A Unified Elementary Approach to the Dyson, Morris, Aomoto, and Forrester Constant Term Identities

Transcription

1 A Unfed Eementary Approach to the Dyson, Morrs, Aomoto, and Forrester Constant Term Identtes Ira M Gesse 1, Lun Lv, Guoce Xn 3, Yue Zhou 4 1 Department of Mathematcs Brandes Unversty, Watham, MA , USA,3,4 Center for Combnatorcs, LPMC-TJKLC Nanka Unversty, Tanjn , PR Chna 1 gesse@brandesedu vun@manankaeducn 3 gn@nankaeducn 4 zhouyue@manankaeducn February 1, 008 Abstract We ntroduce an eementary method to gve unfed proofs of the Dyson, Morrs, and Aomoto denttes for constant terms of Laurent poynomas These denttes can be epressed as equates of poynomas and thus can be proved by verfyng them for suffcenty many vaues, usuay at negatve ntegers where they vansh Our method aso proves some speca cases of the Forrester conjecture Key words Dyson conjecture, Morrs dentty, constant term dentty 1 Introducton In 196, Freeman Dyson [6] conjectured the foowng dentty: Theorem 11 For nonnegatve ntegers a 0, a 1,, a n, 1 ) aj = a 0 + a a n )!, 11) j a 0! a 1! a n! 0 j n where denotes the constant term Dyson s conjecture was proved ndependenty before hs paper was pubshed by Gunson [10] and by Wson [0], and an eegant recursve proof was ater found by Good [9] Smar denttes for constant terms of Laurent poynomas epressed as products are of consderabe nterest, and we sha dscuss severa of them n ths paper Frst s an dentty of Morrs [17] For a, b, k N the nonnegatve ntegers) defne n H 0, 1,, n ; a, b, k) := 1 ) a 1 ) b 0 1 ) k 1) 0 j =1 1 1 j n

2 Morrs proved the foowng resut: Theorem 1 where H 0, 1,, n ; a, b, k) = M n a, b, k), 13) M n a, b, k) := n 1 =0 a + b + k)! k + 1))! 14) a + k)! b + k)! k! Snce H 0,, n ; a, b, k) s homogeneous of degree 0 n the, settng 0 = 1 n 13) gves an equvaent resut, whch s the form stated by Morrs [17] A generazaton of the Morrs dentty was gven by Aomoto [], who etended Seberg s ntegra to obtan a formua equvaent to the foowng constant term dentty [1]: Let A m 0, 1,, n ; a, b, k) := n =1 1 0 ) χ m) H0, 1,, n ; a, b, k), where χs) equas 1 f the statement S s true and 0 otherwse Theorem 13 A m 0, 1,, n ; a, b, k) = n 1 =0 Another generazaton was conjectured by Forrester [7]: Conjecture 14 We have n,j=n 0 +1, j where n = n 0 + n 1 1 ) H 0, 1,, n ; a, b, k) j n 1 1 = M n0 a, b, k) j=0 [a + b + k + χ n m)]! k + 1))! 15) [a + k + χ n m)]! b + k)! k! j + 1)k + 1)j + a + b + kn 0 )! kj + k + j + kn 0 )!, 16) k! k + 1)j + a + kn 0 )! k + 1)j + b + kn 0 )! In [7], Forrester proved the speca case a = b = 0 for a k, n 0, and n 1 ) usng a formua due to Bressoud and Gouden [4], and the case k = 1 for a a, b, n 0, and n 1 ) Kaneko [13, 15] proved the speca cases n 1 =, 3 and n 1 = n 1 Moreover, Forrester and Baker [3] formuated a q-anaog of Conjecture 14, whch was recenty studed by Kaneko [14] Our objectve n ths paper s to ntroduce an eementary method whch eads to new proofs of the Dyson, Morrs, and Aomoto denttes Moreover, our method can be used to obtan some parta resuts on Forrester s conjecture The dea behnd the proofs s the we-known fact that to prove the equaty of two poynomas of degree at most d, t s suffcent to prove that they are equa at d + 1 ponts Ths approach was used by Dyson [6] to prove the case n = 3 of 11) Dyson used Douga s method [5], n whch most of the ponts are obtaned by nducton, makng heavy use of the symmetry of 11) Ths approach does not seem to generaze beyond n = 4 In our

