EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ a1,...,a r

Transcription

1 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r Jia-yi Si Departmet of Matematics, East Cia Normal Uiversity, Sagai, 0006, P R Cia Astract We itroduce two kids of geeralized Catala umers a a, Z ad χa 1,, a r ; a i P, N Tey ca e iterpreted as te umers of certai ouded lattice pats i a lattice poit set 1 Tree explicit formulae for te Breti s polyomials γ,,a r are give i terms of geeralized Catala umers By tese formulae, we study te properties of γ,,a r ad aswer some prolems of Breti Itroductio Let P resp N, resp Z, resp Q e te set of all te positive itegers resp oegative itegers, resp itegers, resp ratioal umers For X {P, N, Z, Q}, deote y Xq te rig of polyomials i a idetermiate q wit X-coefficiets I te paper, Sectio 3, Breti defied te polyomials γ,,a r q Zq, a 1,, a r P, wic descrie te Kazda-Lusztig polyomials of ay Coxeter group, Corollary 46 To do tis, Breti first defied a operator U j, j Q, o Zq y U j a i q i = a i q i i N i j Te for ay a 1,, a r P, e defied 01 γ,,a r q = q1γ 1,,a r q + U +1 1 q 1 1qγ 1,a,,a r q if a 1 were = a a r, Key words ad prases Breti s polyomials, geeralized Catala umers, Kazda Lusztig polyomials Supported y te Natioal Sciece Foudatio of Cia, te Sciece Foudatio of te Uiversity Doctorial Program of CNEC, te Sagai Priority Academic Disciplie ad te Foudatio for te Uiversity Sciece ad Tecology Developmet of Sagai 1 Typeset y AMS-TEX

2 JIAN-YI SHI 0 γ,,a r q = q 1U if a 1 = 1 ad r, 03 γ 1 q = 1 1 q 1 γ a,,a r q For k 0 i N, let k e te iomal coefficiet! k!k! wit te covetio tat k k = 1 We exted te otatio to ay, k Z y settig k = 0 if te coditio k 0 does ot old Breti proposed te followig Cojecture A 4, Cojecture 710 Let P Te γ 1 = γ +1 1 = 1 +1 C, were C = 1 +1 is te t Catala umer Te Breti raised some furter prolems: oe is to ask for a explicit formula for te umer γ,,a r 1 see te commet immediately after 4, Cojecture 710 Havig sow i, Teorem 310 tat 1 a 1+a 3 + γ,,a r q Nq, Breti asked for a comiatorial iterpretatio of 1 a 1+a 3 + γ,,a r q for ay r, a 1,, a r P see 4, Prolem 78 I te preset paper, we first itroduce two kids of geeralized Catala umers a a, Z ad χa 1,, a r ; a i P, N i Sectio 1 Tese umers ave a explicit comiatorial iterpretatio Te i Sectios -4, we deduce tree formulae for te polyomials γ,,a r a 1,, a r P i terms of geeralized Catala umers, wic are for te cases of r = 1, r = ad r 3 respectively Tey are so differet tat we are uale to uify tem ito a sigle oe By tese formulae, we study some properties of γ,,a r ad also aswer te aove prolems of Breti I particular, we verify Cojecture A i Teorem ad give a closed formula for γ,,a r 1 i Corollary 57, te latter geeralizes Teorem We also give a closed formula for γ,,a r 1 i Corollary 55 1 Geeralized Catala umers We itroduce two geeralized catala umers a a, Z ad χa 1,, a r ; a i P, N Tey sall play a crucial role i te formulae of γ,,a r Te lemmas give i te sectio will e used quite frequetly i te susequet sectios 11 For a, Z, defie a = a a +1 Te is te t Catala umer C We call a geeralized Catala umer We 0 a, te umers are a allot umers see 8, Propositio 7103 e We ave te followig simple results a

