies a growth gap anda co-growth gap for non-amenable linear groups, and has consequencesabout the girth and uniform expansion of small sets in finite subgroups ofGL d (F q ) as well as other diophantine properties of non-discrete subgroupsof Lie groups.1. IntroductionThe goal of this paper is to show the following theorem and some consequencesof it.Theorem 1.1. For every d ∈ N there is N(d) ∈ N such that if K is anyfield and F a finite symmetric subset of GL d (K) containing 1, either F N(d)contains two elements which freely generate a non abelian free group, orthe group generated by F is virtually solvable (i.e. contains a finite indexsolvable subgroup).By F N(d) = F · ... · F we mean the set of elements which can be writtenas a product of at most N(d) elements from F, and by symmetric we meanthat if f ∈ F then f −1 ∈ F. This statement is a strengthening of the classicalTits alternative [39], which asserts that any finitely generated subgroup〈F 〉 of GL d (K), where K is any field, either contains a non abelian free subgroupor contains a solvable subgroup of finite index. It also improves earlierstrengthenings of the Tits alternative, due to Eskin-Mozes-Oh [18] (for freesemigroups) and to T. Gelander and the author [14] (for free groups), whichshowed a statement of a similar form, except that the integer N(d) dependedon the group Γ generated by F (not on the generating set) but was not independentof the field of coefficients. Note that N(d) cannot be boundeduniformly in d (see Remark 1.4).The present paper essentially contains the geometric part of the proof ofTheorem 1.1. The arithmetic part is the object of the paper [13]. The readerDate: March 2008.1

2 EMMANUEL BREUILLARDonly interested in the GL 2 case can read a self-contained proof of Theorem1.1 (both arithmetic and geometric parts) and its consequences in this specialcase in [12].The novelty of the above statement resides precisely in the fact that theinteger N(d) can be taken to depend only on d and not on F nor 〈F 〉 . Assuch Theorem 1.1 is a statement of a different nature. What it really assertsis an inclusion of countably many algebraic varieties into another algebraicvariety. Indeed, the condition on a k-tuple of matrices in say GL d (C) thatthey generate a virtually solvable group is an algebraic one (see Prop. 7.1below). On the other hand to say that no two words of length at most N(d)with letters in this k-tuple are generators of a free group is itself a countableunion of algebraic conditions. This way of interpreting the result allows toderive, via an effective Nullstellensatz, several corollaries about the girth infinite simple groups of Lie type, as well as some diophantine properties ofnon-discrete subgroups of GL n (C), in the spirit of the works of Kaloshin-Rodnianski, Helfgott and Bourgain-Gamburd ([26], [22], [9], [10]).Comments on the proof.Tits’ proof of his alternative consists of two parts. In a first arithmeticstep, he exhibits a semisimple element of 〈F 〉 which has some eigenvalue ofabsolute value |λ| > 1 for a clever choice of absolute value on K. Thenin a second geometric step, he studies the action of 〈F 〉 on the projectivespace P(k n ) under some suitably chosen linear representation, where k is thecompletion of K with respect to that absolute value. The free group is thenobtained by building a so-called ping-pong pair acting on P(k n ) (see [39]).The proof of Theorem 1.1 consists in reproducing Tits’ proof almost wordby word while making sure that each step can be done in a uniform way.The arithmetic step is much harder to perform, as we need a uniform gap|λ| > 1 + ε, where ε is allowed to depend on d only. This first arithmeticstep is the content of the paper [13], which shows a height gap theorem fornon amenable linear groups (see Theorem 3.3). The key idea there and alsoin the present paper is to introduce arithmetic heights in order to treat allabsolute values of K on an equal footing. This first arithmetic step is neededonly in characteristic zero. In a second arithmetic step, we find an absolutevalue for which the geometric conditions needed for the ping-pong to workare fulfilled. This is done in Section 6 by estimating the Arakelov heights ofthe characteristic subspaces of the matrices in F in terms of the normalizedheight ĥ(F ) introduced in [13] and by making use of another result from[13] which says that ĥ(F ) can be realized up to a multiplicative factor asthe height of some conjugate of F inside SL d (Q). Once the right absolutevalue has been found, the actual geometric construction of the ping-pong

A STRONG TITS ALTERNATIVE 3pair follows Tits’ geometric step very closely (unlike the argument in [14]) ;the only notable difference is that our estimates need to be uniform over alllocal fields. This requires a bit of care and is performed in Sections 4 and 5.Some consequences.Theorem 1.1 admits several consequences about the structure of nonamenablelinear groups. The first is a gap for the growth exponent, namely:Corollary 1.2. (Uniform exponential growth) For every d ∈ N, there existsa constant ε = ε(d) > 0 such that if K is any field and F is a finite subsetof GL d (C) containing 1 and generating a non amenable subgroup, then forall n ≥ 1|F n | ≥ (1 + ε) nHence1ρ F = limn→+∞ n log |F n | ≥ log(1 + ε) > 0Remark 1.3. It is possible that the assumption “non-amenable” in theabove corollary can be replaced by “of exponential growth”. However weobserved in [11] that this would imply the Lehmer conjecture about theMalher measure of algebraic numbers. We also observed there that althoughevery linear solvable group of exponential growth contains a free semigroup,no analog of Theorem 1.1 holds for solvable groups, namely one may find setsF n in GL 2 (C) containing 1 and generating a solvable subgroup of exponentialgrowth, such that no pair of elements in (F n ) n may generate a free semigroup.Remark 1.4. Examples due to Grigorchuk and de la Harpe [21] (see also [3])show that there is a sequence of groups Γ n with finite generating set F n whichare virtually a direct product of finitely many copies of the free group F 2 suchthat ρ Fn → 0 as n → +∞. Those examples can be embedded in SL m (Z) forsome possibly large m = m(n). Therefore we must have N(d) → +∞ andε(d) → 0 as d → +∞ in Theorem 1.1 and Corollary 1.2.The following corollary says that non-amenable linear groups have fewrelations: there is a co-growth gap.Corollary 1.5. (Co-growth gap) For every d, k ∈ N, there is ε > 0 such thatif K is a field and F = {a 1 , ..., a k } generates a non virtually solvable subgroupof GL d (K), then for every n ∈ N, the proportion of relations w(a 1 , ..., a k ) = 1in the free group F k of word length at most n among all elements in F k ofword length at most n is at most exp(−εn).Von Neumann showed that groups containing a free subgroup are nonamenable, i.e. have a spectral gap in l 2 . The uniformity in Theorem 1.1implies also a uniformity for the spectral gap (see [36] for this observation).More precisely:

4 EMMANUEL BREUILLARDCorollary 1.6. (Uniform Spectral Gap in l 2 ) For every d ∈ N, there isε = ε(d) > 0 with the following property. If K is a field and F is a finitesubset of GL d (K) containing the identity and generating a non amenablesubgroup and if Γ is any countable subgroup of GL d (K) containing F andf ∈ l 2 (Γ), then there is σ ∈ F such that∑∣ f(σ −1 x) − f(x) ∣ 2 ≥ ε · ∑|f(x)| 2x∈ΓIn particular, if F in GL d (K) is a finite subset containing the identityand generating a non amenable subgroup, then for every finite subset Ain GL 2 (K), we have |F A| ≥ (1 + ε)|A|.This shows also that if µ is a uniform probability measure on a set Fof cardinal k in GL d (K), then the Kesten spectral radius of µ (see [24]) isuniformly bounded away from 1 by a bound depending only on k and d.Hence the return probability of the simple random walk on the group 〈F 〉decays exponentially with an exponential rate depending only on k and d.The uniformity in Theorem 1.1 allows to reduce mod p and we obtain astatement giving a lower bound on the girth of subgroups of GL d in positivecharacteristic:Corollary 1.7. (Large girth) Given d, k ∈ N, there is N 0 , N ∈ N and ε 0 , C >0 such that for every prime p and every field K of characteristic p and anyfinite k-element subset F generating a subgroup of GL d (K) which containsno solvable subgroup of index at most N, then F N 0contains two elementsa, b such that w(a, b) ≠ 1 in GL d (K) for any non trivial word w in F 2 oflength at most f(p) = C · (log p) ε 0.Corollary 1.8. (Expansion of small sets) There is ε = ε(d) > 0 suchthat given k, N ∈ N, there is a constant C k,N,d such that for any fieldK of charateristic p > 1 and any subset F of GL d (K) with k elementsgenerating a subgroup which has no solvable subgroup of index at mostN, we have max f∈F |A △ fA| ≥ ε|A| for all subsets A in GL d (K) with|A| ≤ C k,N,d log log log p.It was conjectured in [20] that the statement of Corollary 1.7 holds forgenerating subsets F of GL 2 (F p ) with ε 0 = 1. It was also proved therethat a random k-regular Cayley graph of GL 2 (F p ) has girth at least (1 −o(1)) log k−1 (p).In a similar fashion one can derive the following weak diophantine propertyfor subgroups of GL d (C). Let d be some Riemannian distance on GL d (C).x∈Γ

A STRONG TITS ALTERNATIVE 5Corollary 1.9. (Weak diophantine condition) Given d ∈ N, there is N 0 ∈ Nand ε 1 > 0 with the following property. For every finite set F ⊂ GL d (C) generatinga non virtually solvable subgroup, there is δ 0 (F ) > 0 such that for everyδ ∈ (0, δ 0 ) there are two short words a, b ∈ F N 0such that d(w(a, b), 1) ≥ δfor every reduced word w in the free group F 2 with length l(w) at most(log δ −1 ) ε 1.In [26] Kaloshin and Rodnianski proved that for G = SU(2, R) ≤ GL 2 (C)almost every pair (a, b) ∈ G×G satisfies d(w(a, b), 1) ≥ exp(−C(a, b)·l(w) 2 )for all w ∈ F 2 \{e} and some constant C(a, b) > 0. Besides it is easy to seethat if a, b ∈ GL 2 (Q) then the pair (a, b) satisfies the stronger diophantinecondition d(w(a, b), 1) ≥ exp(−C(a, b) · l(w)). It is conjectured in [34] and[20], that this stronger condition also holds for almost every pair (a, b) ∈SU(2, R).Our result also allows us to estimate the number of words of length ≤ nthat fall in a shrinking neighborhood of 1 in GL d (C). More precisely,Corollary 1.10. (Weak equidistribution) Given d ∈ N, there are τ, ε 1 , C > 0with the following property. For every {a, b} ≤ GL d (C) which generates anon virtually solvable subgroup, there is δ 0 (a, b) > 0 such that for everyδ ∈ (0, δ 0 ) and every n ≤ C(log δ −1 ) ε 1, the proportion of elements w inthe free group F 2 of word length n such that d(w(a, b), 1) ≤ δ is at mostexp(−τn).In [19], Gamburd, Jacobson and Sarnak, showed for G = SU(2, R) that ifa pair (a, b) ∈ G satisfies the conclusion of Corollary 1.10 with ε 1 = 1 andC > C 0 (for some explicit C 0 > 0) then (a, b) has a spectral gap on L 2 (G). In[9], Bourgain and Gamburd showed that if a pair (a, b) ∈ G satisfies the abovecondition with ε 1 = 1 and some C = C(a, b) > 0, then (a, b) has a spectralgap on L 2 (G). This latter condition is automatically satisfied if (a, b) satisfiesthe stronger diophantine condition above, for instance if (a, b) ∈ GL 2 (Q).Remark 1.11. Corollaries 1.7, 1.9 and 1.10, are derived from Theorem 1.1by using we use a standard version of the effective Nullstellensatz due toMasser and Wustholz (see [29]) after reformulating Theorem 1.1 in terms ofinclusion of algebraic varieties. See Section 7.Contents1. Introduction 12. Minimal norm and spectral radius formula 63. Normalized height and Height gap 7

6 EMMANUEL BREUILLARD4. Proximality 105. Ping-Pong 176. Height bounds and proof of Theorem 1.1 267. Applications 37References 382. Minimal norm and spectral radius formulaIn this section, we recall results obtained in [13] about the spectral radiusof a finite set of matrices. Given a local field k, we defined the standardnorm || · || k on k d to be the canonical Euclidean (resp. Hermitian) norm if kis R (resp. C) and the sup norm if k is ultrametric. This induces on operatornorm on the space of d × d matrices M d (k), which we again denote by || · || k .Given a finite subset F of matrices in M d (k), we define its norm ||F || k tobe the maximal norm of any given element of F. We define the followingquantitiesE k (F ) =inf ||gF g −1 || kg∈GL d (k)Λ k (F ) = max{|λ| k , λ eigenvalue of some f ∈ F }where k is an algebraic closure of k and | · | k is the absolute value on kextended (uniquely) to k. We also set the spectral radius of F to be:R k (F ) =limn→+∞ ||F n || 1 nkThese quantities enjoy the following key properties.Lemma 2.1. (Spectral Radius Formula for F, [13], Lemma 2.1.)(a) If k is ultrametric, then for any compact set F containing 1 in M d (k),there is a positive integer q ≤ d 2 such that Λ k (F q ) = E k (F ) q . In particular,E k (F ) = R k (F ) = max 1≤q≤d 2 Λ k (F q ) 1 q .(b) If k is archimedean, there is a constant c = c(d) ∈ (0, 1) such thatfor any compact set F in M d (k), there is a positive integer q ≤ d 2 suchthat Λ k (F q ) ≥ c · E k (F ) q . In particular, c · E k (F ) ≤ max 1≤q≤d 2 Λ k (F q ) 1 q ≤R k (F ) ≤ E k (F ).Remark 2.2. This lemma expresses in a condensed form some ideas presentin the proof of the main result of [18] by Eskin-Mozes-Oh. It is useful toproduce elements with large eigenvalues in F n for some small n.We also record the following:

