Surjectivity and equidistribution of the word x a y b on P SL(2, q) and SL(2, q)

Transcription

1 Surjectivity an equiistribution of the wor x a y b on P SL(2, q) an SL(2, q) Tatiana BANDMAN an Shelly GARION Institut es Hautes Étues Scientifiques 5, route e Chartres Bures-sur-Yvette (France) Juin 2011 IHES/M/11/19

2 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) TATIANA BANDMAN AND SHELLY GARION Abstract We etermine the integers a, b 1 an the prime powers q for which the wor map w(x, y) = x a y b is surjective on the group PSL(2, q) (an SL(2, q)) We moreover show that this map is almost equiistribute for the family of groups PSL(2, q) (an SL(2, q)) Our proof is base on the investigation of the trace map of positive wors 1 Introuction 11 Wor maps in finite simple groups During the last years there has been a great interest in wor maps in groups, for an extensive survey see [Se] These maps are efine as follows Let w = w(x 1,, x ) be a non-trivial wor, namely a nonientity element of the free group F on the generators x 1,, x Then we may write w = x n 1 i 1 x n 2 i 2 x n k i k where 1 i j, n j Z, an we may assume further that w is reuce Let G be a group For g 1,, g we write w(g 1,, g ) = g n 1 i 1 g n 2 i 2 g n k i k G, an efine w(g) = {w(g 1,, g ) : g 1,, g G}, as the set of values of w in G The corresponing map w : G G is calle a wor map Borel [Bo] showe that the wor map inuce by w 1 on simple algebraic groups is a ominant map Larsen [La] use this result to show that for every non-trivial wor w an ɛ > 0 there exists a number C(w, ɛ) such that if G is a finite simple group with G > C(w, ɛ) then w(g) G 1 ɛ By recent work of Larsen, Shalev an Tiep [LS, LST], for every non-trivial wor w there exists a constant C(w) such that if G is a finite simple group satisfying G > C(w) then w(g) 2 = G It is therefore interesting to fin wors w for which w(g) = G for any finite simple non-abelian group G Due to immense work sprea over more than 50 years, it is now known that the commutator wor w = [x, y] F 2 satisfies w(g) = G for any finite simple group G (see [LOST10] an the references therein) On the other han, it is easy to see that if G is a finite group an b is an integer which is not relatively prime to the orer of G then w(g) G for the wor w = x b 2000 Mathematics Subject Classification 14G05, 14G15, 20D06, 20G40 Key wors an phrases special linear group, wor map, trace map, finite fiels 1

3 2 TATIANA BANDMAN AND SHELLY GARION The wors of the form w = x a y b F 2 have also attracte special interest Due to recent work of Larsen, Shalev an Tiep [LST], it is known that any such wor is surjective for sufficiently large finite simple groups (see [LST, Theorem 111 an Corollary 11]), more precisely, Theorem 11 [LST] Let a, b be two non-zero integers Then there exists a number N = N(a, b) such that if G is a finite simple non-abelian group of orer at least N, then any element in G can be written as x a y b for some x, y G By further recent results of Guralnick an Malle [GM] an of Liebeck, O Brien, Shalev an Tiep [LOST11], some wors of the form x b y b are known to be surjective on all finite simple groups Theorem 12 [GM, Corollary 15] Let G be a finite simple non-abelian group an let b be either a prime power or a power of 6 Then any element in G can be written as x b y b for some x, y G Note that in general, the wor x b y b is not necessarily surjective on all finite simple groups Inee, if b is a multiple of the exponent of G then necessarily x b y b = i for every x, y G It is therefore interesting to fin more examples for wors of the form x b y b which are not surjective on all finite simple groups More generally, one can ask whether it is possible to generalize Theorem 11 for other wor maps In particular, the following conjecture was raise: Conjecture 1 (Shalev) [Sh07, Conjectures 28 an 29] Let w 1 be a wor which is not a proper power of another wor Then there exists a number C(w) such that, if G is either A r or a finite simple group of Lie type of rank r, where r > C(w), then w(g) = G It is also interesting to investigate the istribution of the wor map Due to the work of Garion an Shalev [GS], it is known that the wor w = x 2 y 2 is almost equiistribute for the family of finite simple groups, namely, Theorem 14 [GS, Theorem 71] Let G be a finite simple group, an let w : G G G be the map given by w(x, y) = x 2 y 2 Then there is a subset S G with S = (1 o(1)) G such that w 1 (g) = (1 + o(1)) G for all g S Where o(1) enotes a function epening only on G which tens to zero as G Another question which was raise by Shalev [Sh07, Problem 210] is which wors w inuce an almost equiistribute map for the family of finite simple groups In particular, oes wors of the form w = x a y b inuce almost equiistribute maps? 12 The wor w(x, y) = x a y b on the groups SL(2, q) an PSL(2, q) In this paper we analyze the wor map x a y b in the groups SL(2, q) an PSL(2, q) Analysis of Engel wor maps in these groups was carrie out in our previous work [BGG] We start by etermining precisely the positive integers a, b an prime powers q for which the wor map w = x a y b is surjective on SL(2, q) \ { i} an on PSL(2, q)

4 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) Definition 15 Let a, b 1 an let q = p e be a prime power We say that the wor w = x a y b is non-egenerate with respect to q if an only if none of the following conitions hols: p = 2, a is a multiple of 2(q 2 1) an b is not relatively prime to 2(q 2 1); p = 2, b is a multiple of 2(q 2 1) an a is not relatively prime to 2(q 2 1); p is o, a is a multiple of p(q2 1) an b is not relatively prime to p(q2 1) ; 4 4 p is o, b is a multiple of p(q2 1) an a is not relatively prime to p(q2 1) 4 4 Obviously, if the wor map w = x a y b is surjective on PSL(2, q) then it is necessarily non-egenerate with respect to q On the other han, we prove the following proposition Proposition 16 If w = x a y b is non-egenerate with respect to q, then all semisimple elements, namely matrices in SL(2, q) whose trace is not ±2, are in the image of the wor map w Unfortunately, even if w = x a y b is non-egenerate with respect to q, the image of w = x a y b may not contain the unipotent elements, namely, matrices ±i z SL(2, q) satisfying tr(z) = ±2 This phenomenon happens when one of the following obstructions occurs Definition 17 Let a, b 1 an let q be a prime power We efine the following obstructions: Obstruction (i): q = 2 e, e is o, an a, b are ivisible by 2(q2 1) ; Obstruction (ii): q mo 4 an a, b are ivisible by p(q2 1) 8 ; Obstruction (iii): q 11 mo 12 an a, b are ivisible by p(q2 1) 6 ; Obstruction (iv): q 5 mo 12 an a, b are ivisible by p(q2 1) 12 Theorem 18 Let e 1 an let q = 2 e Let a, b 1 Then the wor map w = x a y b is surjective on SL(2, q) = PSL(2, q) if an only if w is non-egenerate with respect to q an obstruction (i) oes not occur Theorem 19 Let p be an o prime number, e 1 an q = p e Let a, b 1 Then the wor map w = x a y b is surjective on SL(2, q) \ { i} if an only if w is non-egenerate with respect to q an none of the obstructions (ii),(iii),(iv) occurs Theorem 110 Let p be an o prime number, e 1 an q = p e Let a, b 1 Then the wor map w = x a y b is surjective on PSL(2, q) if an only if w is nonegenerate with respect to q an obstruction (ii) oes not occur For example, we euce that the wor w = x 42 y 42 is not surjective on the groups PSL(2, 7) an PSL(2, 8) The last theorem implies that for the family of groups PSL(2, q) one can give a precise estimation for the boun N = N(a, b) appearing in Theorem 11

