Optimal Littlewood-Offord inequalities in groups

Transcription

1 Optimal Littlewood-Offord iequalities i groups T. Juškevičius 1, G. Šemetulskis arxiv: v [math.pr] Apr 018 Abstract We prove several Littlewood-Offord type iequalities for arbitrary groups. I groups havig elemets of fiite order the worst case sceario is provided by the simple radom walk o a cyclic subgroup. The iequalities we obtai are optimal if the uderlyig group cotais a elemet of a certai order. It turs out that for torsio-free groups Erdős s boud still holds. Our results stregthe ad geeralize some very recet results by Tiep ad Vu. 1 Itroductio Let V = {g 1,...,g } be a multiset of o-idetify elemets of a arbitrary group G. Cosider a collectio of idepedet radom variables X i that are each distributed o a two poit set {g 1 i,g i } ad defie the quatity ρ(v ) = supp(x 1 X = g). g G IthecaseG = RthelatterquatityisthemaximumprobabilityofthesumX X. Wheever G = R, Z m ad g i = 1 we shall adopt the covetio to write ε i istead of the radom variable X i. Ivestigatig radom polyomials Littlewood ad Offord [9] proved a almost optimal boud for the probability that a sum of radom sigs with o-zero weights hits a poit. To be more precise, usig harmoic aalysis they proved that i the case G = R we have ρ(v ) = O( 1/ log). Erdős[5], usig Sperer s theorem from fiite set combiatorics, showed that, actually, ) ρ(v ) ( /. 1 Vilius Uiversity, Istitute of Mathematics ad Iformatics, Vilius, Lithuaia, - tomas.juskevicius@gmail.com. Uiversity of Vilius, Vilius, Lithuaia, - grazvydas.semetulskis@gmail.com. 1

2 This boud is optimal as ca be see by takig g i = 1 i V. I this case we have ) ( / ρ(v ) = P(ε 1 + +ε {0,1}) =. Aswerig a questio of Erdős, Kleitma [7] used a igeious iductio to show that the latter boud still holds for g i lyig i a arbitrary ormed space. See also [] for a very ice expositio of Kleitma s beautiful argumet. Griggs used a similar approach i [6] as i Erdős s semial paper [5] to obtai the best possible result i Z m. More recetly Tiep ad Vu [10] ivestigated the same questio for certai matrix groups ad obtaied results that are sharp up to a costat factor. To be more precise, let m,k, be itegers ad G = GL k (C). Let V = {g 1,...,g } be a multiset of elemets i G, each of which has order at least m. I this case they have obtaied the boud ρ(v ) 141max{ 1 m, 1 }. (1) Furthermore, they have also established the same boud for GL k (p). Let us explai the meaig of the two terms i the upper boud give i (1). Take some elemet g i G of order m ad cosider the multiset V = {g,...,g}. Let us for the simplicity assume that m is odd. I this setup the radom variable S k = X 1 X k is just the simple radom walk o a subgroup of G that is isomorphic to Z m. It is a well kow fact that the distributio of S is asymptotically uiform, which accouts for the 1 ( term i (1). For < m the poit masses of S m are just the usual biomial probabilities ) / /. Therefore i this regime P(S = g) ( ) / / 1. This shows that the iequality (1) caot be improved apart from the costat factor. It is also very atural that the term 1 is domiat for m m, exactly above the mixig time of S, that is kow to be of magitude m (see [8], page 96). I this paper we shall prove a optimal upper boud for ρ(v ), where the elemets of the multiset V lie i a arbitrary group. It turs out that a boud as i (1) holds for arbitrary groups. Furthermore, for groups with elemets havig odd or ifiite order we shall establish a optimal iequality for P(X 1 X = x) without the requiremet that the radom variables X i are two-valued. Let us remid the reader that we deote by ε (usually supplied with a subscript) a uiform radom variable o { 1, 1}. Sometimes it will be importat to stress that these radom variables are defied o Z m istead of R ad we shall do so o each occasio. We deote by (a,b] m ad [a,b] m the set of itegers i the itervals (a,b] ad [a,b] modulo m. Give a atural umber m, we shall write m for the smallest eve umber such that m m. That is, we have m = m. Theorem 1. Let g 1,...,g be elemets of some group G such that g i m. Let X 1,...,X be idepedet radom variables so that each X i has the uiform distributio o the two poit set {gi 1,g i }. The for ay A G with A = k we have