A STRONG TITS ALTERNATIVE 7Lemma 2.3. ([13], Proposition 2.5.) Suppose k is archimedean (i.e. k isR or C). Then for every n ∈ N and every compact subset F in SL d (k)containing 1, we haveE k (F n ) ≥ E k (F )√ n8d3. Normalized height and Height gapIn this section we recall results obtained in [13] about heights. In [13],we introduced the notion of normalized height ĥ(F ) of a finite subset ofmatrices F in SL d (Q). A similar definition can be made over the algebraicclosure F p (t) of F p (t). Below we recall the relevant definitions and notations.Let Ω be either Q or F p (t) for some prime p > 1. By a global field K, wemean a field isomorphic to a finite algebraic extension of K 0 , where eitherK 0 = Q or K 0 = F p (t) for some prime p > 1. We denote by V K the setof equivalence classes of non trivial absolute values on K. We make thefollowing standard choice of representatives | · | v for each v in V K . Everyv ∈ V K induces on K 0 an absolute value v 0 ∈ V K0 . We first determine aset of representatives of V K0 , then pick in each v ∈ V K the representativewith that normalization. If K 0 = Q then any absolute value is equivalentto either the standard absolute value over R or the p-adic absolute valuenormalized so that |p| p = 1. These form our representatives. If K p 0 = F p (t),then every absolute value is equivalent to either | P | Q 0 = p deg P −deg Q or | P | Q π =(vπ(Q)−vπ(P ))·deg πp where π ∈ F p [t] is a monic irreducible polynomial andv π (P ) is the valuation of π in the prime factor decomposition of P ∈ F p [t].These form our representatives. For background on these issues see Weil’sbook [43].Each v ∈ V K gives rise to a local field K v which is the completion of Kaccording to this absolute value. Let n v be the dimension of K v over theclosure of K 0 in K v . The product formula reads∑(1)n v log |x| v = 0v∈V Kfor every x ∈ K. We can now recall the definition of the standard Weil heightof an algebraic number. Let x ∈ K\{0},1 ∑h(x) =n v log + |x| v ,[K : K 0 ]v∈V Kwhere log + = max{log, 0}.In [13], we introduced the following heights for F a finite subset of M d (K)\{0},1 ∑(2) h(F ) =n v log + ||F || v[K : K 0 ]v∈V K

8 EMMANUEL BREUILLARDWe also defined the normalized height of F as1ĥ(F ) = limn→+∞ n h(F n 1 ∑) =n v log + R v (F )[K : K 0 ]v∈V Kand the minimal height of F as1 ∑e(F ) =n v log + E v (F )[K : K 0 ]v∈V Kwhere we have denoted by E v (F ) (resp. R v (F )) the quantity E Kv (F ) (resp.R Kv (F )) defined above. Observe that the height h(F ) depends on the choiceof basis in K d , while the normalized height ĥ(F ) and minimal height e(F )do not. We will often write h = h f + h ∞ to distinguish the finite part andthe infinite part of the height in the obvious way.In [13] we proved the following results:Lemma 3.1. ([13] Proposition 2.18) There is a constant c 1 = c 1 (d) > 0 suchthat for every finite subset F in M d (Ω)e(F ) ≥ ĥ(F ) ≥ e f(F ) + c 1 · e ∞ (F ) min{1, e ∞ (F )}It is easy to verify that ĥ(F ) = 0 if and only if e(F ) = 0 if and only if〈F 〉 is virtually unipotent. Note in particular that if char(Ω) > 0, there areno infinite places so the normalized height and the minimal height coincide,andLemma 3.2. If char(Ω) = p > 0 then ĥ(F ) = 0 if and only if 〈F 〉 finite.Proof. The if part follows from the definition of ĥ(F ). Suppose now thatĥ(F ) = 0. Then for every eigenvalue λ of an element g ∈ F, h(λ) = 0,hence λ is of finite order and belongs to F p . Hence g also is of finite orderas both the semisimple part g s and the unipotent part g u are of finite order.But Shur’s theorem (see [16]) says that any finitely generated torsion lineargroup is finite.□The main theorem of [13] is the following.Theorem 3.3. (Height Gap, [13] Theorem 1.1) There is a constant ε =ε(d) > 0, such that if F is a finite subset of SL d (Q) generating a non virtuallysolvable subgroup, thenĥ(F ) > εGiven a Chevalley group G, there is a special choice of basis of the Liealgebra g = Lie(G) which is made of weight vectors of a maximal split torusand defines a Z-structure on G (see Steinberg’s notes [38], and Paragraph 6.3below). With respect to this basis and viewing G as a subgroup of SL d (g)we may define the height h(g) for any g ∈ G(Ω) as in (2). We then have:

A STRONG TITS ALTERNATIVE 9Theorem 3.4. ([13] Proposition 3.3) If G is a Chevalley group, then thereis a constant C = C(G) > 0 and a Zariski open subset O = O(G) of G × Gsuch that for any choice of Ω and for any pair (a, b) ∈ O(Ω), there is g ∈ G(Ω)such that, setting F = {a, b},(3) h(gF g −1 ) ≤ C · ĥ(F )Unlike Theorem 3.4, there is no analog of Theorem 3.3 for Ω = F p (t). In[13], we proved Theorem 3.4 when Ω = Q because we were only concernedwith characteristic zero. However the proof we gave works the same wordby word in the positive characteristic case, and is even simpler since in thatcase there are no infinite places : in particular the additive constants C ∞and C ′ ∞ that are obtained along the way vanish and the use of Theorem 3.3to get rid of them is not needed (see [13]).3.1. Arakelov Height on Grassmannians. Here we record some wellknownfacts about Arakelov heights. Let K be a global field. The Arakelovheight on the projective space P(K d ) is defined as follows (see [6]) for x =(x 1 : ... : x d ),1 ∑h Ar (x) =n v log ||x|| v[K : K 0 ]v∈V Kwhere ||x|| v is the standard norm on Kv d as defined above. It is well definedthanks to the product formula (1) and always non-negative. This allows todefine the height of a projective linear subspace of P(K d ). Indeed if W ≤P(K d ) is such then we seth Ar (W ) = h Ar (Λ dim W W )where Λ dim W W is the wedge product of W viewed as a projective point in theprojective space P(Λ dim W K d ). By convention we set h Ar ({0}) = 0. Recallthat the following holds for two projective linear subspaces (see [6]) V andWh Ar (V ) + h Ar (W ) ≥ h Ar (V + W ) + h Ar (V ∩ W ).Moreover for every linear form f, seen as a point in the dual space (K d ) ∗ ,h Ar (f) makes sense as we have (see [6]),h Ar (ker f) = h Ar (f)and more generally, h Ar (W ) = h Ar (W ⊥ ), where W ⊥ is the orthogonal of Win (K d ) ∗ .Also note that if g ∈ SL d (K) and W is a subspace of K d , thenh Ar (gW ) ≤ d · h(g) + h Ar (W )where h(g) = h({g}) as defined in the last paragraph.h(g −1 ) ≤ (d − 1)h(g).Note also that

10 EMMANUEL BREUILLARDDefinition 3.5. Given A ∈ SL d (K), we will say that a vector subspaceW (or its projectivization) is A-admissible if it is a sum of generalizedeigenspaces of A. We also denote by W c its complementary subspace, i.e.the sum of the remaining generalized eigenspaces.Lemma 3.6. Let W be an A-admissible subspace. Thenh Ar (W ) ≤ d 2 · (2h(A) + ε Ω log 2)where ε Ω = 0 if car(K) > 1 and 1 if car(K) = 0.Note that if A, B ∈ M d (K), then h(A + B) ≤ h(A) + h(B) + ε Ω log 2.Moreover, if α is an eigenvalue of A, then h(α) ≤ h(A). Also h Ar (ker A) ≤(rk(A)) · h(A). Indeed h Ar (ker A) = h Ar ((ker A) ⊥ ) = h Ar (ImA t ). But forB ∈ M d (K), h Ar (ImB) ≤ (rk(B)) · h(B), since we may choose a subset ofthe canonical basis, say e 1 , ..., e k such that Be 1 , ..., Be k generates ImB, and||Be 1 ∧ ... ∧ Be k || v ≤ ||B|| k v where k = rk(B). With these observations inhand we can prove Lemma 3.6.Proof. We have W = ⊕ E α for some eigenvalues α of A, where E α is thecorresponding generalized eigenspace. Hence h Ar (W ) ≤ ∑ h Ar (E α ). If n α =dim E α , then E α = ker(A − α) nα . Hence h Ar (E α ) ≤ d · h((A − α) nα ) =dn α · (2h(A) + ε Ω log 2). Hence the result.□4. ProximalityIn this paragraph we recall the well-known notion of a proximal elementin SL d (k), where k is a local field, and we show some precise estimates asto how such elements act on the projective space P(k d ). The results of thisparagraph are contained in Lemma 4.6 and Lemma 4.7 below.A element a ∈ SL d (k) is said to be proximal if there is a unique (multiplicityone) eigenvalue of a with maximum modulus Λ k (a). We will also needto consider almost proximal elements where the eigenvalues which are largerthan, say, some ω are much larger than all other eigenvalues.Lemma 4.6 computes the rate of convergence to the attracting point ofpowers of a given proximal element a in terms of three quantities : its norm||a||, the modulus of its maximal eigenvalue Λ k (a) and the modulus of itssecond to maximal eigenvalue λ k (a). A similar estimate is given for an almostproximal element depending on the choice of the cursor ω. Lemma 4.7 is aconverse statement originally used by Tits in the proof of his alternativewhich gives a sufficient condition for a ∈ SL d (k) to be proximal : it is assoon as a stabilizes some open subset where it contracts distances.We had to be careful in those estimates, and they differ in some non insignificantways from the estimates used in earlier works (as in [1][14]). In

A STRONG TITS ALTERNATIVE 11particular they are uniform over all ultrametric local fields. The multiplicativeconstants C k,i ’s that appears in the estimates always disappears when kis ultrametric. This will turn out to be crucial for us in the sequel.4.1. The Fubini-Study metric on P(k d ). Let k be a local field and k analgebraic closure of k. Recall that we endow the projective space P(k d ) withthe standard (Fubini-Study) distance defined by(4) d([u], [v]) =||u ∧ v||||u|| · ||v||for any u, v ∈ k d \{0} and || · || is the standard norm on k d (i.e. Euclideannorm if k is archimedean and sup norm if k is non archimedean). To avoidheavy notation, we will denote by the same letter a non zero vector, orsubspace of k d and its projectivization in P(k d ). This ambiguity should notlead to any serious confusion.We denote by K k the maximal compact subgroup of SL d (k) equal toSO(d, R) if k = R, SU(d, R) if k = C, and SL d (O k ) is k is ultrametric.Its action on P(k d ) preserves d (in fact this characterizes d up to compositionby some positive function).Lemma 4.1. Let h ∈ SL d (k). Then Lip(h) ≤ (||h||·||h −1 ||) 2 ≤ ||h|| 2d , whereLip(h) is the smallest constant L ≥ 0 such that d(hx, hy) ≤ L · d(x, y) forall x, y ∈ P(k d ).Proof. Writing h in Cartan’s K k AK k decomposition of SL d (k), one sees thatwe can assume that h is diagonal and we are thus reduced to a straightforwardverification.□Recall that if H is a hyperplane in k d , and f a non zero linear form on k dwith kernel H, then if u ∈ k d \{0}, its distance to H is(5) d(u, H) = |f(u)|||f|| · ||u||where ||f|| = sup{|f(x)|, ||x|| ≤ 1, x ∈ k d }. More generally, if V and W aretwo k-subspaces in direct sum, i.e. V ⊕ W = k d , then(6) d(V, W ) =||v ∧ w||||v|| · ||w||where v = v 1 ∧ ... ∧ v l and w = w 1 ∧ ... ∧ w d−l for any basis (v 1 , ..., v l ) of Vand (w 1 , ..., w d−l ) of W. In particular, when k is archimedean, two subspacesare orthogonal if and only if they are at distance 1. Let (e 1 , ..., e d ) be thecanonical basis in k d .