5 4 TATIANA BANDMAN AND SHELLY GARION Corollary 111 For every a, b 1, let Q = Q(a, b) = max{ a, b}, N = N(a, b) = max { 2 a/2, 2 b/2} Then the wor map w = x a y b is surjective on the group PSL(2, q) for any q > Q, an hence whenever PSL(2, q) > N However, the statement of Theorem 11 [LST] no longer hols for the quasi-simple group SL(2, q), as inicate by the following theorem an its corollary { } Theorem 112 Let q be an o prime power, an set K = max k : 2 k ivies q2 1 2 Let a, b 1 Then i x a y b for every x, y SL(2, q) if an only if 2 K ivies both a an b Corollary 11 If q ± mo 8, then x 4 y 4 i for every x, y SL(2, q) In aition, we show that for any a, b 1, the wor map w = x a y b is almost equiitribute for the family of groups PSL(2, q) (an SL(2, q)) Theorem 114 Let q be a prime power an let G be either the group SL(2, q) or the group PSL(2, q) Let a, b 1 an let w : G G G be the map given by w(x, y) = x a y b Then there is a subset S G with S = (1 o(1)) G such that w 1 (g) = (1 + o(1)) G for all g S Where o(1) enotes a function of q which tens to zero as q 1 Organization an outline of the proof For the convenience of the reaer, we escribe the organization of the paper, as well as give a bir s eye view of the proofs In Section 2 we compute the trace map of the wor w(x, y) = x a y b (Lemma 2), an more generally, of any positive wor in F 2 (Theorem 25) For any wor w = w(x, y) F 2, the trace map tr(w) is a polynomial P (s, u, t) in s = tr(x), t = tr(y) an u = tr(xy) In Section we collect basic facts on the surjectivity of w = x a y b on finite groups in general, an in Section 4 we escribe some properties of the groups SL(2, q) an PSL(2, q) that are use later on By Lemma 2, tr(x a y b ) is a linear polynomial in u We euce in Section 5 that if neither a nor b is ivisible by the exponent of PSL(2, q), then any element in F q can be written as tr(x a y b ) for some x, y SL(2, q) (Corollary 5) This immeiately implies Proposition 16, stating that if w = x a y b is non-egenerate with respect to q, any semisimple element (namely, z SL(2, q) with tr(z) ±2) can be written as z = x a y b for some x, y SL(2, q) However, when z is unipotent (namely, z ±i an tr(z) = ±2) one has to be more careful, an a etaile analysis is one in Section 8 Inee, it may happen

6 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 5 that w = x a y b is non-egenerate with respect to q, but nevertheless the image of the wor map w = x a y b oes not contain any unipotent (see Propositions 64 an 65) These are the ingreients neee for the proofs of Theorems 18, 19 an 110, on the surjectivity of the wor w = x a y b on PSL(2, q) an SL(2, q) \ { i}, which are presente in Section 6 In aition, we etermine in Section 8 when i can be written as x a y b for some x, y SL(2, q), thus proving Theorem 112 In Section 7 we prove Theorem 114 an show that the wor map w = x a y b is almost equiistribute for the family of groups PSL(2, q) (an SL(2, q)) The basic iea is to show that for a general α F q, the surface S α (F q ) = {P (s, u, t) = tr(z) = α} A (F q ) is birational to a plane A 2 s,t As a result, we are able not only to fin points on S α (F q ), but even to estimate their number Acknowlegement Banman is supporte in part by Ministry of Absorption (Israel), Israeli Acaemy of Sciences an Minerva Founation (through the Emmy Noether Research Institute of Mathematics) Garion is supporte by a European Post-octoral Fellowship (EPDI), uring her stay at the Institut es Hautes Étues Scientifiques (Bures-sur-Yvette) The authors are grateful to A Shalev for iscussing his questions an conjectures with them They are also thankful for A Mann an M Larsen for useful iscussions 2 The trace map 21 The trace map The trace map metho is base on the following classical Theorem (see, for example, [Vo, Fr, FK] or [Mag, Go] for a more moern exposition) Theorem 21 (Trace map) Let F = x, y enote the free group on two generators Let us embe F into SL(2, Z) an enote by tr the trace character If w is an arbitrary element of F, then the character of w can be expresse as a polynomial tr(w) = P (s, u, t) with integer coefficients in the three characters s = tr(x), u = tr(xy) an t = tr(y) Note that the same remains true for the group SL(2, q) The general case, SL(2, R), where R is a commutative ring, can be foun in [CMS] The following theorem is originally ue to Macbeath [Mac] an was use by Banman, Grunewal, Kunyavskii an Jones to investigate verbal ynamical systems in the group SL(2, q) (see [BGKJ, Theorem 4]) Theorem 22 [Mac, Theorem 1] For any (s, u, t) F q there exist two matrices x, y SL(2, q) satisfying tr(x) = s, tr(y) = t an tr(xy) = u

7 6 TATIANA BANDMAN AND SHELLY GARION 22 Trace map of the wor w(x, y) = x a y b The following Lemma shows that the trace map of the wor w(x, y) = x a y b is a linear polynomial in tr(xy) Lemma 2 Let w(x, y) = x a y b where a, b 1 an x, y SL(2, q) Let s = tr(x), u = tr(xy), t = tr(y) Then where are polynomials satisfying: tr w(x, y) = u f a,b (s, t) + h a,b (s, t), f a,b (s, t), h a,b (s, t) F q [s, t] the highest egree summan of f a,b (s, t) (of egree a + b 2) is: s a 1 t b 1 ; the highest egree summan of h a,b (s, t) (of egree a + b 2) is: (s a t b 2 + s a 2 t b ) Proof We nee to prove that the polynomials f(s, t) = f a,b (s, t) an h(s, t) = h a,b (s, t) satisfy the following properties: (i) the coefficient of u, f(s, t), has precisely one monomial summan s a 1 t b 1 with coefficient 1; (ii) for all other monomial summans c i,j s i t j, c i,j F q, of f(s, t), the following inequalities hol: i a 1, j b 1, an i + j < a + b 2; (iii) h(s, t) contains the summan (s a t b 2 + s a 2 t b ) with coefficient 1; (iv) for all other monomial summans c i,j s i t j, c i,j F q, of h(s, t), the following inequalities hol: i a 2, j b 2, an so i + j a + b 4 We prove these properties by inuction on a + b, using the well-known formula (1) tr(ab) + tr(ab 1 ) = tr(a) tr(b) Inuction base a, b In these cases, tr(xy) = u, tr(x 2 y) = us t, tr(xy 2 ) = ut s, tr(x 2 y 2 ) = ust s 2 t 2 + 2, tr(xy ) = (t 2 1)u st, tr(x y) = (s 2 1)u st, tr(x 2 y ) = (st 2 s)u s 2 t t + t, tr(x y 2 ) = (s 2 t t)u s st 2 + s, tr(x y ) = (s 2 t 2 s 2 t 2 + 1)u s t st + 4st Inuction hypothesis Assume that the Lemma is vali for a+b < n for some n 5 Inuction Step We prove the claim for a + b = n, by consiering the following cases: Case 1 w(x, y) = x a y b, a b