3 P(X 1 X A) P(ε 1 + +ε ( k,k] m ), () where ε i are idepedet uiform radom variables o the set { 1,1} Z m. Note that Theorem 1 is optimal i the sese that if G cotais a elemet of order m, the boud i () ca be attaied. For istace, i the case G = GL k (C) the upper boud i () is achieved by takig two poit distributios cocetrated o the diagoal matrix e πi m I k ad its iverse. Theorem 1 implies a iequality of the same type as the oe by Tiep ad Vu, but with a much better costat. Corollary 1. Let V = {g 1,...,g } be elemets i some group G satisfyig g i m. The ρ(v ) m + 1 3max{ 1 π m, 1 }. (3) The sequece of sums appearig o the right had side of () is a periodic Markov chai ad so does ot coverge to a limit as. Noetheless, it is well kow that it does coverge to a limit if we restrict the parity of. Let us ow express the quatity i the right had side of () i the case A = 1 i asymptotic terms. Propositio. Let m N ad assume that. The for ay l Z m of the same parity as we have P(ε 1 + +ε = l) = +o(1). m The o(1) term is actually expoetially small i terms of. For such sharp quatitative estimates see [3] pages Note that Propositio implies that i (3) the costat after the last iequality caot be smaller tha. Let us also ote that both costats i the expressio m + 1 π are sharp. The term is domiat i the case m, m ad m ad so Propositio shows that the costat caot be reduced. I the case m, ad < m the therm elemet g of order m we have π ρ(v ) = P(ε 1 + +ε ( 1,1] m ) = 1 is domiatig. For V = {g,...,g} for some ( ) / 1 = (1+o(1)). π The simple radom walk o Z m for m odd coverges to the uiform distributio o Z m ad so all probabilities coverge to 1. It should ow be usurprisig that the simple m radom walk o Z m+1 is a much better cadidate for a maximizer of the left had side i (), as by Propositio we gai a extra factor of asymptotically. From this poit our prime focus will be o the particular case G = Z l m for m odd. I this case Theorem 1 does ot provide the optimal boud. The approach we have for 3

4 this case also works for certai groups other tha Z l m ad therefore we will state it i a geeral form. For k 1 we defie [ k +1 +k 1 ] I,k m =,...,. The latter set is a iterval of k poits i Z m. We shall use the covetio that I m,0 =. Theorem 3. Let X 1,...,X be idepedet discrete radom variables takig values i some group G such that for each i we have supp(x i = g) 1 g G. (4) Furthermore, assume that all o-idetity elemets i G have odd or ifiite order ad that the miimal such order is at least some odd umber m 3. The for ay set A G of cardiality k we have P(X 1 X A) P ( τ 1 + +τ I m,k), where τ i are idepedet uiform radom variables o the set {0,1} Z m. The distributio of τ 1 + +τ is asymptotically uiform i Z m ad thus we have P(X 1 X = g) 1 m +o(1). Remark 1. Note that P ( τ 1 + +τ I m,k) = P ( ε1 + +ε I m,k ). We formulated the result i terms of {0,1} radom variables τ i for the sake of coveiece oly - i this formulatio the set of maximum probability is a iterval. As oe otices, it is ot so i formulatig it i terms of { 1,1} distributios ε i. Remark. The reaso we restrict the elemets to have odd order i Theorem 3 is as follows. If there is a elemet of eve order i the uderlyig group, the the group cotais a elemet of order, say h. The by takig idepedet uiform radom variables X i o the set {1,h} we obtai sup g G P(X 1 X = g) = 1. I the case whe G is torsio-free we ca actually prove that Erdős s boud still holds eve i this geeral settig. Propositio 4. Uder the otatio of Theorem 3 ad assumig that G is torsio-free for ay set A G of cardiality k we have P(X 1 X A) P(ε 1 + +ε ( k,k]), m 4