3 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 3 Lemma Let a, Z Te 1 a a = a1 a = a1 + a1 1 3 a > 0 if ad oly if 0 a 4 We a, N, we ave a = 0 if ad oly if eiter a < or a = = for a P a a a1 a 1 Arrage all te lattice poits a, N wit 0 a i a staircase order as follows: 16,8 14,7 15,8 1,6 13,7 14,8 10,5 11,6 1,7 13,8 8,4 9,5 10,6 11,7 1,8 6,3 7,4 8,5 9,6 10,7 11,8 4, 5,3 6,4 7,5 8,6 9,7 10,8,1 3, 4,3 5,4 6,5 7,6 8,7 9,8 0,0 1,1, 3,3 4,4 5,5 6,6 7,7 8,8 Fig 1 were a, ad c, d are i te same row resp i te same colum if ad oly if a = c d resp = d Eac poit of te form a, a is located o te top of te colum cotaiig it:, is to te orteast of d, d if ad oly if > d A ouded lattice pat from a, a to 0, 0 i te aove staircase diagram is y defiitio a sequece of poits a 0, 0 = a,, a 1, 1,, a r, r = 0, 0 i tur suc tat i a i i, a i = a i1 1 ad i { i1, i1 1} for 1 i r Te we see y Lemma 11 tat for ay a i N, a is exactly te umer of suc ouded lattice pats from a, to 0, 0 We also see from te diagram tat = for ay a P a a a1 a 13 Let us geeralize te aove otio a little it Let e te set cosistig of all te - tuples a, = a 1, 1, a,,, a, of lattice poits a i, i wit 0 i a i i I particular, deote y 0, 0 te elemet of wit all te compoets eig te poit 0, 0 Give a, = a 1, 1, a,,, a, By a ouded lattice pat from

4 4 JIAN-YI SHI a, to 0, 0 i, we mea a -tuple p = p 1,, p, were for ay 1 i, p i is a ouded lattice pat from a i, i to 0, 0 i Fig 1 Clearly, te product just te umer of ouded lattice pats from a, to 0, 0 i 14 For ay c R, deote y c te largest iteger ot greater ta c, ad also deote y c te smallest iteger ot less ta c Te we ave te followig simple results Lemma Let a Z Te 1 a = a + a+1 a = a1 + 1 By te lemma, we see tat te U-operator occurrig i 01 ca e writte as U +1, ad tat i 0 ca e writte as U Fix a 1,, a r P ad r 1 For 1 i r, deote w i = a i + a i a r ad, deote x i = a i + a i+ + + a r if r i mod, or x i = a i + a i+ + + a r1 if r i mod For 0 N, defie χa 1,, a r ; 0 to e ak k is 151 1,, r r i=1 wi+1 i + a i 1 i1 wr1 1 r wi+ + 1 ar, + a i i1 a r1 + r were te sum rages over all te 1,, r wit 0 j a j+1 1, 1 j r, ad + a 1 0 we stipulate tat χa 1, a ; 0 = a ad χa 1 ; 0 = 1 We see tat ay a geeralized Catala umer a occurrig i 151 satisfies te iequalities 0 ad a ad ece te iequality 0 always olds y Lemma 11 So we ave te followig results Lemma 16 For a 1,, a r P, 0 N ad r 1, we ave 1 χa 1,, a r ; 0 0 a χa 1,, a r ; 0 = χa 1 + 1, a,, a r ; χa 1,, a r ; 0 = a a =0 w w3 +1 χa +,, a r ; 1 for r 3 0 a 1 4 χa 1, a ; a 1 = a is equal to te a -t Catala umer Owig to Lemma 16 1, 4, χa 1,, a r ; 0 ca e regarded also as a geeralized Catala umer I te defiitio of χa 1,, a r ; 0 i 151, te rage of te sum ca

5 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 5 e restricted to oly tose 1,, r wit w j+1 wj j a j+1 1, 1 j r y Lemma 11 3 Tus χa 1,, a r ; 0 ca e iterpreted as te umer of certai ouded lattice pats i te set r1 i te sese of For a i N, defie a a a a = Note tat = + 1 ad a a for ay a i N Lemma Let a > e i N Te a a 1 a 1 1 = + 1 a = c a 1 c 1 c c=0 a a a + 1 a 0 = for ay N I geeral, we ave, wit te covetio tat a = 0 1 Proof 1 follows y te idetity m k = m1 k + m1 k1 For, te equatio clearly + 1 olds we = 0 First assume a = + 1 Te = Deote 3 1 y te rigt-ad-side of It is clear tat + 1 equatio =, we wat to sow tat, tis is equivalet to sowig = = 1 = 5 Assumig te = + 1 Sice + 1 = + or = But te last equatio is easy to ceck Next assume ma, = a + 1 > 0, > 0, ad tat as ee sow for all a + 1 i N wit eiter < or ma, < ma, Te y 1, we ave a a1 a1 = + = 1 c ac 1 c ac = 1c c c=0 c=0 c a1c 1c c=0 So is sow