8 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 7 Using (1) we get: tr(x a y b ) = tr(x) tr(x a 1 y b ) tr(xy b x a+1 ) = tr(x) tr(x a 1 y b ) tr(x a 2 y b ) = s(u f a 1,b (s, t) + h a 1,b (s, t)) (u f a 2,b (s, t) + h a 2,b (s, t)) = u(s f a 1,b (s, t) f a 2,b (s, t)) + (s h a 1,b (s, t)) h a 2,b (s, t)) = u f a,b (s, t) + h a,b (s, t) By the inuction hypothesis, the resulting polynomial: (i) is linear in u; (ii) the highest egree summan of f a,b (s, t) is: ss (a 1) 1 t b 1 = s a 1 t b 1 ; (iii) for all other monomial summans c i,j s i t j of f(s, t) the following inequalities hol: i (a 1) = a 1, j b 1, an i + j < (a 1) b 2 = a + b 2; (iv) the highest egree summan in h(s, t) is: s(s a 1 t b 2 + s (a 1) 2 t b ) = (s a t b 2 + s a 2 t b ) (v) for all other monomial summans c i,j s i t j of h(s, t) the following inequalities hol: i (a 1) = a 2, j b 2, an so Case 2 w(x, y) = x a y b, a < b Similarly we get: i + j (a 1) b 4 = a + b 4; tr(x a y b ) = tr(x a y b 1 ) tr(y) tr(x a y b 1 y 1 ) = tr(y) tr(x a y b 1 ) tr(x a y b 2 ) = t(u f a,b 1 (s, t) + h a,b 1 (s, t)) (u f a,b 2 (s, t) + h a,b 2 (s, t)) = = u(t f a,b 1 (s, t) + f a,b 2 (s, t)) + (t h a,b 1 (s, t) h a,b 2 (s, t)) = = u f a,b (s, t) + h a,b (s, t) Similarly to Case 1 we get a polynomial satisfying the esire properties Remark 24 Assume that a, b 0 but not necessarily positive Since tr(xy 1 ) = st u, we euce from Lemma 2 that tr(x a y b ) = u f a,b (s, t) + h a,b (s, t), where the highest egree summan of f a,b (s, t) (of egree a + b 2) is ±s a 1 t b 1

9 8 TATIANA BANDMAN AND SHELLY GARION 2 Trace map of positive wors We can moreover compute the trace map for any positive wor in F 2, namely for any wor of the form w = x a 1 y b 1 x a k y b k where a 2,, a k, b 1,, b k 1 1 an a 1, b k 0 We note that we can consier only wors of the form w = x a 1 y b 1 x a k y b k where a 1, b 1,, a k, b k 1, an then we call k the length of this wor Inee, if b k = 0 then tr(x a 1 y b 1 x a k 1 y b k 1x a k) = tr(x a 1 +a ky b 1 x a k 1 y b k 1), is the trace map of a positive wor of length k 1 Theorem 25 Let G = SL(2, q) an let w = x a 1 y b 1 x a k y b k, a 1, b 1,, a k, b k 1 Denote s = tr(x), u = tr(xy), t = tr(y), an A = k i=1 a i, B = k i=1 b i Then tr(w) = P (s, u, t) = k r=0 ur p r (s, t), where p k (s, t) = s A k t B k + Φ(s, t) is a polynomial in s, t an for all r < k, eg s Φ(s, t) A k, eg t Φ(s, t) B k, eg s Φ(s, t) + eg t Φ(s, t) < A + B 2k; eg s p r (s, t) A k, eg t p r (s, t) B k Proof The proof is by inuction on k The case k = 1 was treate in Lemma 2 We may always assume that a 1 a k Let (2) an Then, by (1), tr(w) = tr(x a 1 y b 1 x a k y b k ) w 1 (x, y) = x a 1 y b 1 x a k 1 y b k 1, w 2 (x, y) = x a k y b k, w (x, y) = x a 1 a k y b 1 x a k 1 y b k 1 b k = tr(x a 1 y b 1 x a k 1 y b k 1 ) tr(x a k y b k ) tr(x a 1 a k y b 1 x a k 1 y b k 1 b k ) = tr(w 1 (x, y)) tr(w 2 (x, y)) tr(w (x, y)), By the inuction assumption we have k 1 tr(w 1 ) = P 1 (s, u, t) = u r p r (s, t), r=0 tr(w 2 ) = uf + h, tr(w ) = Q(s, u, t), p k 1 (s, t) = s A a k k+1 t B b k k+1 + Φ 1 (s, t); eg s Φ 1 (s, t) A a k k + 1, eg t Φ 1 (s, t) B b k k + 1, an eg s Φ 1 (s, t) + eg t Φ 1 (s, t) < A + B a k b k 2k + 2;

10 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 9 eg s f = a k 1, eg t f = b k 1 We want to show that () eg u Q < k, eg s Q A k, eg t Q B k Since tr(w) = P (s, u, t) = P 1 (s, u, t)(uf + h) Q(s, u, t), the theorem woul follow from () Consier the following cases Case 1 w (x, y) is a positive wor Then its length is either k 1, if both a = a 1 a k > 0 an b = b k 1 b k > 0, or k 2 if a = 0 or b = 0 Anyway, eg u Q < k, eg s Q A 2a k k + 2 A k, an eg t Q B 2b k k + 2 B k, as neee Case 2 a 1 a k > 0, b k 1 b k < 0 Then tr(w ) = Q(s, u, t) = tr(w 4 (x, y)) tr(y b k b k 1 ) tr(w 5 (x, y)), where w 4 (x, y) = x a 1 a k +a k 1 y b 1 x a k 2 y b k 2, w 5 (x, y) = x a 1 a k y b 1 x a k 2 y b k 2 x a k 1 y b k b k 1 Thus, w 4 is a wor of length k 2 an w 5 has length k 1, an both are positive wors Let Q 1 (s, u, t) = tr(w 4 (x, y)), Q 2 (s, u, t) = tr(w 5 (x, y)) By the inuction assumption, eg u Q 1 = k 2, eg s Q 1 A 2a k k + 2 A k, eg t Q 1 B b k b k 1 k + 2; eg u Q 2 = k 1, eg s Q 2 A 2a k k+1 A k, eg t Q 2 B 2b k 1 k+1 B k Moreover, T 1 (t) = tr(y b k b k 1) is a polynomial in t of egree bk b k 1, thus eg t Q 1 T 1 (t) B b k b k 1 k b k b k 1 = B 2b k 1 k + 2 B k Hence, for Q = Q 1 T 1 (t) Q 2 conition () is vali Case a 1 a k = 0, b k 1 b k < 0 In this case w (x, y) = y b 1 x a k 1 y b k 1 b k an Q(s, u, t) = tr(x a 2 x a k 1 y b k 1 b k +b 1 ) If b k 1 b k + b 1 0 then the wor is positive of length k 2 an () is vali If b k 1 b k + b 1 < 0 then we perform the proceure escribe in Case 2 for computing Q(s, u, t) = tr(x a 2 x a k 1 y b k 1 b k +b 1 ) = tr(w 4 (x, y)) tr(y b k b k 1 b 1 ) tr(w 5 (x, y)), where w 4 (x, y) = x a 2+a k 1 y b 2 x a k 2 y b k 2, w 5 (x, y) = x a 2 y b 2 x a k 2 y b k 2 x a k 1 y b k 1+b k b 1

11 10 TATIANA BANDMAN AND SHELLY GARION The Length of w 4 is k, an the length of w 5 is k 2 Hence, eg u Q = k 2 Moreover, eg s Q A 2a k k + 2 A k an eg t Q B b 1 b k 1 b k b k 1 + b k b 1 k + 2 B k Once more, we fin out that conitions () are met by Q Remark 26 Assume that a i, b i 0 but not necessarily positive In view of Remark 24, for the wor w = x a 1 y b 1 x a k y b k equation (2) implies that (4) tr(w) = k u r G r (s, t) an G k (s, t) = 0 k f ai,b i i=1 Basic facts on the wor w(x, y) = x a y b an finite groups In this section we present some elementary facts regaring the surjectivity of the wor map w = x a y b on finite groups Proposition 1 Let G be a finite group an let a be an integer Then the wor map corresponing to w = x a is surjective on G if an only if gc( G, a) = 1 Proof Let = gc( G, a) If > 1 then there exists some prime p which ivies both a an G Thus, G contains some element g i of orer p, an moreover, g a = (g p ) a/p = i Hence the wor map w = x a is not 1 to 1, an cannot be surjective on G If = 1 then there exists some integer l st l a 1 (mo G ) Let g G an take x = g l, then x a = (g l ) a = g l a = g 1 = g, as neee Proposition 2 Let G be a finite group an let a, b be two relatively prime integers Then the wor map corresponing to w = x a y b is always surjective on G Proof Since a, b are relatively prime, there exist integers k, l st k a + l b = 1 Let g G an take x = g k an y = g l, then x a y b = g k a g l b = g k a+l b = g 1 = g Proposition Let G be a finite group an let a, b be two integers If either a or b is relatively prime to G then the wor map corresponing to w = x a y b is surjective on G Proof Assume that a is relatively prime to G Then there exists some integer l st l a 1 (mo G ) Then for every g G, take x = g l an y = i Thus, x a y b = (g l ) a i b = g l a i = g 1 = g