12 EMMANUEL BREUILLARDLemma 4.2. Let f be a non-zero linear form on k d and H = ker f. Let Va k-subspace in k d and V ∗ the orthogonal of V in the dual of k d . Then forevery v ∈ V,(7) d(v, H) = d(v, V ∩ H) · d(f, V ∗ )Proof. Observe that as K k permutes transitively the k-subspaces of givendimension and preserves d, we may assume that V = 〈e 1 , ..., e p 〉 for some p ∈[0, d]. Then we may write f in the dual canonical basis f = ∑ f i e ∗ i = f < +f >where f < is the part of the sum involving indices i ≤ k and f > the other part.Let e > = e ∗ p+1 ∧ ... ∧ e ∗ d . Then ||f|| · d(f, V ∗ ) = ||f ∧ e|| = ||f < ∧ e|| = ||f < ||.On the other hand note that f < coincides with f on V. Hence for v ∈ V ,d(v, V ∩ H) =do obtain (7).|f(v)|||f < ||·||v||. As d(v, H) =|f(v)|||f||·||v||, combining these relations we□Lemma 4.3. Let V ⊕ W = k d and H a hyperplane in k d with V H. Let πbe the linear projection onto V with kernel W . Then for every u ∈ P(k d )\Wwe haved(πu, V ∩ H) ≥ d(u, W + V ∩ H) · d(V, W )Proof. Write u = πu + πu ⊥ ∈ V ⊕ W . If v 1 , ..., v k−1 is a basis of V ∩ H andw 1 , ..., w d−k a basis of W we set v = v 1 ∧...∧v k−1 and w = w 1 ∧...∧w d−k . Wehave d(πu, V ∩H) ≥ d(πu, W +V ∩H) = ||πu∧v∧w||||u||= d(u, W +V ∩H)· .||πu||·||v∧w|| ||πu||We may assume u /∈ V , then on the other hand d(V, W ) ≤ d(πu, πu ⊥ ) =||u∧πu ⊥ ||≤ ||u||||πu||·||πu ⊥ || ||πu||. We are done.□4.2. Contraction properties of proximal and almost proximal elements.For a ∈ SL d (k) we set E λ its generalized eigenspace with eigenvalueλ. In this paragraph, we will assume that eigenvalues of a belong to k. We letΛ k (a) = max{|µ| k , µ eigenvalue of a} and λ k (a) the modulus of the secondheighest eigenvalue of a. An element a ∈ SL d (k) is said to be proximal ifλ k (a) < Λ k (a).To deal with non proximal elements we introduce some positive real numberω > 0, such that Λ k (a −1 ) −1 < ω ≤ Λ k (a). We set Λ ω k (a) = min{|µ| k, µeigenvalue of a, |µ| k ≥ ω} and λ ω k (a) = max{|µ| k, µ eigenvalue of a, |µ| k < ω}.Lemma 4.4. Suppose a ∈ SL d (k) and let A = Λ k (a)Λ k (a −1 ) ≥ 1. For everyε > 0 there is η = η(ε, d) > 0 and ω such thatand(8) A η ·Λ k (a −1 ) −1 < ω ≤ Λ k (a)( ) 1Λk (a)εΛ ω k (a)≤Λ ω k (a)λ ω k (a)

A STRONG TITS ALTERNATIVE 13Proof. Let λ 1 , ..., λ d be the eigenvalues of a ordered as |λ 1 | k ≥ ... ≥ |λ d | k . Letl i = log |λ i||λ i+1≥ 0. Fix ε > 0 and take some η > 0. We claim that for η small|enough, there exists i 0 ∈ [0, d−2] such that l i0 +1 −η log A ≥ 1 ·(l ε 1 +...+l i0 ).Indeed, otherwise we would have l 1 < η log A, l 2 < η log A + 1l ε 1, etc, untilwe get log A = l 1 + ... + l d−1 ≤ C(ε, d)η log A for some computable constant1C(ε, d), a contradiction if η is smaller than say . Let ω = |λ 2C(ε,d) i 0 +1| k . Weare done.□When ε is fixed and ω so given by Lemma 4.4, we will refer to a as beingalmost proximal for ω.We will let Haω be the vector subspace equal to the sum of the E λ ’s forwhich |λ| k ≤ λ ω k (a). Similarly, we denote its complementary subspace byVaω = ⊕ E λ , the sum being over those λ’s such that |λ| k ≥ Λ ω k (a). We let πω abe the linear projection onto Vaω with kernel Ha ω . We also set l ω = dim Va ω .If a is proximal, we will drop the superscript ω (and set it to be Λ k (a)) andsimply denote by V a , H a , and π a the corresponding quantities.Remark 4.5. Note that if a ∈ GL d (k), then its eigenvalues belong to theextension of k generated by all algebraic extensions of k in a given algebraicclosure k of degree at most d (there are finitely many such). So this extensionis also a local field. Hence up to passing to this finite extension one mayalways assume that the eigenvalues of a belong to k.Lemma 4.6 below is the main result of this section. Its proof will occupythe subsequent two paragraphs. When k is archimedean, let C k = 2 andC k,1 = d. When k is ultrametric set C k = C k,1 = 1. Let also p(d) = 10 d(these are only given as crude estimates, we made no attempt at findingsharp constants in this statement). Finally let L ω k (a) = 1 if k is ultrametricwhile L ω k (a) =Λ ω k (a)if k is archimedean.(a)Λ ω k (a)−λω kLemma 4.6. Let a ∈ SL d (k) whose eigenvalues belong to k and assume ωis a real number such that Λ k (a −1 ) −1 < ω ≤ Λ k (a). Let l ω = dim Vaω and πaωthe projection on Vaω with kernel Ha ω . Then for any u ≠ v ∈ P(k d ), and anyinteger n ∈ N(9)d(a n u, π ω a (a n u)) · d(u, H ω a ) ≤ (C k · ||a|| k ) p(d) ·Furthermore(10)d(a n u, a n v) · d(v, H ω a ) · d(u, H ω a )d(u, v)(C lωk,1 ·≤ (C k·L ω k (a)·||a|| k ) p(d)·( ) ) lω−1 Λk (a)Λ ω k (a) · λω k (a)nΛ ω k (a)(C 2lω+2k,1·( ) ) 2lω−1Λk (a)Λ ω k (a) · λk(a)nΛ k (a)

14 EMMANUEL BREUILLARDObserve that (9) says nothing if the quantity inside the bracket is not < 1.The following Tits Converse Lemma is useful when one needs to build anelement x such that both x and x −1 are proximal.Lemma 4.7. (Tits Converse Lemma [39]) Let a ∈ SL d (k). Assume thatthere exists a point v ∈ P(k d ) and an open neighborhood U of v such thataU ⊂ U and such that Lip(a |U ) < 1, where Lip(a |U ) is the smallest constantL > 0 such that d(ax, ay) ≤ L·d(x, y) for every x, y ∈ U. Then a is proximal,V a ∈ U and λ k(a)Λ k (a) ≤ Lip(a |U).Proof. The compact subset aU is stable under a and on it a contracts distances.It follows immediately that all orbits (a n u) n≥0 converge to the uniquefixed point v a of a in aU. Let α be the eigenvalue of a with eigenvector v a .Let β be another eigenvalue of a (if α has multiplicity higher than 1, we maytake β = α). There is a non zero vector w such that aw = βw + κv a forsome κ ∈ k. Let ε ∈ k\{0} with |ε| k arbitrarily small. Then one computesfrom (4) lim |ε|→0d(a(v a+εw),v a)d(v a+εw,v a)thus |β||α| ≤ Lip(a |U) < 1. We are done.= |β||α| . If |ε| k is small enough, v a + εw ∈ U and4.3. Four intermediary geometric lemmas. In this paragraph, we stateand prove four intermediary results needed in the proof of Lemma 4.6. Unlessotherwise stated a ∈ GL d (k) and its eigenvalues belong to k.Lemma 4.8. Let a ∈ GL d (k) and α an eigenvalue of a. Then there is someh ∈ K k such that hah −1 is a lower triangular matrix with top left entry equalto α.Proof. Since eigenvalues of a belong to k, a and hence also its transpose a tare triangularizable over k, i.e. a t stabilizes a full k-flag F. We may alsoassume that F starts with the line kv, where v is an eigenvector of a t witheigenvalue α. But full k-flags are conjugate under GL d (k). Hence F = gF 0where F 0 is the standard flag generated by the canonical basis of k d andge 1 = v. The Iwasawa decomposition reads GL d (k) = K k B 0 where B 0 isthe Borel stabilizing F 0 . Thus we may assume that g ∈ K k . Thus g −1 a t gstabilizes F 0 and is upper triangular. Hence h = g t ∈ K k will do. □Let C k,1 be equal to d if k is archimedean and equal to 1 if k is ultrametric.Lemma 4.9. Let a ∈ GL d (k). Then there exists an h ∈ SL d (k) such that‖hah −1 ‖ ≤ C k,1 Λ k (a) and max{‖h‖ , ||h −1 ||} ≤ ‖a‖ d−12.Proof. By Lemma 4.8, one may assume that a ∈ GL d (k) is lower triangular.Let h = t d+12 diag(t −1 , ..., t −d ) ∈ SL d (k). Choose t ∈ k such that |t −1 | k =||a|| k . Then max{‖h‖ , ||h −1 ||} ≤ ‖a‖ d−12and the off-diagonal coefficients ofhah −1 are of modulus ≤ 1. As ||a|| ≤ C k,1 max |a ij |, we are done. □□

A STRONG TITS ALTERNATIVE 15Remark 4.10. Note that we also get ||Λ 2 (hah −1 )|| ≤ C 4 k,1 Λ k(a)λ k (a).Recall that C k is 2 if k is archimedean and 1 if k is ultrametric.Lemma 4.11. Let a ∈ SL d (k) and ω with Λ k (a −1 ) −1 < ω ≤ Λ k (a). Letl ω = dim Va ω and L ω k (a) = 1 if k is ultrametric while Lω k (a) = Λ ω k (a)if kΛ ω k (a)−λω k (a)is archimedean. Then d(Va ω , Ha ω ) −1 ≤ (C k · L ω k (a)||a||l ) ((d l)−1) .Proof. First let us assume a is proximal (and ω = Λ k (a)) with eigenvalueof maximal modulus α 1 , and let α 2 , ..., α d be the remaining eigenvalues. ByLemma 4.8, one may assume that a ∈ SL d (k) is lower triangular with α 1 inthe upper left corner. Let V a ∈ k d \{0} be such that aV a = α 1 V a . As V a /∈H a = 〈e 2 , ..., e d 〉 , we may assume that V a = (1, x 2 , ..., x d ) in the canonicalbasis. Then d(V a , H a ) = 1/||V a ||. Decomposing aV a = α 1 V a in coordinates,we obtain (α 1 − α 2 )v 2 = a 21 , (α 1 − α 3 )v 3 = a 31 + a 32 v 2 , etc. This allowsto recursively estimate each v i and at the end we get that ||V a || ≤ (1 +L 2 k||a|| 2Λ k (a) 2 ) (d−1)/2 ≤ 2 (d−1)/2 · L d−1(k· ||a||d−1Λ k (a))when k is archimedean while( d−1 ||a||Λ k (a)).when k is ultrametric ||V a || ≤ L d−1kWe now explain how to reduce the general case to the proximal case.Let (v 1 , ..., v l ) and (w 1 , ..., w d−l ) be respective basis of Va ω and Ha ω . Let v =v 1 ∧ ... ∧ v l and w = w 1 ∧ ... ∧ w d−l . From (6) we have d(Va ω , Ha ω ) = ||v∧w|| . ||v||·||w||The canonical map Λ l k d × Λ d−l k d → k establishes an isomorphism betweenΛ d−l k d and the dual of Λ l k d . Under this identification w is a linear form onΛ l k d and formulae (6) and (5) coincide, i.e. d(Va ω , Ha ω ) = d(Λ l Va ω , ker w). Onthe other hand Λ l a is proximal on Λ l k d with V Λ l a = Λ l Va ω and H Λ l a = ker w.Hence by the above d(Λ l V a , ker w) −1 ≤ (C k · L ω k (Λl a) ·where||Λ l a||Λ k (Λ l a) )D−1D = dim Λ l k d = ( dl)and the result follows as Λk (Λ l a) ≥ 1. □Recall that C k,1 is d if k is archimedean and 1 if k is ultrametric.Lemma 4.12. Let a ∈ SL d (k) with Λ k (a −1 ) −1 < ω ≤ Λ k (a) and l ω =dim Va ω . Set V = 〈e 2 , ..., e lω 〉, H = 〈e lω+1, ..., e d 〉 . There exists h ∈ SL d (k)with hVa ω = V, hHa ω = H such that if we set ã = hah −1 then ∥ ∥∥ã|H ≤C k,1 λ ω k (a) and ∥ (∥√ ) d(d+1)∥ã|V ≤ Ck,1 Λ k (a) and ‖h −1 2Ck,1‖ ≤‖a‖ d−1d(Va ω,Hω a ) 2.Proof. First note that applying Lemma 4.8 we can assume that a is lowertriangular and that Ha ω = H. Then observe that for any subspace F of k d ,one may find a basis f 1 , ..., f p of F such that ||f 1 ∧ ... ∧ f p || = 1 and ||f i || = 1for each i = 1, ..., p. Choose such a basis, say v 1 , ..., v l of Vaω and, for µ ∈ k tobe defined later, denote by h 1 ∈ GL d (k) the map h 1 v i = e i if i < l, h 1 v l = µe land h 1 e i = e i for i > l. Then compute h −11 e 1 ∧ ... ∧ h −11 e d = det(h −11 )e =