5 where ε i are idepedet. I particular, for ay g G we have ) ( / P(X 1 X = g). The latter propositio immediately follows by takig m large eough i Theorem 3 so that τ 1 + +τ is cocetrated i a proper subset of Z m. For istace, assume that m = +. I this case the latter sum is strictly cotaied i Z m ad its probabilities are exactly the largest k probabilities of ε 1 + +ε ad we are doe. Our proofs are similar i spirit to Kleitma s approach i his solutio of the Littlewood- Offord problem i all dimesios. Actually, it is closer to a simplificatio of Kleitma s proof i dimesio 1 obtaied i [4]. The proofs thus proceed by iductio o dimesio, takig ito accout a certai recurrece relatio satisfied by the worst-case radom walk. A ope problem Theorem 1 gives a optimal iequality if a elemet with a give order exists. To be more precise, if a elemet of order m exists. For groups i which all elemets have odd or ifiite order, Theorem 3 gives the best possible result. It is thus atural to ask what happes if we have full kowledge of the orders of the elemets of the uderlyig group G ad we are ot i the aforemetioed cases. The asymptotics of the cases whe we do kow the exact aswer suggest the followig guess. Cojecture. Let G be ay group ad fix a odd iteger m 3. Suppose that all possible eve orders of elemets i G greater tha m are give by the sequece S = {m 1,m,...} i icreasig order. Cosider a collectio of idepedet radom variables X 1,...,X i G such that each X i is cocetrated o a two poit set {g i,g 1 i } ad g i m. The if m 1 < m for ay A G with A = k we have P(X 1 X A) P(ε 1 + +ε ( k,k] m1 ), where ε i are idepedet uiform radom variables o the set { 1,1} Z m1. O the other had, if m 1 m we have P(X 1 X A) P ( τ 1 + +τ I m,k), where τ i are idepedet uiform radom variables o the set {0,1} Z m. If true, the latter cojecture would settle the remaiig cases. 3 Proofs I order to prove Theorems 1-3, we shall require a simple group theoretic statemet cotaied i the followig lemma. 5

6 Lemma 1. Let G be a group ad g G be a elemet of order greater the or equal to m. The for ay fiite set A G ad a positive iteger s such that s < m we have A A Ag s. Proof of Lemma 1. Suppose there is a oempty set A G ad a positive iteger s such that A = k < m ad A = Ag s. Take some a A ad cosider elemets s ag si, i = 0...k. All these k + 1 elemets are i the set A hece at least two of them must be equal. Let us say ag si = ag sj for some itegers 0 i < j k. But this immediately gives a cotradictio sice the g s(j i) is equal to the group idetity elemet ad m s(j i) sk. Proof of Theorem 1. If = 1 the iequality () is trivial. For k m ad all the right had side of () becomes 1 sice i this case ( k,k] m covers the support of the sum ε 1 + +ε ad so there is othig to prove. We shall heceforth assume that > 1 ad k < m. ByLemma1wehavethatAg Ag 1. Takesomeh Ag \Ag 1addefieB = Ag \{h} ad C = Ag 1 {h}. We the have P(X 1 X A) = P(X 1 X 1 Ag )+P(X 1 X 1 Ag 1 ) = P(X 1 X 1 B)+P(X 1 X 1 C) (5) P(ε 1 + +ε 1 ( k 1,k +1] m )+P(ε 1 + +ε 1 ( k +1,k 1] m ) (6) = P(ε 1 + +ε 1 ( k 1,k 1] m )+P(ε 1 + +ε 1 ( k +1,k+1] m ) (7) = P(ε 1 + +ε ( k,k] m ). This completes the proof. Remark 3. Notethati(6)-(7)weusedthefactthatfork < m thesets( k+1,k 1] m ad (k 1,k +1] m are disjoit i Z m. I the proof of Theorem 3 we shall make use of the followig simple lemma which will allow us to switch from geeral distributios satisfyig the coditio (4) to two-poit distributios. Lemma. Let X be a radom variable o some group G that takes oly fiitely may values, say x 1,...,x. Suppose that p i = P(X = x i ) are ratioal umbers ad that p i 1. The we ca express the distributio of X as a covex combiatio of distributios that are uiform o some two poit set. Proof of Lemma. DeotebyµthedistributioofX. Sicethep i sareallratioal, we have p i = k i K i for some k i,k i Z. We shall ow view µ as a distributio o a multiset M made from the elemets x i i the followig way - take x i exactly k i j i K i times ito M. This way µ has the uiform distributio o M. We thus have that M = {y 1,...,y N } for the appropriate N. Costruct a graph o the elemets o M by joiig two of them by a edge if ad oly if they are distict. Sice we had p i 1, each vertex of this graph 6