6 6 JIAN-YI SHI Te formula for te polyomial γ t I te preset sectio, we deduce te formula for te polyomial γ t wit t P By applyig te formula, we are ale to verify Cojecture A Teorem 1 For ay P, we ave 1 γ = 1 1 q + 1 k 1 γ 1 = 1 k+1 1 1k q k k q k + q k Proof We sall sow te equatios 1 ad simultaeously Sice γ 1 = 1 y 03, te result is oviously true i tis case Now assumig te result for γ 1 Te y 01, we ave q γ = q 1γ 1 q + U 1 +1 q 1 γ 1 q 1 =q1 1 k+1 1 q k + 1 k q 1k + 1 k 1 1k 1k 1k Sice =1 1 1 = 1 q + 1 k q k + q k k, te result is true for γ Next we cosider te polyomial γ +1 By 01, we ave q γ +1 = q 1γ q + U +1 q +1 1 γ q 1 k q k+1 +q +1k +1 1 q +1 + k 1 = 1 k+1 k q k +q k q + 1 k q k q k+1 k k = 1 k q k + 1 k q k q +1k So te result olds for γ +1 also By iductio, tis proves our result Now we ca sow

7 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 7 Teorem Cojecture A is true Proof By Teorem 1, equatio γ 1 = 1 +1 C is 1 1 wic is equivalet to k = 1 +1 k k = k, Also, equatio γ +1 1 = 1 +1 C is 3 1 k = k Now equatios ad 3 follow immediately from Lemma 11 Remark 3 We uderstad tat D Zeilerger proved Cojecture A see 9 Our proof is more direct comparig wit tat of Zeilerger, as we ave Teorem 1 3 Te formula for te polyomial γ m,t I tis sectio, we give te explicit formula for te polyomial γ m,t wit m, t P 31 Defie a operator D j, j Q, o f = k 0 a kq k Zq y settig D j f = k j a kq k Te from Teorem 1, we see tat 311 D t1 γ t = t1 1 t+k t t k Te followig is te formula for te polyomial γ m,t Teorem 3 For m, t P, γ m,t is equal to t1 k= +1 m1 q k 1 t+k m + t q m+tk + 1 m + t k +m+k t mk m m1 + 1 t +m+k mk k+1 t χm, t; δ 1,1 m χm, t; q +k t q t +k,

8 8 JIAN-YI SHI were δ i,j is te Kroecker delta, wic is equal to 1 if i = j, ad 0 if i j Proof By 0 ad 311, we ave γ 1,t =q 1U t +1 = t1 1 q t γ t = q 1 q 1 t+k t + 1 t + 1 k t1 q t+1k + 1 t 1 t+k t t k t q tk t q +1 t + 1 So te result is true i tis case Now assumig te result for γ m,t, we sall sow it for γ,t For m, t P, let F 1 m, t = F m, t = F 3 m, t = t1 k= +1 1 t+k m + t m + t k m1 q m+tk, χm, t; q +k t, 1 +m+k t m k m m1 1 t +m+k mk k+1 t δ 1,1 m χm, t; q t +k We must sow tat γ,t = F 1 m + 1, t + F m + 1, t + F 3 m + 1, t By 01, we ave 31 γ,t = q1γ m,t +U m+t1 + q m+t q m+t+1 γ m,t 1 = A+B+C+D, q were A = q 1F 1 m, t, B = q 1F m, t, C = q 1F 3 m, t ad D = m+t t + 1 t ++k m+t 1 t m1 1 t +m+k mk m1 χm, t; q t1 +m+k mk χm, t; q t1 +k