16 EMMANUEL BREUILLARDµ −1 v ∧ w where e = e 1 ∧ ... ∧ e d , v = v 1 ∧ ... ∧ v l and w = e l+1 ∧ ... ∧ e d .Now choose µ ∈ k so that det(h 1 ) = 1, then |µ| k = ||v ∧ w|| = d(Va ω , Ha ω ).Then ||h −11 || ≤ |µ −1 | k when k is ultrametric and ||h −11 || ≤ √ d|µ −1 | k when kis archimedean.So h 1 ah −11 stabilizes V and H. Now applying Lemma 4.9 on V and on H,we can find h 0 ∈ SL d (k), stabilizing V and H such that ||h 0 h 1 ah −11 h −10|V || ≤C k,1 Λ k (a) and ||h 0 h 1 ah −11 h −10|H || ≤ C k,1λ ω k (a) and ||h−1 0 || ≤ ||h 1 ah −11 || d+12 . Seth = h 0 h 1 we are done.□4.4. Proof of Lemma 4.6. For l ∈ [1, d − 1] set as above V = 〈e 2 , ..., e l 〉and H = 〈e l+1 , ..., e d 〉 . Let b ∈ SL d (k) be such that bV = V and bH = Hand let π b be the linear projection onto V with kernel H. We first claim thatfor every u ∈ P(k d )(11) d(u, H) · d(bu, π b (bu)) ≤ ||b |H || · ||(b |V ) −1 ||Indeed, note that d(u, H) = ||u∧w||||u||≤ ||π b(u)||||u||where w = e l+1 ∧...∧e d and writingu = π b (u) + π b (u) ⊥ we have d(bu, π b (bu)) = ||b |Hπ b (u) ⊥ ∧π b (u)||||bu||Combining both inequalities we get (11).Now we claim that for any u ≠ v ∈ P(k d ) we claim that:(12)d(u, H) · d(v, H) · d(bu, bv)d(u, v)≤ ||b |H||·||u||||bπ b (u)|| .≤ max{||Λ 2 b |V ||, ||Λ 2 b |H ||, ||b |H ||·||b |V ||}·||(b |V ) −1 || 2Indeed, using Cartan’s K k AK k decomposition on V and H separately, wemay assume that b is diagonal diag(α 1 , ..., α l ). Then write bu ∧ bv = bu |V ∧∑bv |V + bu |H ∧ bv |V + bu |V ∧ bv |H + bu |H ∧ bv |H . Since bu |H ∧ bv |V + bu |V ∧ bv |H =1≤i≤l α ie i ∧(u i v |H −v i u |H ) we get ||bu∧bv|| ≤ max{||Λ 2 b |V ||, ||Λ 2 b |H ||, ||b |H ||·||b |V ||} · ||u ∧ v||. On the other hand ||bu|| ≥ ||u |V ||(resp. ||bv|| ≥ ||v |V ||||b −1|V || ||b −1 ||)and|Vd(u, H) ≤ ||u |V ||(resp.d(v, H) ≤ ||v |V ||). This shows (12).||u|| ||v||We now prove (9) and (10). For n ∈ N and a ∈ SL d (k), we may applyLemma 4.12 and Remark 4.10 to a and get h ∈ SL d (k) with ã = hah −1 ∈SL d (k) satisfying ãV = V (resp. ãH = H) and ||ã |H || ≤ C k,1 λ ω k (a) (resp.∥ ∥||ã |V || ≤ C k,1 Λ k (a) and ||Λ 2 ã |V || ≤ C 4 k,1 Λ k(a)λ k (a)). Note that∥ ∥1det ã |Vã |V l ω−1≤1translate asΛ ω k (a) ·d(u, H ω a ) · d(a n u, π ω a (a n u)) ≤ Lip(h −1 ) 2 ·∥ ≤( ) lω−1C k,1 · Λk(a)Λ . Let b = (ã) n . Then (11) and (12)ωk (a)(C lωk,1 ·∥ã −1|V( ) ) lω−1 Λk (a)Λ ω k (a) · λω k (a)nΛ ω k (a)

A STRONG TITS ALTERNATIVE 17d(u, H ω a )·d(v, H ω a )·d(a n u, a n v) ≤ Lip(h −1 ) 3 Lip(h)·(C 2lω+2k,1·( ) ) 2lω−1Λk (a)Λ ω k (a) · λk(a)nΛ ω k (a)Recall that by Lemma 4.1, Lip(h) and Lip(h −1 ) are at most ||h −1 || 2d . FromLemma 4.11 we have d(Va ω , Ha ω ) −1 ≤ ( C k · L k (a) · ||a|| l) (( d l)−1) . Then fromLemma 4.12||h −1 || ≤( √Ck,1· (L k (a)C k ) ((d l)−1) ) d(d+1)2‖a‖ d−12 +l((d l)−1) d(d+1)2Note that, when k is archimedean, inequality (9) is trivial if λ ω k (a) ≥ 1 2 Λω k (a).Therefore we may assume in the archimedean case that L k (a) ≤ 2. As8d( d−1+l(( ) )d2 l − 1 d(d+1)) ≤ 10 d estimating the constant we do indeed obtain2(10) and (9).5. Ping-PongIn this technical section, we work with a fixed local field k and we explainhow to construct two short words x and y with letters in some finite setF in SL d (k) such that x and y form a ping-pong pair and thus generate afree subgroup. The goal of this introductory paragraph is to give a list ofseveral conditions of geometric nature (i) to (vi) on F and state two lemmas,Lemma 5.1 and 5.3 below, which assert precisely that these conditions aresufficient to construct the ping-pong pair. These two statements are the onlyones which will be used in further sections.As in Tits [39], the construction of the ping-pong pair follows two steps.First, starting from a proximal element a lying in F or in a bounded powerof F, we need to build a short word with letters in F , say x, such that bothx and x −1 are proximal elements (Lemma 5.1). Second, we need to find aconjugate of it, say y = cxc −1 such that x and y together play ping-pong(Lemma 5.3).The construction presented here follows verbatim that of Tits. But whileTits needed only asymptotic statements which held for sufficienlty high powersof group elements, no matter how high, we need to have control on thelength of the words. We thus have to give a quantified version of Tits’ argumentand give precise estimates at each step. More importantly, while Titsdid not need to care about the choice of a distance on P(k d ) (any one insidethe “admissible” class he defined was good for his purposes), it is crucialfor us that we work with the Fubini-Study distance introduced in Section4. The reason is that all constants then disappear and are equal to 1 forall ultrametric local fields, hence giving to us the possibility to bound thelength of the generators of the free subgroup independently of the choiceof the local field.

18 EMMANUEL BREUILLARDLet (k i ) 1≤i≤4 be four positive integers and ε 0 , T 0 , T 1 , T 2 > 0 be positivereal numbers. Let ε > 0 with ε ≤ ε 0 /12d 2 . Let k 0 be a local field. SupposeF ⊂ SL d (k 0 ) is a finite set containing 1. All eigenvalues and eigenspaces ofelements in the group generated by F are defined over a fixed finite extensionof k 0 of degree at most d!. Let k be this extension. For a subspace V ink d we denote by V ⊥ its orthogonal in the dual space of k d . We say that anon-trivial subspace W of k d is a-admissible for a ∈ SL d (k 0 ) if it is a sum ofgeneralized eigenspaces of a. We also denote by W c its complementary, i.e.the sum of the remaining generalized eigenspaces, so that k d = W ⊕ W c .List of Conditions for ping-pong (i)-(vi):Let a ∈ F k 1, b ∈ F k 2, t ∈ F k 3. Assume(i) a is proximal(ii)(13) ||F || k > C 2dk,1(14)(iii)( ) 1Λk (a)ε 0≥ Λk (a) ≥ ||F || T 1kλ k (a)(iv) For any a-admissible subspace W (see Def. 3.5) we have(15) d( t b ±1 · H ⊥ a , W ⊥ ) > ||F || −T 0k(16)(17)(v) For any a-admissible subspace W we haved(tV a , W c + W ∩ b −1 H a ) ≥ ||F || −T 0kd(t −1 V a , W c + W ∩ bH a ) ≥ ||F || −T 0kNote that condition (15) on b implies that W c + W 1 ∩ b ±1 H a are hyperplanes,so these distances are computable via (5).Lemma 5.1. There is τ 1 (d, ε) ∈ N and τ 3 = τ 3 (d, k 1 , k 2 , k 3 , ε 0 , ε, T 0 , T 1 ) ∈ Nsuch that if T 1 ≥ τ 1 and T 3 ≥ τ 3 , there is l = l(d, k 1 , k 2 , k 3 , ε 0 , ε, T 0 , T 1 , T 3 ) ∈N such that for some l 0 , l 1 ∈ [0, l] the element x = a l 0ba −l 1t is proximal aswell as x −1 and Λ k (x) ≥ Λ k (a) T 3λ k (x) and Λ k (x −1 ) ≥ Λ k (a) T 3λ k (x −1 ) andd(V x , H x ) ≥ Λ k (a) −2T 3and d(V x −1, H x −1) ≥ Λ k (a) −2T 3.We let k 4 = 2k 1 l + k 2 + k 3 so that x ∈ F k 4.Remark 5.2. As Y. Benoist observed in [4] (see also J-F. Quint [32]) itis possible to construct Zariski-dense semi-groups, say in SL 3 (Q p ) whichare made exclusively of proximal elements whose inverses are not proximal.Hence our method does not allow in general (the SL 2 case is fine however)to construct the generators of a free subgroup as positive words in F .

A STRONG TITS ALTERNATIVE 19Assume T 1 and T 3 satisfy the assumptions of Lemma 5.1 let x be theelement we get. Assume that there is c ∈ F k 5such that(vi)(18) d(c ±1 (V x ∪ V x −1), H x ∪ H x −1) ≥ 1||F || T 2kLemma 5.3. Then there is l 2 = l 2 (d, (k i ) 1≤i≤5 , ε, ε 0 , (T i ) 0≤i≤3 ) ∈ N such thatfor every n ≥ l 2 , x n and y = cx n c −1 play ping-pong on P(k d ) and generate afree subgroup of SL d (k).5.1. Cayley-Hamilton trick. In [39] Tits used the fact that if a 0 ∈ GL d (k)has all eigenvalues of the same modulus and if a vector v lies far from ahyperplane H then for a set of positive density of n ∈ N the vectors a n · v liefar from H. In [18], Eskin-Mozes-Oh made clever use of the Cayley-Hamiltontheorem in order to show a statement of a similar nature which also gives abound on the smallest appropriate n. The following lemma is a reformulationof the same trick.Recall that C k,2 is d2 d when k is archimedean and 1 when k is ultrametric.The following lemma, which we will use in the proof of Claim 0 below,expresses the same idea.Lemma 5.4. Let a 0 ∈ GL d (k), let H be a hyperplane in k d and let v ∈ P(k d ).Then there is some integer j 0 ∈ [1, d − 1] such that(19) d(a j 00 v, H) ≥ 1 ( ) j0 Λk (a 0 )·· | det a 0| k · d(v, H)C k,2 ||a 0 || k Λ k (a 0 )dProof. Let Λ ∈ k such that |Λ| = Λ k (a 0 ) and set ã 0 = a 0Λ. According tothe Cayley-Hamilton theorem, there are coefficients (c j ) 1≤j≤d−1 in k suchthat ∑ d−1j=1 c jã j 0 = det ã 0 . Moreover |c j | k ≤ ( dj)≤ 2 d when k is archimedean,when |c j | k ≤ 1 when k is ultrametric. Let f be a linear form on k d with||f|| k = 1 and ker f = H. There must exist some j 0 ∈ [1, d − 1] such thatj|c j0 f(ã 0 0 v)| k ≥ | det ã 0| kC k,1· |f(v)| k where C k,1 is d if k is archimedean and 1 ifk is ultrametric. Hence |f(a j 00 v)| k ≥ 1C k,2· Λ k (a 0 ) j0 · | det a 0| kΛ k (a 0· |f(v)|) d k and (19)follows.□5.2. Proof of Lemma 5.1. Recall that a is proximal but a −1 may not be.However, as we have fixed ε > 0, Lemma 4.4 gives us some ω for whicha −1 is almost proximal. Let α = λω k (a−1 ). It also give η = η(d, ε) > 0.Λ ω k (a−1 )Recall that ε 0 , T 0 and T 1 are defined in (13) to (16). We assume here thatT 1 ≥ τ 1 := max{2/η, 3/ηε, 4/ε 0 } and ε ≤ ε 0 /12d 2 and let ε 1 = ε 04.