12 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 11 Remark 4 Let G be a finite group an let w = x a y b We can always assume that 0 a, b < exp(g) Moreover, if G is of even orer, we can assume that 0 a, b exp(g)/2 Inee, let a 1 = a mo exp(g) an b 1 = b mo exp(g), then for every x, y G, x a y b = x a 1 y b 1 Hence, x a y b is surjective on G if an only if x a 1 y b 1 is surjective on G If exp(g) is even, let a 2 = { a 1 if a 1 exp(g)/2 exp(g) a 1 if a 1 > exp(g)/2 an b 2 = { b 1 if b 1 exp(g)/2 exp(g) b 1 if b 1 > exp(g)/2 Then, for every x, y G, x a 1 y b 1 = x ɛ 1a 2 y ɛ 2b 2, where ɛ 1, ɛ 2 {±1}, an {z = x a 2 y b 2 : x, y G} = {z = x a 2 y b 2 : x, y G} ={z = x a 2 y b 2 : x, y G} = {z = x a 2 y b 2 : x, y G} 4 Properties of the groups SL(2, q) an PSL(2, q) In this section we summarize some well-known properties of the groups SL(2, q) an PSL(2, q) (see for example [Do] an [Su]) Let q = p e, where p is a prime number an e 1 Recall that GL(2, q) is the group of invertible 2 2 matrices over the finite fiel with q elements, which we enote by F q, an SL(2, q) is the subgroup of GL(2, q) comprising the matrices with eterminant 1 Then PGL(2, q) an PSL(2, q) are the quotients of GL(2, q) an SL(2, q) by their respective centers Also recall{ that PSL(2, q) is simple for q 2, 1 if q is even Denote = gc(2, q 1) = 2 if q is o Then the orers of SL(2, q) an PSL(2, q) are q(q 1)(q + 1) an 1 q(q 1)(q + 1) respectively, an their respective exponents are 1 p(q2 1) an 1 p(q 2 1) 2 One can classify the elements of SL(2, q) accoring to their possible Joran forms The following Table 1 lists the three types of (non-trivial) elements, accoring to whether the characteristic polynomial P t (λ) := λ 2 tλ + 1 of the matrix A SL(2, q) (where t = tr(a)) has 0, 1 or 2 istinct roots in F q Table 1 shows that there is a eep connection between traces of elements in SL(2, q), their orers an their conjugacy classes, as is expresse in the following Lemmas Lemma 41 If q is o, then x SL(2, q) has orer 4 if an only if tr(x) = 0 x SL(2, q) has orer if an only if tr(x) = 1 If p 5, then x SL(2, q) has orer 6 if an only if tr(x) = 1 Moreover, for any x SL(2, q) satisfying tr(x) 0, ±1, ±2, there exists some y SL(2, q) such that tr(x) tr(y), but the orers of x an y are the same

13 12 TATIANA BANDMAN AND SHELLY GARION element roots canonical form in orer in orer in conjugacy classes type of P t (λ) SL(2, F p ) SL(2, q) PSL(2, q) in SL(2, q) 1 0 i 1 root 1 1 one element ( 0 1 ) 1 0 i 1 root 1 one element unipotent 1 root p p conjugacy classes 0 1 t = 2 each of size q root p p conjugacy classes 0 1 t = 2 each of size q2 1 α 0 semisimple 2 roots 0 α 1 ivies ivies for each t: split where α F q 1 q q 1 one conjugacy class an α + α 1 = t of size q(q + 1) α 0 semisimple no roots 0 α q ivies ivies for each t: non-split where α F q \ F q+1 2 q q + 1 one conjugacy class α q+1 = 1 of size q(q 1) an α + α q = t Table 1 Elements in the groups SL(2, q) an PSL(2, q) Proof If m > 2 is an integer iviing q 1 then F q \{0, 1, 1} contains φ(m) elements of orer m, where φ enotes Euler s phi function Similarly, if m > 2 ivies q + 1 then F q 2 \ F q contains φ(m) elements α of orer m satisfying α q+1 = 1 Hence, if m > 2 ivies either q 1 or q + 1, then #{t F q : t = tr(x), x SL(2, q), x = m} = φ(m) 2 The claim follows from the fact that φ(m) 4 if an only if m 1, 2,, 4, 6 Lemma 42 Assume that q is o, take λ F q 2 \ F q satisfying λ 2 F q, an let λ 0 g = 0 λ 1 Then for any x SL(2, q), gxg 1 SL(2, q), an moreover, 1 1 If tr(x) = 2 then exactly one of x, gxg 1 is conjugate in SL(2, q) to ; If tr(x) = 2 then exactly one of x, gxg 1 is conjugate in SL(2, q) to 0 1

14 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 1 α β α βλ Proof Let x = SL(2, q), then gxg γ δ 1 2 = γλ 2 SL(2, q) δ λ Moreover, if x = then gxg = is not conjugate to x in SL(2, q), 0 1 since λ 2 is not a square of some element in F q Corollary 4 Let w F 2 be some non-trivial wor, let z ±i be some matrix in SL(2, q), an assume that z can be written as z = w(x, y) for some x, y SL(2, q) Then for any matrix ±i z SL(2, q) with tr(z ) = tr(z) there exist x, y SL(2, q) such that z = w(x, y ) Proof If q is even, or if q is o an tr(z) ±2, then necessarily z = hzh 1 for some h SL(2, q), so one can take x = hxh 1 an y = hyh 1, an then w(x, y ) = w(hxh 1, hyh 1 ) = hw(x, y)h 1 = hzh 1 = z Assume that q is o, an let z SL(2, q) be some element with tr(z ) = 2 = tr(z), then by Lemma 42, z is either conjugate in SL(2, q) to z or to gzg 1 (where g SL(2, q 2 )) If z = hzh 1 for some h SL(2, q), take x = hxh 1 an y = hyh 1, an then w(x, y ) = w(hxh 1, hyh 1 ) = hw(x, y)h 1 = hzh 1 = z If z = hgzg 1 h 1 for some h SL(2, q), take x = hgxg 1 h 1 an y = hgyg 1 h 1, an then x, y SL(2, q) an moreover, w(x, y ) = w(hgxg 1 h 1, hgyg 1 h 1 ) = hgw(x, y)g 1 h 1 = hgzg 1 h 1 = z Similarly, if tr(z ) = 2 = tr(z), then z = w(x, y ) for some x, y SL(2, q) 5 Surjectivity of the trace map of w(x, y) = x a y b on F q Recall that by Lemma 2, the trace map of w(x, y) = x a y b can be written as tr w(x, y) = u f a,b (s, t) + h a,b (s, t) The following proposition shows that if neither a nor b is ivisible by the exponent of PSL(2, q), then the polynomial f a,b (s, t) oes not vanish ientically on A 2 s,t(f q ) Proposition 51 Let a, b 1 an assume that neither a nor b is ivisible by p(q2 1) 2 Then f a,b (s, t) oes not vanish ientically on A 2 s,t(f q ) In particular, the following table summarizes the possible nine cases q 1 q+1 p a a a p b f a,b (2, 2) 0 f a,b (s 1, 2) 0 f a,b (s 2, 2) 0 q 1 b f a,b (2, t 1 ) 0 f a,b (s 1, t 1 ) 0 f a,b (s 2, t 1 ) 0 q+1 b f a,b (2, t 2 ) 0 f a,b (s 1, t 2 ) 0 f a,b (s 2, t 2 ) 0