9 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 9 a First assume tat ot m ad t are eve Te F 1 m + 1, t = F m + 1, t = F 3 m + 1, t = t 1 m +1 k= m + 1 k m + t + 1 m + t + 1 k 1 m +1+k t 1 m +1+k t q m+t+1k, m + 1 k k k χm + 1, t; q +k t, χm + 1, t; q t +k Now i 31, we ave A = F 1 m + 1, t + α 1, B = α + α 3, C = α 4 + α 5 ad D = α 6, were α 1 =1 t m + t q + t, + 1 m + t m α = 1 +1+k t α 3 =1 t + m α 4 = k= m + α 5 =1 t + m α 6 = m k= m +1 m1 1 t +1+k m1 1 t +k m1 m m1 + 1 m + 1 k χm, t; χm, t; q +k t, + 1 k m 1 m1 q t + m +1, +1 χm, t; q t +k, k+1 m χm, t; q t + m +1, χm, t; q k +k t Te α 4 + α 6 = α 7 + α 8, were m1 α 7 = 1 +1+k t + χm, t; q +k t, k k k= m + m1 α 8 =1 t + m χm, t; q +1 t + m +1 m We ave F 3 m + 1, t = α 1 + α 7 O te oter ad, α 3 + α 5 + α 8 is α 9 = 1 m + t m m χm + 1, t; q + t m +1

10 10 JIAN-YI SHI Clearly, F m + 1, t = α 9 + α Te result is sow Next assume tat ot m ad t are odd Sice te case of m = 1 is easy to ceck, we may assume m > 1 Te t F 1 m + 1, t = 1 k+1 m + t + 1 m + t + 1 k F m + 1, t = F 3 m + 1, t = m +1 k= m + 1 m +k t q m+t+1k, m + 1 k 1 m +k t k χm + 1, t; q +k t, χm + 1, t; q +k t k1 Now i 31, we ave A = F 1 m + 1, t + α 1, B = F m + 1, t + qα 3, C = β 1 + qα 5 ad D = qα 6, were β 1 = k= m +3 k= m +3 1 t +k m1 + χm, t; q +k t k k We ave β 1 + qα 6 = β + qα 8, were m1 β = 1 +k t + χm, t; q +k t k k1 k= m +3 Let us write F 3 m + 1, t = F + F, were F = 1 m +k t χm + 1, t; q +k t, k k1 F =1 m + t m m χm + 1, t; q t + m + Te F = α 1 + β ad F = qα 3 + α 5 + α 8 We get te result c Next assume tat m is odd ad t is eve Te 1 t F 1 m + 1, t = 1 k m + t + 1 q m+t+1k, m + t + 1 k m +1 F m + 1, t = 1 m +k t χm + 1, t; q m + 1 k +k t,

11 F 3 m + 1, t= EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 11 k= m + 1 m +k t χm + 1, t; q +k t k k+1 Now i 31, we ave A = F 1 m + 1, t + α 1, B = F m + 1, t + qα 3, C = β 1 + qα 5 ad D = γ 1, were γ 1 = m k= m + Te β 1 + γ 1 = γ + γ 3, were γ = k= m +3 γ 3 =1 t + m γ + α 1 is equal to γ 4 = k= m +3 1 t +k m1 1 t +k m1 m1 + 1 m + 1 k + 1 m χm, t; χm, t; q k +k t + 1 χm, t; q +k t, k + 1 q t + m + 1 m +k t χm + 1, t; q +k t, k k + 1 ad qα 3 + α 5 + γ 3 is γ 5 = 1 t + m m1 + 1 m Sice F 3 m + 1, t = γ 4 + γ 5, te result follows d Fially assume tat m is eve ad t is odd Te t F 1 m + 1, t = 1 k+1 m + t + 1 q m+t+1k, m + t + 1 k m +1 F m + 1, t = 1 m +1+k t m + 1 k F 3 m + 1, t= 1 m +1+k t k k= m m χm, t; q + t m + +3 χm + 1, t; q +k t, χm + 1, t; q +k t k