20 EMMANUEL BREUILLARDClaim 0: There is n 0 = n 0 (k 1 , T 0 , T 1 , ε, d) ∈ N such that for all n ≥ n 0there exists j 0 (n), j 1 (n) ∈ [1, d − 1] such thatmin { d(a −nj 0(n) t −1 V a , bH a ), d(a −nj 1(n) tV a , b −1 H a ) } ≥||F || −rkC k,2 · (C k,1 · α −ε ) dn · C p(d)+1kwhere r = 2T 0 + k 1 (2p(d) + d(d − 1)) and p(d) = 10 d .Claim 1: Let ε 1 = ε 04. There exists n 1 = n 1 (k 1 , k 2 , k 3 , T 0 , T 1 , ε, ε 0 , d) ∈ Nsuch that for every n ≥ n 1 the ball B n = B(V a , Λ k (a) −ε1n ) (resp. B n ′ =B(t −1 V a , Λ k (a) −ε1n ) is mapped into Bn − = B(V a , Λ k (a) −2ε1n ) (resp. B n ′− =B(t −1 V a , Λ k (a) −2ε1n )) by x n = a nj0(n) ba −nj1(n) t (resp. x −1n ).Claim 2: Under the assumptions of Claim 1, there is n 4 ∈ N dependingonly on k 1 , k 2 , k 3 , T 0 , T 1 , ε, ε 0 and d such that for any n ≥ n 4 we also haveLip(x n|Bn ) ≤ Λ k (a) −ε1n (resp. Lip(x −1 ) ≤ Λn|Bn− k (a) −ε1n ).The proofs of these claims are straightforward once we have at our disposalthe Lemmas proved in Section 4 and in particular Lemma 4.6. Neverthelesswe provide full details in the next paragraph below.With these claims in hands we can quickly prove Lemma 5.1. Indeedlet n = T 3 /ε 1 . If T 3 ≥ ε 1 · max{n 0 , n 1 , n 4 } we get by Claim 2 and 3 thatx n sends B n into itself and x −1n sends Bn − into itself, while the Lipschitzconstants are ≤ Λ k (a) −ε1n . We are thus in a position to apply Tits ConverseLemma, Lemma 4.7, which says that x n and x −1n are proximal andsatisfy λ k(x n), λ k(x −1n )≤ Λ Λ k (x n) Λ k (x −1n )k(a) −T 3. Finally by Claim 2, x n maps B n into thesmaller ball Bn − , which must then contain V xn while B n cannot intersectH xn . It follows that d(V xn , H xn ) ≥ d(B n , (Bn − ) c ). But we see that in both thearchimedean and the ultrametric case:d(B n , (B − n ) c ) ≥ 1 C kΛ k (a) −T 3≥ Λ k (a) −2T 3as soon as Λ k (a) −T 3< 1/C k , which holds if T 3 ≥ 1 for instance. A similarargument takes place for x −1n . This ends the proof of Lemma 5.1.5.2.1. Proof of Claim 0. Let W = V ωa −1 and hence W c = H ω a −1 and π theprojection on W with kernel W c . Recall that (14) gives(20) α ≤ Λ k (a) −η ≤ ||F || −ηT 1When k is archimedean this together with (13) forces L ω k (a−1 ) ≤ 2 if ηT 1 ≥ 2.Indeed, we have α ≤ ||F || −ηT 1≤ 1 2 i.e. λω k (a−1 ) ≤ 1 2 Λω k (a−1 ).We do the calculation for u = tV a , keeping in mind that an entirely analogouscalculation can be done for u − = t −1 V a at the same time at each step.Let n ∈ N be arbitrary.

A STRONG TITS ALTERNATIVE 21Since as d(u, W c ) ≥ d(u 0 , W c + W ∩ b −1 H a ) ≥ ||F || −T 0we can combineLemmas 4.4, 4.6 (a) and (16) to get(d(a −n u, π(a −n (21) u)) ≤ (C k · ||a −1 || k ) p(d) ·≤C p(d)k· ||F || T 0+dk 1 p(d)kOn the other hand according to Lemma 4.3,C lω 1k,1 ·( ) )Λk (a −1 lω−1n)· λω k (a−1 )· d(u, W c ) −1Λ ω k (a−1 ) Λ ω k (a−1 )· (C d k,1 · α 1−dε) n(22) d(πu, W ∩ b −1 H a ) ≥ d(u, W c + W ∩ b −1 H a ) · d(W, W c )But by Lemma 4.11(23) d(W, W c ) −1 ≤ (C 2 k · ||a −1 || lω ) (( dlω )−1) ≤ (C k · ||a −1 || k ) p(d)because when k is archimedean L ω k (a−1 ) ≤ 2 as explained above. Hence (22)(23) and (16) give(24) d(πu, W ∩ b −1 H a ) −1 ≤ C p(d)k· ||F || T 0+dk 1 p(d)kWe may now apply Lemma 5.4 to a 0 = a −n restricted to W. We find j 1 ∈[1, d − 1] such that(25)(d(a −nj 1 ||aπu, W ∩b −1 H a ) −1 −1 || k≤ C k,2·Λ k (a −1 )) nj1·(Λk (a −1 )Λ ω k (a−1 )) lωn·d(πu, W ∩b −1 H a ) −1But Lemma 4.9 applied to a −1 gives an h ∈ SL d (k) such that ||ha −1 h −1 || ≤C k,1 · Λ k (a −1 ) and max{||h||, ||h −1 ||} ≤ ||a −1 || d−12 . Hence ||a −n || k ≤ ||h|| ·||h −1 || · (C k,1 · Λ k (a −1 )) n and ||a−n || k≤ Λ k (a −n ) ||a−1 || d−1 · Ck,1 n . Thus combining (24)and (25) and bearing in mind that(26)Λ k (a −1 )Λ ω k (a−1 ) ≤ α−ε(this is 8), we get(27)d(a −nj 1πu, W ∩ b −1 H a ) −1 ≤ C k,2 · (C k,1 α −ε ) nd · C p(d)k· ||F || T 0+dk 1 p(d)+k 1 d 2 (d−1)kCompare (21) and (28). When k is ultrametricd(a −n u, π(a −n u)) ≤ ||F || T 0+dk 1 p(d)kα n(1−dε)< ||F || −T 0−dk 1 p(d)−k 1 d 2 (d−1)kα εnd< d(a −nj 1πu, W ∩ b −1 H a )

22 EMMANUEL BREUILLARDas soon as α n(1−2dε) < ||F || −rkwhere r = r(T 0 , d) := 2T 0 +dk 1 (2p(d)+d(d−1)).As α −1 ≥ Λ k (a) η by Lemma 4.4, this happens as soon asrn >T 1 η(1 − εd)Similarly, if k is archimedean, d(a −n u, π(a −n u)) ≤ 1 2 d(a−nj 1πu, W ∩ b −1 H a )as soon as n > n 0 (T 0 , T 1 , ε, d) for some computable constant n 0 . Finallywhether k is archimedean or ultrametric we get:(28)d(a −nj 1u, W ∩ b −1 H a ) ≥ C −1k,2 · (C−1 k,1 · αε ) dn · C −p(d)−1k· ||F || −T 0−k 1 p(d)−k 1 d 2 (d−1)k·Finally applying Lemma 4.2 and (15) we getd(a −nj 1u, b −1 H a ) ≥ d(a −nj 1u, W ∩ b −1 H a ) · d( t b −1 · H ⊥ a , W ⊥ )≥ C −1k,2 · (C−1 k,1 · αε ) dn · C −p(d)−1k· ||F || −rkThis ends the proof of Claim 0.5.2.2. Proof of Claim 1. First recall as in Claim 0 that L ω k (a−1 ) ≤ 2 whenk is archimedean (since ηT 1 ≥ 2, which we assume). We give the proof forx n and B n keeping in mind that the same arguments are being performed atthe same time and at each step for x −1n and Bn − .We first justify the following:Claim 1.1: There is m 0 = m 0 (d, ε, k 3 , T 0 , T 1 ) ∈ N such that for n ≥ m 0and u ∈ B(V a , α 3dεn )1(29) d(tu, W c ) ≥ 1 C kd(tV a , W c ) ≥C k ||F || T 0kIndeed, the second inequality is just (16), while to get the first, it is enough1that d(tu, tV a )

which by Claim 0 reduces to showA STRONG TITS ALTERNATIVE 23(32) d(a −nj1(n) tu, a −nj1(n) tV a ) < C −1k,2 · (C−1 k,1 · αε ) dn · C −p(d)−2k· ||F || −rkBut bearing in mind (26) Lemma 4.6 (10), we have for n ≥ m 0d(a −nj 1(n) tu, a −nj 1(n) tV a )d(u, V a )Hence we get (32) as soon asC 3p+3k· C k,2 · ||F || k 1dp+k 3 2d+2T 0 +rk·Since u ∈ B(V a , α 3dεn ) this holds as soon as(33) C 3p+3k· C k,2 · ||F || k 1dp+k 3 2d+2T 0 +rk·≤ Lip(t) · d(a−nj 1(n) tu, a −nj1(n) tV a )d(tu, tV a )(C2d+2≤ Lip(t) · (Ck 2 · ||a −1 ||) p k,1α −ε(2d−1)) n·d(tu, W c ) · d(tV a , W c )≤ C 2p+1k· ||F || k 1dp+k 3 2d+2T 0(C2d+2k·k,1α −ε(2d−1)) n(C3d+2k,1α −ε(3d−1)) n· d(u, Va ) < 1(C3d+2k,1α ε) n< 1Since α ε ≤ ||F || −ηεT 1kby (20) and C 3d+2k,1≤ ||F || 2 k by (13) while we assumedηεT 1 ≥ 3, we get the existence of n 2 = n 2 (ε, d, T 0 , T 1 , k 1 , k 3 ) ∈ N for which(33) holds for n ≥ n 1 . Hence (30) holds and Claim 1.2. is proved.With (30) in hand we can apply Lemma 4.6 (9) to positive powers of athis time and get:Claim 1.3.: Suppose ε 0 ≥ 12εd 2 and fix ε 1 = ε 0 /4. There is n 3 ∈ N dependingon ε, ε 0 , d, T 0 , T 1 , k 1 , k 2 , k 3 such that for n ≥ n 3 and u ∈ B(V a , Λ k (a) −ε1n )we have for x n = a nj0(n) ba −nj1(n) td(x n u, V a ) < Λ k (a) −ε 1nProof of Claim 1.3.: First note that Λ k (a) −ε 1≤ α 3dε because α −1 ≤Λ k (a)Λ k (a −1 ) ≤ Λ k (a) d and ε 1 = ε 0 /4 ≥ 3εd 2 . Lemma 4.6 (9) translates as(d(a nj0(n) ba −nj1(n) tu, V a ) ≤ (C k · ||a|| k ) p(d) · Ck,1 4 · λk(a) ) n· d(ba −nj1(n) tu, H a ) −1Λ k (a)(≤ C p k · ||F ||k 1p(d)+2dk 2 C4) nk,1k·· d(a −nj1(n) tu, b −1 HΛ k (a) ε a ) −10(≤ C 2p+2kC k,2 · ||F || k 1p(d)+2dk 2 +r Cd+4k,1k· · ) nα−εdΛ k (a) ε 0where we have used successively (14) and Claim 1.2. NowC d+4k,1 · α−εd· ΛΛ k (a) ε k (a) ε 1≤0C d+4k,1Λ k (a) ε 0−ε 1≤ Cd+4 k,1−εd 2 Λ k (a) ε 0/2

24 EMMANUEL BREUILLARDbecause α −1 ≤ Λ k (a)Λ k (a −1 ) ≤ Λ k (a) d and we have assumed ε 0 ≥ 4εd 2 andε 0 = 4ε 1 . Then the existence of n 3 follows from (13) and (14). Thus Claim1.3. is proved.Working out the same three claims for x −1n and B n ′ in place of x n and B nwe get Claim 1.5.2.3. Proof of Claim 2. We apply Lemma 4.6 (10) to a nj0(n) and pointsba −nj1(n) tu and ba −nj1(n) tv for u, v ∈ B n . Recall that L k (a) ≤ 2 when k isarchimedean as λ k(a)≤ ||F Λ k (a) ||−T 1ε 0≤ 1 by (13) and (14) and since T 2 1ε 0 ≥ 1.We getd(x n u, x n v)≤ (C 2d(u, v)k||a|| k ) p(d)·( C4)k,1 λ k (a) n·d(ba −nj1(n) tu, H a ) −1·d(ba −nj1(n) tv, H a ) −1Λ k (a)Since Λ k (a) −ε 1≤ α 3dε Claim 1.2. applies and we get(d(x n u, x n v)≤ Lip(b −1 ) 2 p(d) C4) nk,1 λ k (a)· ||a|| k ·· Ck,2 2 · (C k,1 · α −ε ) 2dn · C 4p(d)+4kd(u, v)Λ k (a)()≤ C 4p+4kCk,2 2 · ||F || 4dk 2+2r+k 1 p(d)k· C 4+2d α −2dε nk,1Butα −2dεΛ k (a) ≤ 1ε 0 Λ k (a) 2ε 1Λ k (a) ε 0Hence for some computable n 4 , for all n ≥ n 4 and u, v ∈ B nd(x n u, x n v)d(u, v)≤ Λ k (a) −ε 1nA similar argument proves the claim about x −1nproved.and B ′ n. Thus Claim 2 is5.3. Proof of Lemma 5.3. Let n, k ∈ N, k = T 2 + 3dk 5 . Let B k (x) =B(V x , Λ k (a) −kT 3) and B − k (x) = B(V x −1, Λ k(a) −kT 3). Similarly, let B k (c) =B(cV x , Λ k (a) −kT 3) and B − k (c) = B(cV x −1, Λ k(a) −kT 3). Note that d(u, c ±1 V x ) C k ||F || T 2kandd(u, cV x ) ≤ Λ k (a) −kT 3< 1 C kd(cV x , H x )· ||F || 2rkHenced(u, H x ) ≥1C k ||F || T 2k≥1||F || T 2+1k