15 14 TATIANA BANDMAN AND SHELLY GARION where: s 1 = tr(x 1 ), x 1 is any element of orer q 1; s 2 = tr(x 2 ), x 2 is any element of orer q + 1; t 1 = tr(y 1 ), y 1 is any element of orer q 1; t 2 = tr(y 2 ), y 2 is any element of orer q + 1 Proof If f a,b (s, t) vanishes ientically on A 2 s,t(f q ), then tr w(x, y) = h a,b (s, t) oes not epen on u We have to show that it is not the case for every F q Take λ c µ 0 x(λ, c) = 1, y(µ, ) = 0 1 λ µ Then for any m, n, λ x(λ, u) n n ch = n (λ) µ 1, y(µ, v) 0 m m 0 = 1 h λ n n (µ) µ m Lemma 52 h n (ζ) = ζ2n 1 ζ n 1 (ζ 2 1) Proof We use inuction on n For n = 1 we have h n (ζ) = 1 Assume that for n > 1 it is prove Then, by computing x(ζ, c) n+1 = x n x an y(ζ, ) n+1 = y n y from the inuction assumption we obtain, respectively, (5) h n+1 (ζ) = ζ n + h n(ζ) ; h n+1 (ζ) = ζh n (ζ) + 1 ζ ζ n Both relation lea to the same result: h n+1 (ζ) = ζ n + ζ2n 1 ζ n (ζ 2 1) = ζ2n (ζ 2 1) + ζ 2n 1 ζ n (ζ 2 1) ζ 2n 1 h n+1 (ζ) = ζ ζ n 1 (ζ 2 1) + 1 ζ = ζ2 (ζ 2n 1) + (ζ 2 1) n ζ n (ζ 2 1) Now, a irect computation shows that tr(x(λ, c) a y(µ, ) b ) = λ a µ b + ch a (λ)h b (µ) + 1 λ a µ b = ζ2n+2 1 ζ n (ζ 2 1) ; = ζ2n+2 1 ζ n (ζ 2 1) We have to show that for every fiel F q there are x(λ, c) an y(µ, ) such that h a (λ)h b (µ) = (λ 2a 1)(µ 2b 1) 0 (λ 2 1)(µ 2 1)λ a 1 µ b 1 Note that h n (1) = n Let α F q be an element such that α q 1 = 1, α m 1 for any m < q 1, an let β F q 2 \ F q be an element satisfying that β q+1 = 1, β m 1 for any m < q + 1

16 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 15 Since p(q2 1) 2 either p or q 1 a it follows that either p or q 1 or q+1 oes not ivie b or q+1 oes not ivie a Similarly, Thus, we nee to consier nine cases, an in each case we have to fin x(λ, c) an y(µ, ) such that h a (λ)h b (µ) 0 The following table shows that this is possible q 1 q+1 p a a a p b λ = µ = 1 λ = α, µ = 1 λ = β, µ = 1 q 1 b λ = 1, µ = α λ = α, µ = α λ = β, µ = α q+1 b λ = 1, µ = β λ = α, µ = β λ = β, µ = β We can now euce that if neither a nor b is ivisible by the exponent of PSL(2, q), then the trace map of the wor w(x, y) = x a y b is surjective onto F q Corollary 5 Let a, b 1 an assume that neither a nor b is ivisible by p(q2 1) 2 Then every α F q can be written as α = tr(x a y b ) for some x, y SL(2, q) Proof Accoring to Lemma 2 the trace of w = x a y b can be written as tr(w) = u f(s, t) + h(s, t), where s = tr(x), u = tr(xy), t = tr(y) Namely, it is linear in u an the coefficient of u is a non-trivial polynomial f(s, t) in s an t By Proposition 51, f(s, t) oes not vanish ientically on A 2 s,t(f q ), an hence for every α F q there is a solution (s, u, t) F q to the equation u f(s, t) + h(s, t) = α 6 Surjectivity of w(x, y) = x a y b on SL(2, q) \ { i} an PSL(2, q) In this Section we prove Theorems 18, 19 an 110 on the surjectivity of the wor map w(x, y) = x a y b on SL(2, q) \ { i} an PSL(2, q) By Remark 4, throughout this section we can assume that 1 a, b p(q2 1) 2 The following two Corollaries follow from the general arguments presente in Section Corollary 61 If either a is relatively prime to p(q2 1) p(q 2 1) 2 2 or b is relatively prime to, then the wor w(x, y) = x a y b is surjective on SL(2, q), an hence on PSL(2, q) Proof The claim follows immeiately from Proposition Corollary 62 If either a = p(q2 1) an b is not relatively prime to p(q2 1) ; or 2 2 b = p(q2 1) an a is not relatively prime to p(q2 1) ; then the wor map w(x, y) = x a y b 2 2 is not surjective on PSL(2, q), an hence not on SL(2, q) \ { i}

17 16 TATIANA BANDMAN AND SHELLY GARION Proof The claim follows immeiately from Proposition 1 Remark 6 It follows that the only interesting cases to consier are when a, b < p(q 2 1), an both a an b are not relatively prime to p(q2 1) 2 2 We can now euce Proposition 16, stating that if w = x a y b is non-egenerate with respect to q, then any semisimple element z (namely, when tr(z) ±2) can be written as z = x a y b for some x, y SL(2, q) Proof of Proposition 16 Assume that w = x a y b is non-egenerate with respect to q Without loss of generality, we may assume that 1 a, b p(q2 1) If a, b < p(q2 1) then 2 2 the result immeiately follows from Corollary 5 an Section 4 Otherwise, either a or b is relatively prime to p(q2 1), an the result follows from Corollary 61 2 Unfortunately, a similar result fails to hol when z is unipotent, namely when z ±i an tr(z) = ±2 This case will be iscusse in etail in Section 8, where we shall prove the following two propositions Proposition 64 Let 1 a, b < p(q2 1) Then the image of the wor map w = x a y b 2 contains any non-trivial element z SL(2, q) with tr(z) = 2, if an only if none of the following obstructions occurs: (i) q = 2 e, e is o an a, b { 2(q2 1), 4(q2 1) }; (ii) q mo 4 an a = b = p(q2 1) ; 8 (iii) q 11 mo 12 an a = b = p(q2 1) ; 6 (iv) q 5 mo 12 an a, b { p(q2 1), p(q2 1) } 6 12 Proposition 65 Assume that q is o an let 1 a, b < p(q2 1) 4 Then the image of the wor map w = x a y b contains any element z i with tr(z) = 2, unless q mo 4 an a = b = p(q2 1) 8 We can now prove the main theorems Proof of Theorem 18 Let q = 2 e If w = x a y b egenerates with respect to q, then by Corollary 62 the wor map w is not surjective on SL(2, q) If obstruction (i) occurs then by Proposition 64(i), the image of w = x a y b oes not contain any non-trivial element z SL(2, q) with tr(z) = 0, an hence it cannot be surjective on SL(2, q) On the other han, if w = x a y b is non-egenerate with respect to q an obstruction (i) oes not hol, then by Proposition 16 an Proposition 64(i), any element z SL(2, q) is in the image of the wor map w = x a y b Proof of Theorem 19 Let q be an o prime power If w = x a y b egenerates with respect to q, then by Corollary 62 the wor map w is not surjective on SL(2, q)\{ i} If one of obstructions (ii), (iii), (iv) occurs then by Proposition 64, the image of w = x a y b oes not contain any non-trivial element z SL(2, q) with tr(z) = 2, an hence it cannot be surjective on SL(2, q) \ { i}