12 1 JIAN-YI SHI Now i 31, we ave A = F 1 m + 1, t + α 1, B = α + α 3, C = δ 1 + δ ad D = qα 6, were δ 1 = k= m + δ =1 t + m Te δ 1 + qα 6 is δ 3 = k= m + 1 t +1+k m1 m1 m + 1 m + 1 k χm, t; q + t m +1 1 =1 + 1 χm, t; q +k t, k 1 1 m +1+k t χm + 1, t; q +k t k k Clearly, F 3 m + 1, t = δ 3 + α 1 O te oter ad, α 3 + δ is δ 4 = 1 m + t m m χm + 1, t; q + t m +1 We ave F m + 1, t = α + δ 4 Hece te result follows 4 Te formula for te polyomial γ,,a r, r 3 Fix a 1,, a r P wit r 3 Recall te otatios w i, x j, 1 i, j r, defied i 15 I tis sectio, we give te formula for γ,,a r Teorem 41 γ,,a r wit r 3, were F 1 a 1,, a r = F a 1,, a r = F 3 a 1,, a r = w 1 k= w a 1 +3 = F 1 a 1,, a r + F a 1,, a r + F 3 a 1,, a r for a 1,, a r P 1 x +k 1 x 1+ w +k a 1 1 x 1+ w +k 1 +a 1 χa,,a r ; q w1k, +a 1 +a k w3 1 a 1 k a 1 k χa 1,,a r ; q w +k, k+1 w δ1,1 a 1 χa 1,,a r ; q w +k Comparig wit Teorems 1 ad 3, we see tat te formula i Teorem 41 does ot old for r Te proof of te teorem will proceed i several steps

13 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 13 Lemma 4 Suppose tat te formula i Teorem 41 olds for γ,,a r wit some r 3 ad a 1,, a r P Te tis formula also olds for γ 1,,,a r Proof By 0 ad our assumptio, γ 1,,,a r =q1u w 1 q +1 w 1 1 γ,,a r q is equal to q1 w 1 1 w 1 x 1+ w By sustitutig k = k + w, it ecomes q1 w 1 1 k = w +1 Multiplyig it out, we get γ 1,,,a r = w 1 1 k = w +1 a x 1+k a 1 1 +k 0 a x 1+k + 1 x 1+ 1+w 1 a 11 χa a 1 k 1,,a r ; 0 q w 1 w 0 a 1 + w k a + w k k χa 1,, a r ; 0 q w 1k χa 1,, a r ; 0 q w 1+1k 0 w1 +1 w +1 χa 1,, a r ; 0 q w 1 +1 By Lemma 16, te last term is equal to 1 1+x + w 1 +1 χ1, a 1,, a r ; 0q w 1 +1 So te result olds for γ 1,,,a r Lemma 43 Te formula i Teorem 41 olds for γ 1,,a wit a 1, a P Proof Te proof is similar to tat of Lemma 4 except tat we use Teorem 3 istead of te assumptio o γ,,a r Lemma 44 Suppose tat te formula i Teorem 41 olds for γ,,a r wit some r 3 ad a 1,, a r P Te tis formula also olds for γ +1,a,,a r Proof I tis proof, we temporarily use F i a 1 for F i a 1,, a r, i = 1,, 3, to simplify te otatio By 01, we ave γ +1,a,,a r = q 1γ,,a r + U w q w 1 q w 1+1 γ,,a r 1 q

14 14 JIAN-YI SHI Te y te assumptio o γ,,a r, we get 441 γ +1,a,,a r = A + B + C + D were A = q 1F 1 a 1, B = q 1F a 1, C = q 1F 3 a 1 ad 1 w 1 w 1 D = 1 x 1+ w +k w 1 w + 1 x 1+ w 1 +1+k 0 0 χa a 1 k 1,, a r ; 0 a 1 k q w 1 w k χa 1,, a r ; 0 q w 1+1 w k We must sow tat γ +1,a,,a r = F 1 a F a F 3 a a First assume tat ot a 1 ad w are eve Te F 1 a = w 1 k= w 3 +1 a 1 +1 F a = F 3 a 1 + 1= a + 1 x +k a 1 1 x 3+ w +1+k 1 x 3+ w +1+k 0 + a w a k 0 0 a k a k χa,, a r ; 0 q 1+w 1k, χa 1 + 1, a,, a r ; 0 q w +k, 0 k χa,a,,a r ; 0 q w +k I 441, we ave A = F 1 a α 1, B = α + α 3, C = α 4 + α 5 ad D = α 6, were α 1 = 1 x 3+ w χ + 1, a,, a r ; 0q w +a, a 1 α = 1 x 3+ w +1+k α 3 = 1 x 3+ w + a 1 α 4 = a + 1 x 3+ w +1+k α 5 = 1 x 3+ w + a 1 a 1 α 6 = 1 1 x 3+ w +1+k χa a k 1,, a r ; 0 q w +k, χa 1,, a r ; a k q w + a 1 +1, 0 +1 χa 1,,a r ; 0 q w +k, k χa 1,,a r ; 0 q w + a, χa a k 1,, a r ; 0 q w +a k