A STRONG TITS ALTERNATIVE 25Similarly if u ∈ B k (c) then d(u, H x −1) ≥ ||F || −T 2−1kd(u, H x ∪ H x −1) ≥ ||F || −T 2−1k. Finally:and if u ∈ B − k(c), then(34) d(B − k (c) ∪ B k(c), H x ∪ H x −1) ≥ ||F || −T 2−1kSimilarly we check thatd(B − k (x) ∪ B k(x), cH x ∪ cH x −1) ≥ ||F || −T 2−2dk 5 −1kWe know that for each n ≥ 1, x n maps B k into itself and x −n maps B − kintoitself. Similarly we check that cx n c −1 maps B k (c) into itself and cx −n c −1maps B − k(c) into itself.We now check that x n maps B − k (c) ∪ B k(c) into B k and x −n maps B − k (c) ∪B k (c) into B − k . From Lemma 5.1, we have λ k(x)≤ Λ Λ k (x) k(a) −T 3. By Lemma 4.6(9) applied to x, for u ∈ P(k d ),(d(x n u, V x ) · d(u, H x ) ≤ (C k · ||x|| k ) p(d) · C k,1 · λk(x) ) nΛ k (x)≤ ||F || p(d)(1+ld+k 2+k 3 )k· Λ k (a) − T 3 n2Hence if u ∈ B k (c) ∪ B − k (c), then d(u, H x) ≥ ||F || −T 2−1kby (34) andd(x n u, V x ) ≤ ||F || p(d)(1+ld+k 2+k 3 )+T 2 +1k≤ Λ k (a) −kT 3as soon as n ≥ n 5 = n 5 (l, d, (k i ) i , (T i ) i , k). Similarlyd(x −n u, V x −1) · d(u, H x −1) ≤ (C k · ||x −1 || k ) p(d) ·≤ ||F || p(d)d(1+ld+k 2+k 3 )k· Λ k (a) − T 3 n2(C k,1 · λk(x ) −1 n)Λ k (x −1 )· Λ k (a) − T 3 n2and hence d(x −n u, V x −1) ≤ Λ k (a) −kT 3if n ≥ n 6 = n 6 (l, d, (k i ) i , (T i ) i ).We check that similarly, cx n c −1 maps B − k ∪B kinto B k (c) and cx −n c −1 mapsB − k ∪ B k into B − k(c) as soon as n is larger that some fixed number dependingonly on the data (l, d, (k i ) i , (T i ) i ).Finally we check that all balls B − k , B k B − k (c), B k(c) are disjoint, sinced(V x , V x −1) ≥ d(V x , H x ) ≥ Λ k (a) −2T 3and d(V x , cV x −1) ≥ d(H x , cV x −1) ≥||F || −T 2≥ Λ k (a) −k and similarly d(cV x , V x −1) ≥ Λ k (a) −k and d(cV x , cV x −1) ≥||F || −2dk 5Λ k (a) −2T 3≥ Λ k (a) −(k−1)T 3.It follows that x n and cx n c −1 play ping-pong on P(k d ), hence generate afree subgroup. This ends the proof of Lemma 5.3.

26 EMMANUEL BREUILLARD6. Height bounds and proof of Theorem 1.16.1. A Product formula for subspaces. In this paragraph we define theadelic distance δ(V ; W ) between two projective subspaces and we give aproduct formula (36) relating it to the Arakelov heights of V, W and V + W.In Paragraph 4.1 we recalled the Fubini-Study metric on P(k d ), where kis a local field. In particular, we had formula (6), which gives the distancebetween two projective linear subspaces. If K is a global field with primefield K 0 and V and W are disjoint projective linear subspace of P(K d ), wecan put together the local distances (i.e. at each place of K) in a way similarto the way the height of an algebraic number is defined. Namely we set:(35) δ(V ; W ) =1[K : K 0 ]∑v∈V Kn v · log1d v (V, W )where d v (·, ·) is the Fubini-Study metric on P(K d v ). Each term in this sum isnon negative. In fact, we see from (6) that δ(V ; W ) is linked to the Arakelovheights (see Paragraph 3.1) in the following simple way:(36) δ(V ; W ) = h Ar (V ) + h Ar (W ) − h Ar (V + W ) ≤ h Ar (V ) + h Ar (W )This can be seen as a product formula for subspaces, since when V and Ware points in P 1 (Q) it reduces to the classical product formula on Q.Note moreover that we can similarly define δ(V ⊥ , W ⊥ ) just as δ(V, W ) inthe projective space of the dual vector space (K d ) ∗ . Since h Ar (V ) = h Ar (V ⊥ )(see [6]), we also haveδ(V ⊥ , W ⊥ ) ≤ h Ar (V ) + h Ar (W )We will often denote by δ v (V ; W ) the term of the sum in (35) correspondingto the place v, so that1 ∑δ(V ; W ) =n v · δ v (V ; W )[K : K 0 ]v∈V K6.2. The Eskin-Mozes-Oh Escape Lemma. In this paragraph we recall acrucial Lemma due Eskin-Mozes-Oh, which allows to “escape from algebraicsubvarieties in bounded time”.Recall Bezout’s theorem about the intersection of finitely many algebraicsubvarieties (see for instance [35]), namely:Theorem 6.1 (Generalized Bezout theorem). Let K be a field, and letY 1 , . . . , Y p be pure dimensional algebraic subvarieties of K n . Denote by W 1 , . . . , W qthe irreducible components of Y 1 ∩ . . . ∩ Y p . Thenq∑p∏deg(W i ) ≤ deg(Y j ).i=1j=1

A STRONG TITS ALTERNATIVE 27Let K be a field and let X be an algebraic variety over K. We set s(X)to be the sum of the degree and the dimension of each of its geometricallyirreducible components. The following result was shown in [18], Lemma 3.2:Lemma 6.2. [18] Given an integer m ≥ 1 there is N = N(m) such that forany field K, any integer d ≥ 1, any K–algebraic subvariety X in GL d (K)with s(X) ≤ m and any (not necessarily symmetric) subset F ⊂ GL d (K)which contains the identity and generates a subgroup which is not containedin X(K), we have F N X(K).6.3. Irreducible representations of Chevalley groups. In this paragraphwe define the linear irreducible representations (ρ α , E α ) which are thepossible candidates for the projective representation where we will play pingpong.We also set a particular basis in each E α , which we use to define theheight h(ρ α (g)) and then show Lemma 6.3.Let G be a Chevalley group of adjoint type and g its Lie algebra withZ-structure g Z . Let T be a maximal torus and t the corresponding Cartansubalgebra in g. Let Λ R be the lattice of roots in the dual of g which weidentify with the space X(T ) of characters of T. Let Λ W be the lattice ofweights. We fix a set of positive roots Φ + and inside a base of simple rootsΠ. Since G is of adjoint type, to every dominant weight λ ∈ Λ R , therecorrespond a finite dimensional absolutely irreducible representation E ofG. Let {π α } α∈Π ⊂ Λ W be the fundamental weights. For each α ∈ Π, thereis a smallest integer k α ∈ N such that k α π α ∈ Λ R . Let χ α = k α π α be thecorresponding dominant weight and (ρ α , E α ) the corresponding absolutelyirreducible representation of G.For background on Chevalley groups and their representations, see Steinberg’snotes [38]. Let also (ρ 0 , E 0 ) be the adjoint representation. Accordingto [38] Section 2 Theorem 2, given an absolutely irreducible representation(ρ, E) of G, one may find in each E a lattice Λ invariant under the actionof ρ(G(Z)) and a basis of Λ which is made of weight vectors. Let us choosethis basis. It defines a standard norm || · || k on Λ ⊗ Z k and it also defines aheight h in SL(E α ) as in Section 3. Recall that Ω is either Q or F(t) andε Ω = 1 in the first case 0 otherwise. We have:Lemma 6.3. There exists a constant C 0 > 0 such that for every finite subsetF ∈ G(Ω) and every α ∈ Π, h(ρ α (F )) ≤ C 0 · (h(ρ 0 (F )) + ε Ω ).Let χ ρ be the heighest weight of ρ, which belongs to the root lattice. LetL be the maximal coefficient appearing in the decomposition of χ ρ as a sumof simple roots (let L 0 the corresponding integer for ρ 0 = Ad). Let M be thesmallest positive integer such that Mχ ρ ≥ α for every α ∈ Π (for the orderdefined by Π). Then Lemma 6.3 follows from:

28 EMMANUEL BREUILLARDLemma 6.4. For every local field k, there is a constant c 0 = c 0 (ρ, k) > 0such that for every g ∈ G(k), we have11L(37)||Ad(g)|| 0 Mk≤ ||ρ(g)|| k ≤ c 0 ||Ad(g)|| L kc 0and c 0 = 1 unless k is Archimedean.Proof. Let K k = G(O k ) when k is ultrametric. Then ρ(K k ) and Ad(K k ) preservethe norm. By Cartan’s K k T K k decomposition, it suffices to prove theinequalities for g in the maximal torus T. But then ||ρ(g)|| k = |χ ρ (g)| k andmax{|α(g)| v , α ∈ Π} 1 M ≤ |χ ρ (g)| k ≤ max{|α(g)| v , α ∈ Π} L . And max{|α(g)| v , α ∈Π} ≤ ||Ad(g)|| k ≤ max{|α(g)| v , α ∈ Π} L 0. Hence (37).When k is Archimedean, K k stabilizes another norm || · || k,new on Λ ⊗ Z k(resp. Λ R ⊗ Z k). For this new norm the same argument gives (37). Since thetwo norms are equivalent, this gives us the constant c 0 .□6.4. Combined adelic distance. In this paragraph, we define the combinedadelic distance δ(F ) = δ 1 (F ) + δ 2 (F ) of all adelic distances δ(V ; W )where V and W range over the relevant projective subspaces involved in theping-pong conditions from Section 5.Let K be a global field. Let (q i ) 1≤i≤5 be five positive integers. Givena ∈ G(K) and α ∈ Π ∪ {0}, let B a,α be the set of elements b ∈ G(K) suchthat t ρ α (b)(V c ) ⊥ W ⊥ and t ρ α (b −1 )(V c ) ⊥ W ⊥ for every (V, W ) ∈ A α (a),where A α (a) is the set of couples (V, W ) of ρ α (a)-admissible (see def. 3.5)non-trivial linear subspaces of E α such that dim(V ) = 1. Given a, b ∈ G(K)with b ∈ B a,α , let T a,b,α be the set of elements t ∈ G(K) such that ρ α (t)V W c +W ∩b −1 V c and ρ α (t −1 )V W c +W ∩bV c for every V, W ∈ A α (a) (notethat since b ∈ B a,α , W c + W ∩ b −1 V c and W c + W ∩ bV c are hyperplanes).Recall from Paragraph 6.2 that given an algebraic variety Z over thealgebraically closed field Ω, we denote by s(Z) the sum of the dimensionand degree of its irreducible components. Given two non-trivial subspacesV and W in E α the set of all g ∈ GL(E α ) such that gW ⊂ V or g −1 W ⊂ Vis a Zariski closed subset Z V,W of GL(E α ). Moreover s(Z V,W ) is boundedindependently on V and W since the one can pass from one Z V,W to theother by multiplying on the left and right by some automorphism in GL(E α ).From these remarks and Lemma 6.2 we obtain:Lemma 6.5. There is a positive integer q 0 such that for any field K andany finite subset F of G(K) containing 1 and generating a Zariski-densesubgroup, any α ∈ Π ∪ {0} and any a ∈ G(K) and b ∈ B a,α , the set F q 0intersects B a,α non trivially and the set F q 0intersects T a,b,α non trivially.We now fix the values of q 2 , q 3 and q 5 to be equal to this q 0 . The valuesof q 1 and q 4 will be specified later. Let Q α be the set of 3-tuples (a, b, t)