3 approach we use the fact that, as a poynoma n a 0, the rght sde of 11) vanshes for a 0 = 1,,, a a n ), and we show that the same s true of the eft sde The same dea was used by Gesse and Xn [8] n provng a q-anaog of Theorem 11, whch was conjectured by George Andrews [1] n 1975, and frst proved by Zeberger and Bressoud [] n 1985 In a of the proofs, t s routne to show that after fng a but one parameter, the constant term s a poynoma of degree at most d n the remanng parameter, say a, and that the eft sde agrees wth the rght sde when a = 0 The proofs then dffer n showng that both sdes vansh at d addtona ponts In 13), 15) and 16), these poynomas may have mutpe roots We use the poynoma approach to prove the cases n whch the roots are dstnct, and use another argument, based on the form of the constant term as a functon of a the parameters Proposton 4), to etend the resut to the genera case Poynomas, vanshng coeffcents, and a ratonaty resut In ths secton we prove severa emmas that w be needed n the proofs of the constant term denttes Frst, we show n Lemma 1 that the constant terms n these denttes can be epressed as poyomas Net, Lemma s usefu n showng that these poynomas vansh at certan negatve ntegers Lemma 3, whch appes Lemma to coeffcents of the Dyson product, gves Dyson s conjecture and s aso needed n the proof of Proposton 4, whch aows us to dea wth poynomas wth mutpe roots 1 A poynoma characterzaton Fundamenta to our approach s the foowng emma, whch shows that the constant terms we study are poynomas Lemma 1 Let a 0,, a n be nonnegatve ntegers, d := a a n, and et L 1,, n ) be a Laurent poynoma ndependent of a 0 Then for fed a 1,, a n, the constant term n Qa 0, a 1,, a n ) := k 0 0 L 1,, n ) 1 ) a0 1 ) a 0 1) 0 s a poynoma n a 0 of degree at most d + k 0 for any nteger k 0 d Proof We can rewrte Qa 0, a 1,, a n ) n the foowng form Qa 0, a 1,, a n ) = 1) n a =1 0 ) a 0+a n L =1 a 1,, n ) Epandng each 0 ) a 0+a Qa 0,a 1,, a n ) = = 1,, n 1) 1,, n 1) k0 as =0 1) ) a a0 + + a 1 a0 + a 1 1 ) 1 a0 + a n n =1 na 0 k 0 0 a0 +a a0 + a n n ) 1,, n ) a 0 +a 0, we get ) k 0+ a ) 0 n =1 n =1 a L 1,, n ) a L 1,, n ), ) 3

4 where the sum ranges over a nonnegatve ntegers 1,, n such that n = d+k 0 To show that the degree of Qa 0, a 1,, a n ) n a 0 s at most d + k 0, t suffces to show that every term has degree n a 0 at most d+k 0 Ths foows from the fact that n =1 a L 1,, n ) s a Laurent poynoma ndependent of a 0, and the fact that the degree of a 0 +a 1 1 n a 0 s d + k 0, snce a 0 +a ) s a poynoma n a0 of degree Some coroares of Lemma 1 are gven n the Append Secton 6) ) a0 +a n Snce Qa 0, a 1,, a n ) as defned n 1) s a poynoma n a 0, we can etend t to a ntegers a 0, not just nonnegatve ntegers It s usefu to etend the meanng of the rght sde of 1) so that 1) hods for negatve ntegers a 0 Snce 1 / 0 ) a 0 for 1 s not a Laurent poynoma uness a 0 s a nonnegatve nteger, we must epand t as a Laurent seres, but snce 1 ) a0 = ) a0 1 ) a we mght concevaby epand ths epresson ether n powers of / 0 or of 0 / To make the epanson we-defned, we need to specfy the rng n whch we work We reca that for a rng R, the rng R 1,,, n )) of forma Laurent seres n 1,, n wth coeffcents n R s the set of a forma seres n these varabes n whch ony fntey many negatve powers of j appear for each j Then t s suffcent to work n the rng C 0 )) 1,, n )) of forma Laurent seres n 1,, n wth coeffcents n C 0 )) Informay, we may thnk of 0 as arger than a the other varabes, so that / 0 s sma for 1 Thus we have n ths rng the epanson 0 ) a 0 = a 0 0 for a ntegers a 0 ; the aternatve epanson 1 ) a0 = 0 =0 0 ) a 0 = 1) a 0 0 ) a 0 = 1) a 0 s not vad n ths rng uness a 0 s a nonnegatve nteger a0 =0 ) a 0 0 ) a0 ) a 0 0 ) n ) Vanshng coeffcents Our goa s to evauate speca cases of the poynoma Qa 0, a 1,, a n ) gven by Lemma 1 by fndng some of ther zeroes The foowng emma heps us to accompsh ths Lemma Let u j for 1 < j n be nonnegatve ntegers and et m and v for 1 n be ntegers If the coeffcent of m 1 1 m mn n n 1 <j n j ) u j n =1 1 ) v 3) s nonzero then for some subset T of [n] := {1,,, n} we have u j v T 4) T,j T <j 4