15 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 15 Let α 7 e te term i α 6 wit k = a 1 : α 7 = 1 x 3+ w + a 1 +1 Te α 3 + α 5 + α 7 is α 8 = 1 x 3+ w + a χa +1 1,, a r ; 0 q w + a 1 Clearly, we ave F a = α + α 8 Rewrite te differece α 6 α 7 as α 9 = Te α 4 + α 9 is α 10 = a + a x 3+ w +k 1 x 3+ w +1+k χa 1,, a r ; 0 q w + a χa k 1,, a r ; 0 q w +k Sice F 3 a = α 1 + α 10, tis sows our result χa 1,,a r ; 0 a k k Next assume tat a 1 is eve ad w is odd Te F 1 a = w k= w x +k a 1 a 1 +1 F a = 1 x 3+ w +1+k F 3 a 1 + 1= a + 1 x 3+ w +1+k 0 +a w a k 0 a k 0 a k q w +k χa,,a r ; 0 q 1+w 1k, χa 1 + 1, a,, a r ; 0 q w +k, 0 k χa,a,,a r ; 0 q w +k I 441, we ave A = F 1 a α 1, B = α + α 3, C = β 1 + β ad D = qα 6, were a 1 β 1 = 1 x 3+ w +1+k χa 1,,a r ; 0 q w +k, a k k1 + β = 1 x 3+ w + a χa 1 1,, a r ; 0 q w + a 1 +1

16 16 JIAN-YI SHI We see tat α 3 + β = α 8 ad F a = α + α 8 Rewrite qα 6 as a 1 So β 1 + qα 6 is β 3 = + a + 1 x 3+ w +1+k 1 x 3+ w +1+k 0 = χa k 1 1,, a r ; 0 a k 0 Sice F 3 a = α 1 + β 3, our result follows i te case c Next assume tat ot a 1 ad w are odd Te F 1 a = w k= w 3 +1 a 1 +1 F a = F 3 a = a + 1 x +k a 1 1 x 3+ w +k 1 x 3+ w +k k 0 + a w a k 0 0 a k a k q w +k χa,a,,a r ; 0 q w +k χa,,a r ; 0 q 1+w 1k, χa 1 + 1, a,, a r ; 0 q w +k, 0 χa,a,,a r ; 0 k1 q w +k I 441, we ave A = F 1 a α 1, B = F a qα 3, C = γ 1 + qα 5 ad D = γ, were γ 1 = γ = a +3 a + 1 x 3+ w +k 1 x 3+ w +1+k a k k 1 we ave γ 1 + γ = γ 3 + qα 7, were γ 3 = But α 1 + γ 3 is γ 4 = a +3 a +3 1 x 3+ w +k 1 x 3+ w +k k1 χa 1,,a r ; 0 χa 1,,a r ; 0 q w +k q w +k, χa 1,,a r ; 0 q w +k a k k1 0 χa,a,,a r ; 0 a k k1 0 q w +k

17 ad qα 3 + α 5 + α 7 is EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 17 γ 5 = 1 x 3+ w + a 1 0 χa,a,,a r ; 0 q w + a 1 + we a 1 > 1 Clearly, F 3 a = γ 4 + γ 5 ad so te result olds for a 1 > 1 We a 1 = 1, we ave F 3 a = α 1 ad qα 3 + γ = γ 1 = α 5 = 0 So te result olds also i tis case d Fially assume tat a 1 is odd ad w is eve Te F 1 a = w 1 k= w 3 +1 a 1 +1 F a = F 3 a = a + 1 x +k a 1 1 x 3+ w +k 1 x 3+ w +k 0 + a w a k 0 0 a k a k χa,,a r ; 0 q 1+w 1k, χa 1 + 1, a,, a r ; 0 q w +k, 0 χa,a,,a r ; 0 k+1 q w +k I 441, we ave A = F 1 a α 1, B = F a qα 3, C = δ 1 + qα 5 ad D = δ, were δ 1 = δ = a +3 a x 3+ w +k 1 x 3+ w +k a k k k χa 1,,a r ; 0 χa 1,,a r ; 0 q w +k q w +k, We ave δ 1 + δ = δ 3 + δ 4, α 1 + δ 3 = δ 5 ad qα 3 + α 5 + δ 4 = δ 6, were a 1 δ 3 = 1 x 3+ w +k χa 1,,a r ; 0 q w +k, a k k+1 δ 4 = 1 x 3+ w + a 1 δ 5 = a x 3+ w +k χa 1,,a r ; 0 q w + a 1 +, a k 0 χa,a,,a r ; 0 q w +k, k+1