18 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 17 On the other han, if w = x a y b is non-egenerate with respect to q an none of the obstructions (ii), (iii), (iv) hols, then by Proposition 16, Proposition 64 an Proposition 65, any element z SL(2, q) \ { i} is in the image of the wor map w = x a y b Proof of Theorem 110 Let q be an o prime power Observe that the wor map w = x a y b is surjective on PSL(2, q) if an only if for every z SL(2, q) either z or z can be written as x a y b for some x, y SL(2, q) If w = x a y b egenerates with respect to q, then by Corollary 62 the wor map w is not surjective on PSL(2, q) If obstruction (ii) occurs then by Proposition 64 an Proposition 65 the image of w = x a y b oes not contain any element ±i z SL(2, q) with tr(z) = 2 or tr(z) = 2, an hence it cannot be surjective on PSL(2, q) On the other han, if w = x a y b is non-egenerate with respect to q an obstruction (ii) oes not hol, then by Proposition 16, Proposition 64 an Proposition 65, for any element z SL(2, q), either z or z can be written as x a y b for some x, y SL(2, q), as neee Proof of Corollary 111 If q is o, then by Theorem 110, the wor map w = x a y b is surjective on PSL(2, q) whenever Thus, one has to prove that p(q 2 1) 8 > max{a, b} (6) q > a = p(q2 1) 8 an (7) PSL(2, q) = q(q2 1) 2 > a > 2 a/2 = p(q2 1) 8 Assume that q > a Since q then q q2 Therefore, > a Moreover, the inequality p(q 2 1) 8 (q2 1) 8 8 q2 9 8 = q2 > a 2 a/2 < PSL(2, q) = q(q2 1) 2 implies that q > a This estimate is sharp Inee, if a = p = q =, we have q = = a, p(q 2 1) 8 = = a < q 2,

19 18 TATIANA BANDMAN AND SHELLY GARION If q is even, then by Theorem 18, the wor map w = x a y b is surjective on PSL(2, q) whenever 2(q 2 1) > max{a, b} Thus, in this case one has to prove that (8) q > a = 2(q2 1) an > a (9) PSL(2, q) = q(q 2 1) > 2 a/2 = 2(q2 1) If q > a then 2(q 2 1) 2(a 1) = 2a 2 > a Let us prove (9) If q = 2, then q(q 2 1) = 6 2 a/2 for any a 2 Hence, we may assume that q 4, an then 2(q2 1) 10 > It follows that (9) is vali for a = 2, On the other han, q(q 2 1) > 2 a/2 = q > 2 a/2 > a if a = q 2 > a 2 2 = 2(q2 1) > 2 1 a 2 a 7 Equiistribution of the wor map w(x, y) = x a y b on PSL(2, q) The goal of this section is to prove Theorem 114 We first consier the case G = SL(2, q) In this case, the Theorem follows from the following Proposition Proposition 71 Denote G = SL(2, q), let a, b 1 an let w : G G G be the map given by w(x, y) = x a y b Then there are a subset S G, an numbers A 1 (a, b), A 2 (a, b) such that: (i) S = ( 1 ɛ(q) ) G, where 0 ɛ(q) A 1(a,b) q ; (ii) M g = q (1 + δ(q)) for any g S, where M g = {(x, y) G 2 w(x, y) = g} an δ(q) A 2(a,b) q Proof We fix a, b an maintain the notation of Section 2 omitting only the inices a, b Let D = a + b 1

20 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 19 Consier the following commutative iagram of morphisms: (10) G G(Fq ) A s,u,t(f q ) ϕ w ψ τ G(F q ) A 1 s(f q ) π In this iagram: A n x 1,x 2, enotes an n imensional affine space with coorinates x 1, x 2, ; π(x, y) = (tr(x), tr(xy), tr(y)) A s,u,t; w(x, y) = x a y b G; ψ(s, u, t) = uf(s, t) + h(s, t); τ(x) = tr(x) A 1 s; ϕ(x, y) = tr(w(x, y)) By efinition, M g = w 1 (g) Let t F q Denote: N t = ϕ 1 (t) G 2, L t = ψ 1 (t) A s,u,t, T t = τ 1 (t) G, p(s, u, t) = s 2 + u 2 + t 2 ust 4, ν i (t), i = 1, 2, are solutions of the quaratic equation x 2 tx + 1 = 0, ωt 2 = t 2 4 Note that ν 1 (t) ν 2 (t) if an only if t 2 For o q the conition ω t F q is equivalent to the conition ν 1,2 F q Recall that if ±2 t F q, then all the elements in T t are conjugate (see Section 4) Thus, if tr(g) = t, then M g = Nt T t From Table 1 in Section 4 we euce that (11) T t = q 2 (1 + δ 1 (t)), where 0 if ω t = 0 1 δ 1 (t) = if ω q t 0 an ν 1,2 (t) F q 1 if ω q t 0 an ν 1,2 (t) F q Hence, in all the cases above, δ 1 1 q We ivie the proof into three steps Step 1 Fibers of π Proposition 72 (a) If { t 2 4, then = q (1 + δ π 1 2 (t)) if p(s, u, t) 0 (s, u, t)(f q ) 2q (1 + 1, ) if p(s, u, t) = 0 q where δ 2 for every (s, u, t) q

21 20 TATIANA BANDMAN AND SHELLY GARION (b) π 1 (s, u, ±2)(F q ) { = q q if p(s, u, 2) 0 2q (1 + δ 1 (s)) if p(s, u, 2) = 0 The proof of this Proposition follows from the next two Lemmas For a fixe y G with tr(y) = t, let K s,u (y) = {x G π(x, y) = (s, u, t)} t 1 Lemma 7 Let t 2 4 an y t = Then 1 0 q ± 1 if p(s, u, t) 0 (12) K s,u (y t )(F q ) = 1 if p(s, u, t) = 0, ν 1,2 (t) F q 2q 1 if p(s, u, t) = 0, ν 1,2 (t) F q Proof If ω t 0 then K s,u (y t ) consists of the matrices ( ) α β x =, u + β αt s α such that (1) α(s α) (u + β αt)β = 1 Assume that q is o Then (1) is equivalent to 2 ( st 2u + ω2 t β ωt 2 α βt s ) 2 p(s, u, t) = 0 2 Thus, if p(s, u, t) 0 then K s,u (y t ) is a non-egenerate conic; whereas if p(s, u, t) = 0 then K s,u (y t ) is a pair of intersecting straight lines, which are not efine over F q if ω t F q, or efine over F q if ω t F q Assume that q = 2 e Substituting α = α + u t, (14) α 2 + β 2 + t α β = β = β + s, we reuce (1) to t p(s, u, t) t 2 Thus, since t 0, we have a conic for p(s, u, t) 0 Hence, K s,u (y t )(F q ) = q ± 1 if p(s, u, t) 0 If p(s, u, t) = 0 then (14) provies (15) (ν 1 (t) α + β)(ν 2 (t) α + β) = 0 Thus, if ν 1,2 (t) F q we have two intersecting lines, whereas if ν 1,2 (t) F q we have precisely one point (0, 0)

22 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 21 1 λ Lemma 74 Let t = 2, λ 0, an v λ = Then 0 1 q if p(s, u, 2) 0 (16) K s,u (v λ )(F q ) = 0 if p(s, u, 2) = 0, ν 1,2 (s) F q 2q or q if p(s, u, 2) = 0, ν 1,2 (s) F q Proof If λ 0 then K s,u (v λ ) consists of the matrices α β x =, s α such that u s λ α(s α) β u s λ Thus if p(s, u, 2) = (s u) 2 0 then we have q points; whereas if p(s, u, 2) = (s u) 2 = 0, then we have = 1 either two isjoint lines, if ν 1,2 (s) F q, ν 1 ν 2 (for o q it means that w s 0, w s F q ); or one line, if ν 1,2 (s) F q, ν 1 = ν 2 (for o q it means that w s = 0); no points, if ν 1,2 (s) F q, (for o q it means that w s F q ) Proof of Proposition 72 (a) Inee, π 1 (s, u, t)(f q ) = K s,u (y t )(F q ) T t, thus the claim follows from Lemma 7 an Equation (11) Moreover, δ 2 (t) δ 1 (t) + 1 q + δ 1(t) 1 q q (b) There are two or three conjugacy classes of matrices y with tr(y) = 2 If u s, then y i, thus there are q 2 1 ifferent matrices y to consier, an accoring to Equation (16), π 1 (s, u, ±2)(F q ) = (q 2 1)q If p(s, u, 2) = 0, ie s = u, then summation over the classes yiels π 1 (s, u, ±2)(F q ) 2q(q 2 1) + q 2 (1 + δ 1 (s)) 2q ( q ) For t = 2 the proof is similar