5 and 1 <j n u j u j + m + v ), 5) T,j T <j where T = [n] \ T Proof Appyng the formua j ) u j = 1 j ) 1 ) ) u j = α j +α j =u j 1) α j uj α j ) 1 ) α j 1 j ) α j and epandng, we can wrte 3) as a near combnaton of terms of the form n 1 ) v =1 1 <j n 1 ) α j 1 j ) α j = n 1 ) v =1 1 j n Now et α := n j=1 α j, where α = 0 Then we may wrte ths product as 1 ) α j n 1 ) α v 6) =1 If the coeffcent of m 1 1 mn n n 6) s nonzero, then for each, ether α v < 0 or m α v So f the coeffcent of m 1 1 mn n n 3) s nonzero, there est nonnegatve ntegers α j wth α j + α j = u j for j and α = 0 and a subset T of [n] such that where α = n j=1 α j and T = [n] \ T Then α = α j T T,j T j [n] Smary, α v 1, for T, 7) α m + v, for T, 8) α α n = α j = α j + α j ) = u j 9),j T <j 1 <j n,j T <j u j 10) Summng 7) for T gves so by 9), whch s 4) α v T, T T u j v T, T,j T <j 5

6 Summng 8) for T gves α m + v ) T T Thus by 10) and 9) we have and ths s 5) 1 <j nu j = T α + T α u j + m + v ), T Net, we appy Lemma to prove the vanshng of some coeffcents reated to the Dyson product Lemma 3 Let a 1,, a n be nonnegatve ntegers and et d := a 1 + +a n Let k 0, k 1,, k n be ntegers and et k be the sum of the postve ntegers among k 1,, k n For a subset T [n] we defne σt ) := T a, and we set J := T [n],j T <j {σt ) + 1, σt ) +,, σt ) + k}, where the unon s over proper subsets T of [n] Then for every a 0 wth a 0 [d] \ J, we have k 0 0 k 1 1 kn n 1 ) aj = 0 11) j 0 j n Proof Frst we note that ths coeffcent s we defned for any negatve nteger a 0, as epaned at the end of Secton 1 Net, snce the product n 11) s homogeneous of degree 0 n 0, 1,, n, the constant term does not change f we set 0 equa to 1, as ong as k k n = 0 Otherwse the constant term s 0) Settng a 0 = h and smpfyng, we need to show that 1 <j n j ) a +a j na 1 k 1 1 n nan kn n =1 1 ) h a = 0, for h [d] \ J 1) We prove the contrapostve: Suppose that h [d] but the eft sde of 1) s not 0 We sha show that h J; e, σt ) < h σt ) + k for some T [n] We appy Lemma wth u j = a + a j, m = na k, and v = h a Then for some subset T [n] we have a + a j ) h a ) T 13) T and,j T <j 1 <j n a + a j ) a + a j ) + ) n 1)a + h k 14) T,j T <j 6