A STRONG TITS ALTERNATIVE 29such that a ∈ F q 1, b ∈ F q 2∩ B a,α , and t ∈ F q 3∩ T a,b,α . Let R α be the set ofcouples (x, c) such that x ∈ F q 4, c ∈ F q 5∩ B x,α . Lemma 6.5 ensures that if Fgenerates a Zariski-dense subgroup, then for any a ∈ F q 1there are b, t suchthat (a, b, t) ∈ Q α and also for any x ∈ F q 4there is c such that (x, c) ∈ R α .Now define for any finite symmetric subset F in G(K), and i = 1, 2δ i (F ) =∑δα(F i )andandδ α,(a,b,t) (F ) =∑(V,W )∈A α(a)δ 1 α(F ) =δ 2 α(F ) =α∈Π∪{0}∑δ α,(a,b,t) (F )(a,b,t)∈Q α∑δ α,(x,c) (F )(x,c)∈R αδ( t ρ α (b)(V c ) ⊥ ; W ⊥ ) + δ( t ρ α (b −1 )(V c ) ⊥ ; W ⊥ ) +δ(ρ α (t)V ; W c + W ∩ ρ α (b −1 )V c ) + δ(ρ α (t −1 )V ; W c + W ∩ ρ α (b)V c )∑δ α,(x,c) (F ) = δ(ρ α (c)V ; W ) + δ(ρ α (c −1 )V ; W )(V,W )∈A α(x)6.5. Height bounds for subspace separation. In this paragraph, applyingthe results of Paragraphs 3.1 and 6.1, we obtain (38) and (39) which givebounds for the combined adelic distances δ i (F ) in terms of the height h(F )and the number of elements in F only.Namely, if r = rank(G) and D = (r + 1) max α∈Π∪{0} 24 · 4 dα d 2 α with d α =dim E α , we have for every (a, b, t) ∈ Q α (F ),∑δ α,(a,b,t) (F ) ≤ d 2 α · (h(ρ α (b)) + h(ρ α (t))) + 4(h Ar (W ) + h Ar (V )) +≤V,W ∈A α(a)Hence using Lemma 6.3,+2(h Ar (W c ) + h Ar (V c ))h(ρ α (F q 0)) + 12 · 4 dα · max{h Ar (W ), W a-admissible}≤ 24 · 4 dα d 2 α · (q 0 + q 1 )h(ρ α (F )) + 12 · 4 dα d 2 α · ε Ω log 2δ 1 (F ) ≤ D|F | q 1+2q 0(q 0 + q 1 ) · h(ρ α (F )) + D · ε Ω log 2≤ D|F | q 1+2q 0(q 0 + q 1 )C 0 · (h(Ad(F )) + ε Ω )Note that if char(Ω) = 0, then by the Height Gap Theorem 3.3, we haveh(Ad(F )) ≥ ĥ(Ad(F )) ≥ g > 0 where g is the gap. So at any case for all

30 EMMANUEL BREUILLARDcharacteristic,(38) δ 1 (F ) ≤ D 1( |F |5) q1 +2q 0h(Ad(F )),where D 1 = D5 q 1+2q 0(q 0 + q 1 )C 0 (1 + g −1 ). Similarly one obtains( ) q4 +q |F |0(39) δ 2 (F ) ≤ D 2 h(Ad(F )),5where D 2 = D5 q 4+q 0(q 0 + q 4 )C 0 (1 + g −1 ).6.6. Proof of Theorem 1.1. The proof is done in three steps. Firstwe reduce to the situation when F generates a Zariski-dense subgroup inG(Ω) where G is a simple Chevalley group of adjoint type to be chosenamong a finite list of such. Second we show that we may assume thatF = {1, X, X −1 , Y, Y −1 }, i.e. F is a symmetric set with 4 elements plusthe identity. And finally, in the third and most difficult step, we check thatthere exists a place v of the field K of coefficients for which the sufficientconditions (i) to (vi) stated in Section 5 are fulfilled with some explicit choiceof constants depending only on G, and thus yield the desired ping-pong pair.Remark 6.6. It is not clear whether or not the assumption F symmetricis a necessary condition in Theorem 1.1. Our proof however requires thisassumption (see Remark 5.2). If one needs only a free semi-group instead ofa free group, then it is not necessary.6.6.1. Preliminary reductions.In this paragraph, we prove the first two steps, Claims 1 and 2. Beforethat note that we may assume that the field K is either Q or F p (t), i.e. thatK = Ω. This follows from Proposition 7.3 below when K has characteristic0 and from [14] Lemma 3.1 and the remark following it in general.So we assume K = Ω. We have:Claim 1: In Theorem 1.1, we may assume that F generates a Zariskidensesubgroup in G(Ω) where G is a simple Chevalley group of adjointtype.Proof. Since F generates a non virtually solvable subgroup 〈F 〉, the connectedcomponent G 0 of the Zariski-closure G of 〈F 〉 is not solvable. Modingout by the solvable radical of G 0 , which is a normal subgroup of G, we seethat we can assume that G 0 is a non trivial semisimple algebraic group. Welet G act on G 0 by conjugation we obtain a homomorphism of G in Aut(G 0 ad )where G 0 ad is the adjoint group of G0 whose image contains G 0 ad . However by[7] IV.14.9. Aut(G 0 ad )/G0 ad is a subgroup of the automorphisms of the Dynkindiagram of G. In particular it is a finite group whose order is bounded in

A STRONG TITS ALTERNATIVE 31terms of dim G only, hence in terms of d only. Recall (see for instance [13]Lemma 4.6.),Lemma 6.7. Let F be a finite subset of a group Γ containing 1. Assumethat the elements of F (together with their inverses) generate Γ. Let Γ 0 bea subgroup of index k in Γ. Then F 2k+1 contains a generating set of Γ 0 .Applying this lemma, we may therefore assume that G = G 0 ad is a semisimplealgebraic group of adjoint type. Further projecting to one of the simplefactors, we may assume that G is a simple algebraic group of adjoint typeover Ω. As Ω is algebraically closed, G(Ω) is the group of Ω-points of aChevalley group (see [38]).□Let O be the Zariski-open subset of G × G obtained in Theorem 3.4.Claim 2: In Theorem 1.1, we may assume that F = {1, X, X −1 , Y, Y −1 }for some (X, Y ) ∈ O(Ω).Proof. This claim was already proven in Proposition 4.14 of [13] in the specialcase of characteristic 0 making key use of Jordan’s theorem about finitesubgroups of GL n (C). This argument fails in positive characteristic so wenow give a different (and more involved) argument. Let G be a simpleChevalley group. Following an idea used in [14] Section 7, we have:Lemma 6.8. Then there is a proper closed subvariety W of G × G suchthat, for any choice of Ω, every pair (x, y) /∈ W(Ω) with x of infinite ordergenerates a Zariski-dense subgroup of G.Proof. Let g be the Lie algebra of G (see [7] I.3.5). Let W be the subset ofpairs (x, y) in G(Ω)×G(Ω) such that the associative subalgebra of End(g)generated by Ad(x) and Ad(y) is proper. Note that W is a closed algebraicsubset with equations over Z. It is also proper because one can constructpairs (x, y) for which the group they generate acts irreducibly on g (see forinstance [8] VIII. 2. ex.8. and [2] §3). Suppose (x, y) /∈ W(Ω) and x hasinfinite order. Let H be the Zariski closed subgroup generated by x and y.Then dim H ≥ 1 and the Lie algebra of H is non trivial and invariant underAd(x) and Ad(y), hence equal to g. By [7] I.3.6 we conclude that H = G. □In order to apply this lemma, we show:Lemma 6.9. There is a constant N = N(d) ∈ N such that, for any choiceof Ω, if F is a finite symmetric subset of G(Ω) containing 1 and generatinga Zariski dense subgroup, one may find a subset F 0 of F N such that for allintegers n ≥ 1 the subset F0n is made only of elements of infinite order andthe subgroup generated by F 0 and F0 −1 is Zariski dense in G.Before going into the proof of Lemma 6.9 let us explain how we deduceClaim 2 from this.

32 EMMANUEL BREUILLARDProof of Claim 2. By Lemma 6.9, we can replace F by F 0 . Now accordingto Lemma 6.2 applied to G × G and F 0 × F 0 there is a constant M ∈ Ndepending only on W and O, hence on d only, such that F0M contains a pair(x, y) such that (x, y) ∈ O and (x, y) /∈ W. By Lemma 6.9, x has infiniteorder, hence by Lemma 6.8, x and y generate a Zariski dense subgroup ofG, and Claim 2 is proved.Proof of Lemma 6.9. Let d 0 = dim G. We have:Lemma 6.10. There is a constant N 0 = N 0 (d 0 ) ∈ N and k ≤ d 0 elementsα 1 , ..., α k in F N 0of infinite order and such that the connected components C iof the Zariski closures of each cyclic subgroup generated by each α i togethergenerate G as an algebraic group.Proof. First we check that there is some α of infinite order in a boundedpower of F, say F N 1. This follows from Theorem 3.3 in characteristic 0 (seeCorollary 1.2). In positive characteristic it follows directly from the fact that〈F 〉 is finite as soon as ĥ(F ) = 0 (Lemma 3.2) and Lemma 2.1 (a) whichsays that F d2 0 already contains an element with eigenvalue of absolute value> 1.Let 〈α〉 be the cyclic group generated by α and C 1 the connected componentof its Zariski closure. Then dim C 1 = 1. Set α 1 = α. Suppose j ≥ 1and we have built α 1 , ..., α j and let C i be the connected component of theZariski closure of 〈α i 〉 and H i the algebraic subgroup generated by all C m for1 ≤ m ≤ i. We show by induction that α i = w i−1 αw −1i−1 for some w i−1 ∈ F i−1and dim H i ≥ i. If H j ≠ G, as G is simple and 〈F 〉 Zariski dense, there mustexist some β j ∈ F such that β j H j β −1j ≠ H j , hence some i ≤ j such thatβ j C i β −1j is not contained in H j . Let α j+1 = β j α i β −1j , i.e. w j = β j w i−1 ∈ F j .Then C j+1 = β j C i β −1j and dim H j+1 ≥ dim H j + 1. □We look at G viewed inside SL(g) via the adjoint representation. We knowfrom Theorem 3.3 and Lemma 2.1 that either there is a non archimedeanplace v of Ω for which Λ v (F d2 0 ) > 1 or there is an archimedean place forwhich Λ v (F d2 0 ) > 1 + ε where ε is the Height Gap. Let f ∈ Fd 2 0 be such thatΛ v (f) = Λ v (F d2 0 ). At any case, in one of boundedly many irreducible representationsof G over the local field K v , f acts as a proximal transformationwith a contracting eigenvalue and its action on the associated projectivespace P(Kv D ) is described by Lemma 4.6. Let v f be its attracting pointand H f be the repelling hyperplane. By Lemma 2.1, we may conjugate Finside GL D (K v ) so that ||F || v is less than say Λ v (f) c 0where c 0 is some constantdepending only on d 0 . Up to changing f into f c 0we may assume that||F || v ≤ Λ v (f).

A STRONG TITS ALTERNATIVE 33Let α 1 , ..., α k be the elements from Lemma 6.10. According to Lemma 5.4,there is some n i ∈ [1, d 0 ] such that d(α n ii v f , H f ) −1 is bounded above by somebounded power of ||F || v . If follows from Lemma 4.6 that there is compactsubset C of the projective space P(Kv D ) which is the complement of someneighborhood of H f , such that after replacing f by some bounded power ofit if necessary, the elements f, α n 11 f, ..., α n kkf are all proximal, send C insideitself, and have a Lipschitz constant < 1 on C. Let F 0 = {f, α n 11 f, ..., α n kk f}.We check that F 0 satisfies the desired conditions. It lies in a bounded powerof F , every positive word with letters in F 0 preserves C and is proximal byTits’ converse Lemma 4.7, hence of infinite order. Finally the group 〈F 0 〉generated by F 0 contains each 〈α n ii 〉, hence its Zariski closure contains theconnected component C i of the cyclic group 〈α i 〉 . Since the C i ’s generate Gas an algebraic group by Lemma 6.10, we get that 〈F 0 〉 is Zariski dense, andthis ends the proof of Lemma 6.9.□6.6.2. End of the proof of Theorem 1.1.So from now on we assume that G is a simple Chevalley group of adjointtype over Ω viewed as embedded inside SL(g) (g = Lie(G)) where it actsvia the adjoint representation. We also assume that F = {1, X, X −1 , Y, Y −1 }generates a Zariski-dense subgroup of G and (X, Y ) lies in the Zariski-opensubset O defined in Theorem 3.4.Constants.We now define or recall our constants. All these constants depend onlyon G (equivalently only on dim G) and not on the field of coefficients wechoose. And this is all that matters, so the reader may freely ignore theirprecise definition, all the more so since we did not try at all to give the bestconstants we could. However there dependence and order in which they aredefined are important in the logic of the proof.Recall that the constant C k,1 from Section 5 was defined to be 1 if k is ultrametricand equal to the dimension of the vector space if k is Archimedean.Below we set the value of d to be the max d α where d α = dim E α forα ∈ Π∪{0} (recall that we chose to denote by E 0 the adjoint representation,so d 0 = dim G).D = (rk(G) + 1) max α∈Π∪{0} 24 · 4 dα d 2 α.L is defined to be the maximum coefficient in the expression of the heighestweight χ α (for each α ∈ Π ∪ {0}) as a sum of simple roots.M is the smallest positive integer k such that kχ ρα − β is positive for anychoice of simple roots α, β ∈ Π.c 0 (v) is the maximum of the constants denoted c 0 in Lemma 6.4 for eachρ α , α ∈ Π ∪ {0} for a given place v (c 0 = 1 when v is finite and c 0 (v) is afixed constant c 0 (∞) if v is infinite).