18 18 JIAN-YI SHI δ 6 = 1 x 3+ w + a χa,a,,a r ; 0 q w + a 1 +, +3 we a 1 > 1 Clearly, F 3 a = δ 5 + δ 6 ad so te result olds for a 1 > 1 We a 1 = 1, we ave F 3 a = α 1 + qα 3 ad δ 1 + qα 5 = δ = 0 So te result olds also i tis case Proof of Teorem 41 Tis follows from Lemmas 4, 43 ad 44 5 Some properties of γ,,a r Fix a 1,, a r P wit r 1 Recall te otatios w i, x j, 1 i, j r defied i 15 I te preset sectio, we study some properties of γ,,a r y Teorems 1, 3 ad 41, wic are cotaied i Corollaries 53, 55 ad 57 Te prolems of Breti stated i Itroductio could e aswered terey 51 Write γ,,a r = m c kq k wit c k Z By 0, we ave γ 1,,,a r = q 1 w11 c k q w 1k Tis tells us tat γ 1,,,a r, or more geerally, γ 1,, t,a 1,,a r for ay 1,, t P, oly depeds o te terms of γ,,a r of degrees w 1 1 w 1 1 So we call c k q k te asic part of γ,,a r From Teorems 41, 3 ad 1, we see tat for a 1,, a r, r P, te asic part of γ,,a r is w 1 1 w 1 x 1+ w a a 1+k +k 1 a 1 k χa 1,,a r ; q w +k we r, q k we r = 1 a 1 k For f Zq, deote y deg f te degree of f, ad y ldeg f te lowest degree of te o-zero terms of f Let us state a simple result for later use Lemma 5 Let a,, c N e suc tat 0 < c a ad + c a Te a a c If we ave a < + c i additio, te a > a c Now we deduce tree corollaries of Teorems 1, 3 ad 41

19 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 19 Corollary 53 Let r 3 ad a 1,, a r P 1 If eiter a i = 1, or a i = wit w i+1 eve for some i, 1 < i < r, te γ,,a r = 0 Suppose tat eiter a i >, or a i = wit w i+1 odd, for ay i, 1 < i < r Te deg γ,,a r = w 1 1 w 3 3 ldeg γ,,a r = w Te coefficiet of q i i γ,a,,a r is o-zero for ldeg γ,,a r i deg γ,,a r 5 Te sigs of te coefficiets of γ,a,,a r alterate Proof Uder te assumptio of 1, we ave i1 + a i1 1 i wi wi+ = 0 for 0 i1 < a i +a i1 i Recall te defiitio of χa 1,, a r ; give i 151 Te χa 1,, a r ; = χa,, a r ; = 0 if i >, ad χa 1,, a r ; = 0, F 1 a 1,, a r = 0 if i = I eiter case, 1 follows y Teorem 41 Now suppose tat eiter a i >, or a i = wit w i+1 odd for ay i, 1 < i < r By Lemmas 11 ad 16, we see tat all te i te formula of γ,,a r a 1 1 a 1 k a s ad χa i,, a r ;, i = 1,, occurrig are positive see Teorem 41 Also, we a 1, we ave k+1 w δ1,1 a >0 for k a 1 y Lemma 5 ad y otig tat te term wit = a 1 is always positive Terefore te results -5 follow y Teorem 41 Remark 54 Cocerig Corollary 53, we see from Teorems 1 ad 3 tat for r = 1,, te results 4 ad 5 remai valid ad 1 is vacuously true But te result o loger olds for r = 1, Actually, we ave deg γ,a = w 1, deg γ 1 = 1 ad deg γ = Te result 3 olds for r = ut ot for r = 1 We ave ldeg γ = 0 Next result is cocerig 1 x 1 γ,,a r q Corollary 55 Let a 1,, a r P wit r x 1 γ,,a r q Nq We r 3, 1 x 1 γ,,a r 1 is equal to