7 Let t = T Then 13) may be wrtten as t 1) a th t a, T T and ths mpes that for T, a < h 15) T But 15) aso hods for T =, snce h 1 We note that by 15), T [n], snce h d Smary, 14) gves n n 1) a t 1) a + n 1)a + n t)h k =1 T T T Takng a the terms n the a to the eft sde gves n t) T a n t)h T k so snce T [n], Thus by 15) and 16), whch competes the proof h a + k 16) T a < h a + k, T T Dyson s conjecture s an easy consequence of Lemma 3: Proof of Theorem 11 F a 1,, a n N Denote by D L a 0 ) and D R a 0 ) the eft and rght sdes of 11) It s routne to check that 1 Both D L a 0 ) and D R a 0 ) are poynomas n a 0 of degree at most d by Lemma 1); D L 0) = D R 0) by nducton on n); 3 D R a 0 ) vanshes when a 0 = 1,,, d Now appy Lemma 3 wth k 0 = k 1 = = k n = 0, so k = 0 and J = Then D L a 0 ) aso vanshes when a 0 = 1,,, d The theorem then foows snce two poynomas of degree at most d are equa f they agree at d + 1 dstnct ponts 7

8 3 A ratonaty resut We denote by D n ; a 0, a 1,, a n ) the Dyson product 1 ) aj j 0 j n Good [9] used Lagrange nterpoaton to derve the foowng recurson n hs proof of the Dyson conjecture: for a 0, a 1,, a n 1, we have D n ; a 0, a 1,, a n ) = n D n ; a 0,, a 1, a 1, a +1,, a n ) 17) =0 Usng ths recurson, Ss and Zeberger [19], Ss [18], and Lv et a [16] found epct formuas for some of the other coeffcents of the Dyson product Ther resuts suggest the foowng proposton, whch we w need n our approach to the Morrs, Aomoto, and Forrester constant terms Proposton 4 For any Laurent poynoma L 0,, n ) ndependent of the a, L 0,, n )D n ; a 0,, a n ) = Ra 0,, a n ) a 0 + a a n )! a 0! a 1! a n! for some ratona functon Ra 0,, a n ) of a 0,, a n 18) Proof We proceed by nducton on n The n = 0 case s trva Assume the proposton hods for n 1, e, d! L 1,, n )D n 1 1,, n ; a 1,, a n ) = Ra 1,, a n ) a 1! a n!, where d := a a n, for any Laurent poynoma L 1,, n ) ndependent of a By nearty, t s suffcent to show that 18) hods when L 0, 1,, n ) s a monoma Defne f = fa 0, a 1,, a n ) := k 0 0 k 1 1 kn n D n 0,, n ; a 0, a 1,, a n ) We construct a ratona functon Ra 0, a 1,, a n ) so that fa 0,, a n ) = Ra 0, a 1,, a n ) a 0 + a a n )! a 0! a 1! a n! hods for a nonnegatve ntegers a 0, a 1,, a n 19) Frst we show that for each nonnegatve nteger a 0, there s a ratona functon R a0 a 1,, a n ) of a 1,, a n such that d! fa 0,, a n ) = R a0 a 1,, a n ) a 1! a n! 0) By ), wth = a 0 + a j for = 1,, n we have f = ) ) a0 + a 1 a0 + a n n a 0 j +k 1,, D n 1 1,, n ; a 1,, a n ), n j 1,,j n 1) k0 j 1 j n =1 8