23 22 TATIANA BANDMAN AND SHELLY GARION Step 2 Definition of the set S Let A = {(s, t) A 2 s,t f(s, t) = 0}, B ζ = {(s, t) A 2 s,t h(s, t) = ζ}, C = {(s, u, t) A s,u,t p(s, u, t) = 0} Note that C is absolutely irreucible for every fiel F q We first efine the set Σ F q by the following rules Rule 1 Assume that there exists ζ F q satisfying p ( s, ζ h(s,t) f(s,t), t ) 0 on C Then ζ Σ Rule 2 Assume that there is an irreucible (over F q ) component A A an ζ F q such that h(s, t) ζ on A Then ζ Σ Note that since eg A = D (by Lemma 2) there are at most D such numbers Rule 2 Σ an 2 Σ Remark 75 By the above construction, Σ contains all the values ζ such that A B ζ contains a curve or L ζ C is not a curve Now, we can efine the sets T = τ 1 (Σ) an S = G \ T Lemma 76 S = G (1 ε(q)), where ε(q) +D q 1 Proof Inee, by construction, Σ ( + D) Thus by (11), T ( + D)q q Hence, S = G T = G (1 ε(q)), where ε(q) ( + D)q2 (1 + 1) q = + D q q q 1 Step Estimation of M g Let ζ F q \ Σ Then L ζ = ψ 1 (ζ) = Y ζ R ζ Q ζ, where Y ζ = {(s, u, t) L ζ p(s, u, t) 0, f(s, t) 0}, R ζ = {(s, u, t) L ζ p(s, u, t) = 0, f(s, t) 0}, Q ζ = {(s, u, t) L ζ f(s, t) = 0} In the estimation of the sizes of the above sets, we will use the following fact, which is the case n = 1 of [GL, Proposition 121] Claim 77 Let X P N be a projective curve in the projective space P N of egree D efine over F q Then (17) X(F q ) D(q + 1) Lemma 78 Q ζ (F q ) D 2 q

24 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 2 Proof If f(s, t) = 0 an (s, u, t) L ζ, then h(s, t) = ζ, ie(s, t) A B ζ Since ζ Σ, the set A B ζ is finite Both curves have egree at most D (by Lemma 2), hence, by Bézout s Theorem, A B ζ D 2 On the other han there is no restriction on the value of u Hence, Q ζ (F q ) D 2 q Lemma 79 R ζ (F q ) D(q + 1) Proof Inee, R ζ (F q ) = {p(s, u, t) = 0, uf(s, t) + h(s, t) = ζ} is a curve, since ζ Σ By Bézout s Theorem, eg(r ζ ) D Hence, accoring to Equation (17), R ζ (F q ) D(q + 1) Lemma 710 Y ζ (F q ) = q 2 (1 + δ (q)) where δ (ζ) 2D q Proof Since eg A = D, by Equation (17), A(F q ) D(q + 1) Hence q 2 (A 2 s,t \ A)(F q ) D(q + 1) For every point (s, t) (A 2 s,t \ A)(F q ) there is precisely one point (s, ζ h(s,t) f(s,t), t) Y ζ Thus, Y ζ (F q ) = (A 2 s,t \ A)(F q ) = q 2 (1 + δ (q)), an δ (ζ) D(q + 1) q 2 2D q We can now estimate M g = N ζ T ζ By Proposition 72, we have where an π 1 (Y ζ )(F q ) = q 5 (1 + δ 4 ), δ 4 2D q + q + 2D q q 8D +, q π 1 (R ζ Q ζ )(F q ) ( D 2 q + D(q + 1) ) 2q ( q ) 2(D 2 + 6D)q 4( ) q Therefore N ζ (F q ) q 5 (1 + δ 4 ) = π 1 (L ζ )(F q ) π 1 (Y ζ )(F q ) 2(D 2 + 6D)q 4( q ), implying that where δ 5 8D + q N ζ (F q ) = q 5 (1 + δ 5 ), + 2(D2 + 6D)(1 + 1 q ) q 4D2 + 2D + q

25 24 TATIANA BANDMAN AND SHELLY GARION We conclue that where M g = N ζ T ζ = q5 (1 + δ 5 ) q 2 (1 + δ 1 ) = q (1 + δ 6 (ζ)), δ 6 (ζ) 4D2 + 2D + q + 2 q = 4D2 + 2D + 5 q an For completing the proof of Proposition 71 it is sufficient to take A 1 (a, b) = 2( + D) ( + D)q q 1, A 2 (a, b) = 4D 2 + 2D + 5 We can now prove Theorem 114 for the group G = PSL(2, q) Proof of Theorem 114 for G = PSL(2, q) Assume that q is o, enote G = PSL(2, q) an consier the commutative iagram (18) G G(Fq ) w 1 G(F q ) ρ κ ρ G G(F q ) w 2 G(F q ) where ρ : G G is the natural projection G G/Z( G); ρ : G G G G is the projection inuce by ρ; w 1, w 2 correspon to the map (x, y) x a y b on G G an on G G respectively Define S = ρ( S) Since for any z G, ρ 1 (z) contains precisely two elements of G, then Proposition 71 implies that S = G (1 ε(q)) 2 = G (1 ε(q)) Take z S, then ρ 1 (z) = {z 1, z 2 }, an enote H z = w2 1 (z) Let y G an enote M y = w1 1 (y) Then M z1 M z2 = ρ 1 (w2 1 (z)) = ρ 1 (H z ) Thus, H z = M z 1 + M z2 = 2q (1 + δ(q)) = G (1 + δ (q)), 4 4 where δ (q) 0 when q

26 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 25 8 Non-surjectivity of some wors w(x, y) = x a y b In this section we prove Propositions 64 an 65, an in particular, we show that there are certain fiels q an positive integers a, b such that the trace map corresponing to the wor w = x a y b is surjective on F q, by Corollary 5, however, the wor w = x a y b itself is not surjective on SL(2, q) (or PSL(2, q)), since the image of w oes not contain i or unipotent elements, yieling the obstructions escribe in Definition Proof of Propositions 64 an 65 Proof of Proposition 64 Since p(q2 1) 2 not ivie a Similarly, either p or q 1 a it follows that either p or q 1 or q+1 oes not ivie b or q+1 We may consier the following four cases: Case 1: p( a ) 1 1 Take x = a an y = i Then x 0 1 a y b = x a = from Corollary 4 oes 1 1, an the claim follows 0 1 For the other cases, it is sufficient to fin some x, y SL(2, q) with s = tr(x), t = tr(y) satisfying: (19) f a,b (s, t) 0 an tr(x a ) tr(y b ) Inee, if f a,b (s, t) 0 then one can fin some u F q such that tr(w) = u f(s, t) + h(s, t) = 2 Hence, there exist some matrices x 1, y 1 SL(2, q) satisfying tr(x 1 ) = s, tr(y 1 ) = t an tr(x 1 y 1 ) = u an so tr(x a 1y b 1) = 2 Moreover, x a 1y b 1 i since tr(x a 1) = tr(x a ) tr(y b ) = tr(y b 1) Therefore, by Corollary 4, there exist x 2, y 2 SL(2, q) such that z = x a 2y b 2 as neee Case 2: q 1 a an q+1 b Let x an y be two matrices of orers q 1 an q + 1 respectively, an let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0 Moreover, since x is a split element while y is a non-split element, an since x a ±i, y b ±i, then tr(x a ) tr(x b ), implying (19) Case : q 1 a an q 1 b Let x, y be some matrices of orer q 1 an note that x a ±i an y b ±i Observe that unless either x a = y b = or x a = y b = 4, for all elements x, y of orer q 1, one can fin two matrices x, y of orer q 1 satisfying tr(x a ) tr(y b ) (see Lemma 41) Let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0, an so (19) hols