7 has degree at least N. Thus by Dirac s Theorem, our graph cotais a Hamiltoia cycle, ad, cosequetly - a perfect matchig. Let µ i be the uiform distributio o two vertices of the latter matchig (i = 1,,...,N). We have µ = 1 N N µ i. Proof of Theorem 3. We shall argue by iductio. First otice that the claim of the Theorem is true for = 1. Furthermore, it is also true for k m sice i that case the boud for the probability i questio becomes 1. We therefore shall from ow o assume that > 1 ad 1 k m 1. Deote by µ i the distributio of the radom variable X i. We ca without loss of geerality assume that each X i is cocetrated o fiitely may poits ad that for each g G we have P(X i = g) Q. By Lemma, each µ i ca be writte as a covex combiatio of distributios that are uiform o some two-poit set. Defie the radom variable f i (X i ) = E i 1{X 1 X A}, where E i stads for itegratio with respect to all uderlyig radom variables except X i. The for each i we have P(X 1 X A) = Ef i (X i ). (8) The latter expectatio is liear with respect to the distributio of X i. Therefore we ca assume that it will be maximized by some choice of two-poit distributios comig from the decompositio of µ i. We shall therefore from this poit assume that X takes oly two values, say h 1 ad h, with equal probabilities. Note that the itervals I,k m have recursive structure. Namely, if 1 k m 1 ad we regard them as multisets, we have the relatio I,k m (Im,k 1) = Im 1,k 1 Im 1,k+1. The pairs o itervals appearig o both sides of the latter equality heavily overlap. This meas that we ca take oe edpoit of I 1,k+1 m that does ot belog to Im 1,k 1 ad move it to this shorter iterval. The resultig itervals are both of legth k ad are exactly the itervals I,k m ad Im,k 1. We shall use this after the iductive step. Take a fiite set A G with k elemets. Note that the elemet h 1 h 1 1 G ad so it has order at least m. By Lemma 1 we have that Ah 1 1 Ah 1 as A Ah 1 h 1. Take some h Ah 1 1 \Ah 1 ad defie B = Ah 1 1 \{h} ad C = Ah 1 {h}. We have i=1 P(X 1 X A) = P(X 1 X 1 Ah 1 1 )+P(X 1 X 1 Ah 1 ) = P(X 1 X 1 B)+P(X 1 X 1 C) This completes the proof. P(τ 1 + +τ 1 I m 1,k 1 )+P(τ 1 + +τ 1 I m 1,k+1 ) = P(τ 1 + +τ 1 I m,k 1)+P(τ 1 + +τ 1 I m,k ) = P(τ 1 + +τ I m,k). 7

8 Proof of Corollary 1. We shall use a idetity o evely spaced biomial coefficiets proved i [1]: ( ) ( ) ( ) = 1 t t+s t+s s By Theorem 1 we have s 1 j=0 ( cos iπ s ) cos π( t)j. (9) s ρ(v ) P(ε 1 + +ε ( 1,1] m ). (10) The right had of the equatio (10) is the sum of biomial probabilities ( i) /, where i is such that i is cogruet to 1 { Z+1} modulo m. Let t be the residue of ( 1 { Z+1} )/ modulo m. Usigtheidetity(9)adtheelemetaryiequalitiescosx exp( x /)forx [0, π] ad e x 0 σ dx σ π we obtai P(ε 1 + +ε ( 1,1] m ) = ( ( t) + ) ( t+ m/ + t+ m/) +... m = m 1 ( cos iπ ) cos π( t)j m m < j=0 m + m m 1 cos jπ (11) m m + 4 m m + 4 m m + 4 m m + π j=1 m 4 j=1 m 4 j=1 0 cos jπ m e π j / m e π x / m dx 1 m + 1. π Note that i (11) we replaced cos πj π( m j) by cos whe j > m. This completes m m 4 the proof. 8

9 Refereces [1] A. Bejami, B. Che, ad K. Kidred, Sums of Evely Spaced Biomial Coefficiets, Mathematics Magazie 83 (010), [] B. Bollobás, Combiatorics, Cambridge Uiversity Press, Cambridge, 1986, Set systems, hypergraphs, families of vectors ad combiatorial probability. [3] P. Diacois, Radom walks o groups: characters ad geometry, Lodo Mathematical Society Lecture Note Series, vol. 1, pp , Cambridge Uiversity Press, 003. [4] D. Dzidzalieta, T. Juškevičius, ad M. Šileikis, Optimal probability iequalities for radom walks related to problems i extremal combiatorics, SIAM J. Discrete Math. 6 (01), o., [5] P. Erdős, O a lemma of Littlewood ad Offord, Bull. Amer. Math. Soc. 51 (1945), [6] J. R. Griggs, O the distributio of sums of residues, Bull. Amer. Math. Soc. (N.S.) 8 (1993), o., [7] D. J. Kleitma, O a lemma of Littlewood ad Offord o the distributios of liear combiatios of vectors, Advaces i Math. 5 (1970), [8] D. A. Levi, Y. Peres, ad E. L. Wilmer, Markov chais ad mixig times, America Mathematical Society, 006. [9] J. E. Littlewood ad A. C. Offord, O the umber of real roots of a radom algebraic equatio. III, Rec. Math. [Mat. Sborik] N.S. 1 (1943), o. 53, [10] P. H. Tiep ad V. H. Vu, No-abelia Littlewood-Offord iequalities, Advaces i Mathematics 30 (016),