34 EMMANUEL BREUILLARDc(d 0 ) v is the constant appearing in the Comparison Lemma, Lemma 2.1.It is 1 if v is finite, a fixed constant c(d 0 ) ∞ if v is infinite.g is the Height Gap from Theorem 3.3 in SL d0 (Q).If char(Ω) > 1, then we set n 1 = 1, otherwise we set n 1 to be the firstinteger such that exp( 4√ g n1) ≥ max{c(d 8d 0) −2 , d 4LMd , c 0 (∞) 2LM }.q 0 is the integer obtained by escape in Lemma 6.5.q 1 is n 1 d 2 .ε 0 is 1 . Lε is ε 0 /12d 2 .T 1 is the maximum of the integers τ 1 (d α , ε) obtained in Lemma 5.1 foreach representation ρ α , α ∈ Π ∪ {0}.C is the constant from Theorem 3.4, applied to G inside SL d0 .C 0 is the constant from Lemma 6.3.Let m = 48T 1 n 1 CL 2 .Let D 1 = D5 q 1+2q 0(q 0 + q 1 )C 0 (1 + g −1 )Let T 0 = 24CD 1 LMLet k 1 = d 2 m, k 2 = k 3 = k 5 = q 0 .Let T 3 be the maximum of the integers τ 3 obtained in Lemma 5.1 forthe above values of d α , k 1 , k 2 , k 3 , ε 0 , ε, T 0 and T 1 for each representation ρ α ,α ∈ Π ∪ {0}.Let l be the maximum of the integers l obtained in Lemma 5.1 for theabove values of d α , k 1 , k 2 , k 3 , ε 0 , ε, T 0 , T 1 and T 3 for each representation ρ α ,α ∈ Π ∪ {0}.Let k 4 = 2k 1 l + k 2 + k 3 .Let q 4 = n 1 k 4 .Let D 2 = D5 q 4+q 0(q 4 + q 0 )C 0 (1 + g −1 ).Let T 2 = 24CD 2 LM.Let l 2 be the maximum of each value l 2 (d α , (k i ) 1≤i≤5 , ε, ε 0 , (T i ) 0≤i≤3 ) obtainedin Lemma 5.3 for each representation ρ α , α ∈ Π ∪ {0}.Choice of a place v.Applying Theorem 3.4, we may change F into a conjugate of it by someelement in G(Ω) and hence get, summing (3), (38) and (39),(40) h(Ad(F )) + 1 D 1δ 1 (F ) + 1 D 2δ 2 (F ) ≤ 3C · e(Ad(F ))Let K be the (global) field generated by the coefficients of F .Claim: There is a place v of K such that the following holds: e v > 0 if vis a finite place, e v ≥ g if v is infinite and in both cases4

A STRONG TITS ALTERNATIVE 35(41)log ||Ad(F )|| v ≤ 12C · e vδ 1 (F ) v ≤ 12CD 1 · e vδ 2 (F ) v ≤ 12CD 2 · e v ,where e v = log E v (Ad(F )) and δ i (F ) v is the part of δ i (F ) associated to v,i.e.δ i 1 ∑(F ) =n v · δ i (F ) v[K : K 0 ]v∈V KProof of claim: This is an easy verification. Indeed, splitting the infinitepart and the finite part write e = e(Ad(F )) = e ∞ + e f . If e ∞ < e 2 , thene ≤ 2e f and (40) impliesh f (Ad(F )) + 1 D 1δ 1 f(F ) + 1 D 2δ 2 f(F ) ≤ 6C · e f (Ad(F ))where the subscript f means that we have restricted the sum to the finiteplaces. Then the existence of a finite place v such that e v > 0 and (41) holdsis guaranteed. On the other hand, if e ∞ ≥ e , then we have2h ∞ (Ad(F )) + 1 D 1δ 1 ∞(F ) + 1 D 2δ 2 ∞(F ) ≤ 6C · e ∞ (Ad(F ))Let V + be the set of places v ∈ V ∞ for which e v ≥ e∞ . We have2e ∞2 ≤ 1 ∑n v e v[K : K 0 ]v∈V +Andh ∞ (Ad(F )) + 1 D 1δ 1 ∞(F ) + 1 D 2δ 2 ∞(F ) ≤ 12C ·1[K : K 0 ]∑v∈V + n v e vwhich surely guarantees the existence of a place v ∈ V + such that (41) holds,and as v ∈ V + , e v ≥ e ≥ g . qed.4 4Verification of the Ping-Pong conditions (i) to (vi) from Section 5.We are going to build an element a ∈ F q 1and choose an α ∈ Π for whichall the six conditions of Section 5 are going to be satisfied with a m in placeof a and ρ α (F n 1) in place of F.According to Lemma 2.3,E v (Ad(F n 1)) ≥ E v (Ad(F ))≥√ n18dmax{c(d 0 ) −2v , CK 4LMdv,1 , c 0 (v) 2LM }.

36 EMMANUEL BREUILLARDFrom Lemma 2.1,Λ v (AdF q 1) ≥ Λ v (AdF n 1d 2 0 ) ≥ c(d0 )E v (AdF n 1)≥ E v (AdF n 1) 1 2 ≥ ||AdF ||112Cv > 1Let us choose a ∈ F q 1such that Λ v (Ad(a)) = Λ v (AdF q 1). Let also α ∈ Π besuch that |α(a)| v = max{|β(a)| v , β ∈ Π}. Then the representation ρ α (definedin Subsection 6.3) and Ad satisfy Λ v (Ad(a)) ≤ |α(a)| L v and Λ v (ρ α (a)) ≤|α(a)| L v by definition of L. Hence |α(a)| v > 1. Moreover, by definition of ρ αand ε 0 = 1 we haveL(42)Λ v (ρ α (a))λ v (ρ α (a)) = |α(a)| v > 1It follows that ρ α (a) is proximal. Moreover( ) 1Λv (ρ α (a))ε 10(43)= |α(a)|ε 0vλ v (ρ α (a))and≥ Λ v (ρ α (a))Λ v (ρ α (a)) ≥ |α(a)| v ≥ Λ v (Ad(a)) 1 L ≥ ||AdF ||112CLvOn the other hand, by Lemma 6.4, we have ||ρ α (F n 1)|| v ≤ c 0 (v)||AdF n 1|| L v ≤||AdF n 1|| vL+1 ≤ ||AdF || n 12Lv soAndΛ v (ρ α (a)) ≥ ||ρ α (F n 1)||124CL 2 n 1v(44) Λ v (ρ α (a m )) ≥ ||ρ α (F n 1)|| T 1vRaising a to the power m, (42) gives condition (i), while (43) and (44) givecondition (iii). On the other hand Lemma 6.4 gives(45) ||ρ α (F n 1)|| v ≥ c 0 (v) −1 ||AdF n 1|| 1LMv ≥ ||AdF n 1||12LMv≥ C 2dK v,1.Hence condition (ii) is fulfilled.We now check (iv) and (v). By (41) we have δ 1 (F ) v ≤ 12CD 1 · e v . Sinceδ 1 (F ) v is a sum of positive terms, we get in particular for any (b, t) such that(a, b, t) ∈ Q α (just pick one!)∑δ( t ρ α (b)Ha ⊥ ; W ⊥ ) v + δ( t ρ α (b −1 )Ha ⊥ ; W ⊥ ) v ≤ 12CD 1 · e vWwhere H a the generalized eigenspace of ρ α (a) corresponding to eigenvaluesthat are < Λ v (ρ α (a)) (it is a hyperplane since ρ α (a) is proximal), and the

A STRONG TITS ALTERNATIVE 37sum is made over all non trivial ρ α (a)-admissible subspaces W . This givesd v ( t ρ α (b)Ha ⊥ ; W ⊥ ) ≥ E v (Ad(F n 1)) −12CD 1≥ ||Ad(F n 1)|| −12CD 1v≥ ||ρ α (F n 1)|| −24CD 1LMv≥ ||ρ α (F n 1)|| −T 0vSimilarlyd v ( t ρ α (b −1 )Ha ⊥ ; W ⊥ ) ≥ ||ρ α (F n 1)|| −T 0vThis proves (iv). Condition (v) is derived in exactly the same way.Therefore we are in the situation where we may apply Lemma 5.1. Ityields an element x ∈ F n 1k 4= F q 4such that ρ α (x) is very proximal andsatisfies the conclusions of Lemma 5.1. Pick c ∈ F q 0such that (x, c) ∈ R α(there are such c by Lemma 6.5). The third inequality in (41) gives for everyρ α (x)-admissible subspaces V and W with dim V = 1,δ(ρ α (c)V ; W ) v + δ(ρ α (c −1 )V ; W ) v ≤ 12CD 2 · e vWe may take V = V x or V x −1 and W = H x or H x −1 and this indeed givescondition (v) with T 2 = 24CD 2 LM.Finally Lemma 5.3 yields that x n and cx n c −1 generate a free subgroup assoon as n is larger than the constant l 2 (l 2 is expressible explicitly in termsof all the other constants introduced so far).This ends the proof of the main theorem. Q.E.D.7. ApplicationsIn this section we briefly discuss the corollaries. We shall be brief as eachof them is derived in exactly the same way as in the GL 2 case, so we willrefer the reader to the paper [12] for details. The proofs of Corollaries 1.7,1.9 and 1.10 rely only on the characteristic 0 part of Theorem 1.1 and ona reformulation of that theorem in terms of algebraic varieties. So we willcontent ourselves to give this reformulation and briefly explain below whatmakes this translation possible. The following fact is standard,Proposition 7.1. (see e.g. [13] Proposition 7.4.) Let G = GL d (C). Forevery integer k, let V be the set of k-tuples (a 1 , ..., a k ) ∈ G k which generatea virtually solvable subgroup. Then V is a closed algebraic subvariety of G k .It is proved via the following proposition:Proposition 7.2. There exists N = N(d) such that (a 1 , ..., a k ) ∈ G k generatesa virtually solvable subgroup if and only if they leave invariant acommon finite subset of at most N points on the flag variety G/B, where Bis the subgroup of upper triangular matrices.

38 EMMANUEL BREUILLARDLet N = N(d) be the integer obtained in the statement of Theorem 1.1and let B(n) be the ball of radius n in the free group F 2 on two generators.For n ≥ 1 let W n be the set of couples (A, B) ∈ GL d (C) 2 such that forany words w 1 and w 2 in B(N) there exists a word w ∈ B(n)\{1} such thatw(w 1 (A, B), w 2 (A, B)) = 1. Clearly W n is a closed subvariety of GL d (C) 2 .We obtain:Proposition 7.3. Theorem 1.1 for K = C is equivalent to the statement:W n ⊂ V for every n ≥ 1.This allows to use the following effective version of Hilbert’s Nullstellensatz:Theorem 7.4. ([29]) Let r, d ∈ N, h > 0 and f, q 1 , ..., q k be polynomials inZ[X 1 , ..., X r ] with logarithmic height at most h and degree at most d. Assumethat f vanishes at all common zeros (if any) of q 1 , ..., q k in C[X 1 , ..., X r ]. Thenthere exist a, e ∈ N and polynomials b 1 , ..., b k ∈ Z[X 1 , ..., X r ] such thataf e = b 1 q 1 + ... + b k q kwith e ≤ (8d) 2r , the total degree of each b i at most (8d) 2r +1 and the logarithmicheight of each b i as well as a is at most (8d) 2r+1 +1 (h + 8d log(8d)).Since the polynomial equations defining W n have degree linear in n andheight exponential in n, one can get from Theorem 7.4 the desired boundon the degree and height of the b i ’s and on a and e. This readily allows todeduce Corollaries 1.7, 1.9 and 1.10 from Theorem 1.1 and Corollary 1.5.Corollary 1.8 is derived in a similar fashion. See [12] for more details.Acknowledgments 7.5. I am grateful to J. Tits for his encouraging remarksat an early stage of this project. I also thank J-F. Quint for telling me abouthis results and those of Y. Benoist on proximal maps over the p-adics. I alsothank J. Bourgain and A. Gamburd for the discussions we had about theapplications to finite groups.References[1] H. Abels, G. Margulis, G. Soifer, Semigroups containing proximal linear maps, IsraelJ. Math. 91 (1995), no. 1-3, 1–30.[2] Barnea Y., Larsen M., Random Generation for semisimple algebraic groups over localfields, J. Algebra 271 (2004), no. 1, 1–10.[3] Bartholdi L., de Cornulier Y., Infinite groups with large balls of torsion elements andsmall entropy, to appear in Archiv der Mathematik.[4] Y. Benoist, Propriétés asymptotiques des groupes linéaires, Geom. Funct. Anal. 7(1997), no. 1, 1–47.[5] Bilu, Y, Limit distribution of small points on algebraic tori, Duke Math. J. 89 (1997),no. 3, 465–476.

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