27 26 TATIANA BANDMAN AND SHELLY GARION Since a, b p(q2 1) 2 the only cases left to consier are cases (i), (iii), (v) of Remark 81 In Proposition 84 we will show that in all these cases the image of w = x a y b contains every non-trivial element z SL(2, q) with tr(z) = 2 Case 4: q+1 a an q+1 b Let x, y be some matrices of orer q + 1 an note that x a ±i an y b ±i Similarly to Case, observe that unless either x a = y b = or x a = y b = 4, for all elements x, y of orer q + 1, one can fin two matrices x, y of orer q + 1 satisfying tr(x a ) tr(y b ) (see Lemma 41) Let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0, an so (19) hols Since a, b p(q2 1) the only cases left to consier are cases (ii), (iv), (vi) of Remark 81 In Proposition 8 we will show that in all these cases the image of w = x a y b 2 contains no non-trivial element z SL(2, q) with tr(z) = 2, yieling the obstructions given in the proposition Proof of Proposition 65 Since p(q2 1) a it follows that either p or q 1 or q+1 oes not ivie a Similarly, either p or q 1 or q+1 oes not ivie b 2 2 We may consier the following six cases: Case 1: p ( a an p ) b 1 2/a 1 0 Take: x =, y = 0 1 2/b Then x a =, y 0 1 b =, an so z = x 2 1 a y b = i, 2 1 satisfies that tr(z) = 2, an the result follows from Corollary 4 For the other cases, it is sufficient to fin some x, y SL(2, q) with s = tr(x), t = tr(y) satisfying: (20) f a,b (s, t) 0 an tr(x a ) tr(y b ) Inee, if f a,b (s, t) 0 then one can fin some u F q such that tr(w) = u f(s, t) + h(s, t) = 2 Hence, there exist some matrices x 1, y 1 SL(2, q) satisfying tr(x 1 ) = s, tr(y 1 ) = t an tr(x 1 y 1 ) = u an so tr(x a 1y b 1) = 2 Moreover, x a 1y b 1 i since tr(x a 1) = tr(x a ) tr(y b ) = tr(y b 1) Therefore, by Corollary 4, there exist x 2, y 2 SL(2, q) such that z = x a 2y b 2 as neee Case 2: p a an q 1 2 b Let y be some matrix of orer q 1 an let t = tr(y) Accoring to the table in Proposition 51, f a,b (2, t) 0 Moreover, since y is a non-split element an y b ±i then tr(y b ) ±2, implying (20) Case : p a an q+1 2 b The proof is the same as in Case 2

28 SURJECTIVITY AND EQUIDISTRIBUTION OF THE WORD x a y b ON PSL(2, q) AND SL(2, q) 27 Case 4: q 1 a an q+1 b 2 2 Let x an y be two matrices of orers q 1 an q + 1 respectively, an let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0 Moreover, since x is a split element while y is a non-split element, an since x a ±i, y b ±i, then tr(x a ) ± tr(y b ), implying (20) Case 5: q 1 a an q 1 b 2 2 Let x, y be some elements of orer q 1 an note that x a ±i an y b ±i Observe that unless tr(x a ) = tr(y b ) = 0 for all elements x, y of orer q 1, one can fin two matrices x, y of orer q 1 satisfying tr(x a ) tr(y b ) (see Lemma 41) Let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0, an so (20) hols Now, the only case left is q 1 mo 4 an a = b = p(q2 1) (see Remark 81(iii)) 8 In Proposition 84 we will show that in this case the image of w = x a y b contains every element z i with tr(z) = 2 Case 6: q+1 a an q+1 b 2 2 Let x, y be some elements of orer q + 1 an note that x a ±i an y b ±i Similarly to Case 5, observe that unless tr(x a ) = tr(y b ) = 0 for all elements x, y of orer q+1, one can fin two matrices x, y of orer q+1 satisfying tr(x a ) tr(y b ) (see Lemma 41) Let s = tr(x) an t = tr(y) Accoring to the table in Proposition 51, f a,b (s, t) 0, an so (20) hols Now, the only case left is q mo 4 an a = b = p(q2 1) (see Remark 81(iv)) 8 In Proposition 8 we will show that in this case the image of w = x a y b contains no element z i with tr(z) = 2, yieling the obstruction given in the proposition 82 Obstructions for surjectivity of the wor w(x, y) = x a y b Remark 81 In the course of the proof of Propositions 64 an 65, we nee to consier the following special cases: (i) q is even, x a, y b {1, 2, } for all x, y SL(2, q), an x a = y b = if x = y = q 1: Namely, q = 2 e, e is even, q 1 a, q + 1 a, 2 a (an similarly for b) Hence, a an b are multiples of q 1 lcm, q + 1, 2 = 2(q2 1), namely, a, b { 2(q2 1), 4(q2 1) } By Remark 4, it is enough to consier the case a = b = 2(q2 1) (ii) q is even, x a, y b {1, 2, } for all x, y SL(2, q), an x a = y b = if x = y = q + 1:

29 28 TATIANA BANDMAN AND SHELLY GARION Namely, q = 2 e, e is o, q+1 a, q 1 a, 2 a (an similarly for b) Hence, a an b are multiples of q + 1 lcm, q 1, 2 = 2(q2 1), namely, a, b { 2(q2 1), 4(q2 1) } By Remark 4, it is enough to consier the case a = b = 2(q2 1) (iii) q ia o, x a, y b {1, 2, 4} for all x, y SL(2, q), an x a = y b = 4 if x = y = q 1: Namely, q 1 mo 4, q 1 a = b = lcm a, q ( q 1 4, q + 1 2, p a, p a (an similarly for b) Hence, ) = p(q2 1) 8 (iv) q is o, x a, y b {1, 2, 4} for all x, y SL(2, q), an x a = y b = 4 if x = y = q + 1: Namely, q mo 4, q+1 a, q 1 a, p a (an similarly for b) Hence, 4 2 ( q + 1 a = b = lcm 4, q 1 ) 2, p = p(q2 1) 8 (v) q is o, x a, y b {1, 2, } for all x, y SL(2, q), an x a = y b = if x = y = q 1: Namely, q 1 mo 6, q 1 a, q+1 a, p a (an similarly for b) Hence a an 2 b are multiples of ( q 1 lcm, q + 1 ) { p(q 2 1) 2, p if q 1 mo 12 = 6 p(q 2 1) if q 7 mo (vi) q is o, x a, y b {1, 2, } for all x, y SL(2, q), an x a = y b = if x = y = q + 1: Namely, q 5 mo 6, q+1 a, q 1 a, p a (an similarly for b) Hence a an 2 b are multiples of ( q + 1 lcm, q 1 ) { p(q 2 1) 2, p if q 11 mo 12 = 6 p(q 2 1) if q 5 mo In orer to investigate these obstructions in etail, we nee the following technical result on unipotent elements It follows from Theorem 22 an Section 4 that for any matrix z SL(2, q) with tr(z) ±2 an any two integers m, n > 2 iviing p(q 2 1), one can fin two matrices x an y, such that x m = i = y n an z = xy However, a similar result fails to hol if z is unipotent