20 0 JIAN-YI SHI a 1 + a 1 a + w w3 +1 a 1 1 χa,,a r ;+ a 3 1 a 1 + a γ,a 1= a a a 1 γ 1 = w 1 + w 1 a 1 +a + a 1 Proof First assume r 3 By Teorem 41, 1 x 1 γ,,a r q is equal to χa 1,,a r ; χa 1,a ; 551 w 1 a 1 k= w a a +1 1 a 1 1 χa +a 1 +a k,,a r ;q w1k + w3 a 1 1 a 1 k k+1 w δ1,1 a 1 a 1 k χa 1,,a r ;q w +k χa 1,, a r ; q w +k As sow i te proof of Corollary 53, we see tat all te geeralized Catala umers occurrig i 551 are o-egative Tis implies 1 we r 3 For r, te result 1 ca e sow similarly y Teorem 3 or 1 istead of Teorem 41 a Recall te otatio i 17 For 0 a 1 1, we ave 55 a 1 = a 1 k = k ad 553 a a k + a 1 +3 k+1 w δ1,1 a 1 were we stipulate = 0 for c < 0 For 0 < a c, we ave = w 1 + w 1, 554 w 1 k= w a 1 w3 = + + a k + a 1 a w w3 +1 So te result follows from Similarly, we ca sow 3 ad 4 y Teorems 3 ad 1

21 EXPLICIT FORMULAE FOR THE BRENTI S POLYNOMIALS γ,,a r 1 Remark 56 a Corollary 53-5 ad Corollary 55 1 appear i, Teorem 310 Comparig wit Breti s proof, our proof is more direct as we ave Teorems 41 ad 3 Note tat 55 ca e rewritte i a form closer to 551: 561 a 1 + a 1 + a a a 1 +a + a 1 χa 1, a ; Te comiatorial iterpretatio of te polyomial 1 x 1 γ,,a r q could e got i terms of geeralized Catala umers see 1, 553 ad Lemma 16 Te followig result geeralizes Teorem Corollary 57 Let a 1,, a r P wit r 1 Te we ave γ,,a r 1 = 1 x + w 1 χ,, a r ; x 1+ w 1 1 χ1,,, a r ; 1, were we stipulate tat te first term is zero we r = 1 Proof First assume r 3 Te for 0 a 1, we ave 571 w 1 k= w k + a 1 w3 + + a k = 1 w 1 + a 1 1 w w ad 57 1 x 1+ w a 1 k a 1 k + a k a 1 k +3 =1 x 1+ w a +1 a 1 1 k 1 k 1 k a 1 k a k =1 x 1+ w + a 1 1 =1 x 1+ w 1 1 =1 x 1+ w δ1,1 w a 1 w1 w w δ1,1 a 1 k+1 w δ1,1 a 1 k+1 w δ1,1 a 1

22 JIAN-YI SHI By Teorem 41, tis implies from 571 tat F 1 a 1,,a r 1=1 x + w 1 χ,,a r ;0, ad from 57 tat F a 1,, a r 1+F 3 a 1,, a r 1 = 1 x 1+ w 1 1 χ1,,, a r ; 1 Te result follows We r =, te result ca e sow similarly y Teorem 3 istead of Teorem 41 Fially, we r = 1, te result is just Teorem Refereces 1 F Breti, A comiatorial formula for Kazda-Lusztig polyomials, Ivet Mat , Breti, Fracesco, Comiatorial expasios of Kazda-Lusztig polyomials, J Lodo Mat Soc , F Breti, Comiatorial properties of te Kazda-Lusztig R-polyomials for S, Adv i Mat , F Breti, Kazda-Lusztig ad R-polyomials from a comiatorial poit of view, Discrete Mat , F Breti, Lattice pats ad Kazda-Lusztig polyomials, J AMS , Breti F ad Simio Rodica, Explicit formulae for some Kazda-Lusztig polyomials, J Alg Com , Polo, Patrick, Costructio de polyômes de Kazda-Lusztig aritraires Frec Costructio of aritrary Kazda-Lusztig polyomials, C R Acad Sci Paris Ser I Mat , R P Staley, Eumerative Comiatorics, vol, Camridge Studies i Advaced Matematics, o 6, Camridge Uiv Press, Camridge, D Zeilerger, Catala strikes agai ad agai, preprit