9 where the sum ranges over a nonnegatve ntegers j 1,, j n such that a 0 + a 1 j 1 ) + + a 0 + a n j n ) = d + k 0, e, j j n = na 0 k 0 Therefore, by the nducton hypothess on n, ) ) a0 + a f = 1) k0 1 a0 + a n d! Rj 1,, j n ; a 1,, a n ) j 1 j n a 1! a n!, j 1 + +j n=na 0 k 0 where for each j 1,, j n, Rj 1,, j n ; a 1,, a n ) s a ratona functon of a 1,, a n whch aso depends on a 0 and k 1,, k n ) Then 0) hods wth ) ) a0 + a R a0 a 1,, a n ) = 1) k0 1 a0 + a n Rj 1,, j n ; a 1,, a n ) j 1 j n j 1 + +j n=na 0 k 0 Now et β 1 = β 1 a 1,, a n ),, β r = β r a 1,, a n ), where r = n 1)k, be the near functons of a 1,, a n of the form σt ) + j, for T [n] and 1 j k, where k and σ are as n Lemma 3 By Lemma 1, fa 0, a 1,, a n ), for fed a 1,, a n, s a poynoma n a 0 of degree at most d + k 0 Moreover, by Lemma 3, f = 0 for a 0 [d] \ {β 1,, β r } Thus there s a poynoma pa 0 ) of degree at most r + k 0 dependng on a 1,, a n ) such that fa 0, a 1,, a n ) = a 0 + 1)a 0 + ) a 0 + d) a 0 + β 1 )a 0 + β ) a 0 + β r ) pa 0), 1) snce f vanshes at the zeroes of the numerator factors that are not canceed by denomnator factors Comparng wth 0), we obtan It foows that a 0 + d)! f = a 0! a 0 + β 1 )a 0 + β ) a 0 + β r ) pa d! 0) = R a0 a 1,, a n ) a 1! a n! pa 0 ) = a 0 + β 1 ) a 0 + β r ) a 0! a 1! a n! d + 1) d + a 0 ) R a 0 a 1,, a n ) = for some ratona functon R a0 a 1,, a n ) of a 1,, a n Appyng the Lagrange nterpoaton formua, we obtan that 1 a 1! a n! R a 0 a 1,, a n ) pa 0 ) = r+k 0 =0 p) r+k 0 =0, a 0 = r+k 0 =0 1 a 1! a n! R a 1,, a n ) r+k 0 =0, a 0 where So by 1) we get f = a r+k 0 + 1)a 0 + ) a 0 + d) 0 1 a 0 + β 1 )a 0 + β ) a 0 + β r ) a 1! a n! R a 1,, a n ) =0 = Ra 0,, a n ) a 0 + a a n )!, a 0! a 1! a n! r+k 1 0 Ra 0,, a n ) = R a 1,, a n ) a 0 + β 1 ) a 0 + β r ) s a ratona functon of a 0,, a n Ths competes the nducton 9 =0 r+k 0 =0, r+k 0 =0, a 0 a 0

10 We note that the proof of Proposton 4 shows that the denomnator of Ra 0,, a n ) s a product of near poynomas of the form a a m + j, where j s a postve nteger Ths s consstent wth the epct formuas of [19], [18], and [16] 3 The Morrs constant term dentty The proof of 13) s smar to that of 11), so we omt some of the detas We denote by M na, b, k) the eft sde of 13) Lemma 31 For fed a Z and b N, f M na, b, k) = M n a, b, k) for k b, then M na, b, k) = M n a, b, k) for a k N Proof For fed a and b, by takng the constant term n 0, we can wrte M na, b, k) as LD n 1 1,, n ; k, k,, k), where D n 1 and L are as n Secton By Proposton 4, M na, b, k)/m n0, 0, k) s a ratona functon of k It s straghtforward to check that M n a, b, k)/m n 0, 0, k) s aso ratona n k Note that M n0, 0, k) = M n 0, 0, k) foows from the equa parameter case of the Dyson conjecture Therefore, the hypothess mpes that M na, b, k)/m n0, 0, k) = M n a, b, k)/m n 0, 0, k) for a k The emma then foows Proof of Theorem 1 By settng a 0 = a, a = b for = 1,, n n Lemma 1, we see that M na, b, k) s a poynoma n a of degree at most bn for fed b and k n N To see that M n a, b, k) aso has ths property, we rewrte 14) as M n a, b, k) = n 1 =0 a + k + 1)a + k + ) a + k + b)k + 1))! 31) b + k)! k! Moreover, t s easy seen that M n a, b, k) vanshes f a equas one of the foowng vaues: 1,,, b; k + 1), k + ),, k + b); [n 1)k + 1], [n 1)k + ],, [n 1)k + b] 3) Note that these vaues are dstnct f k b The theorem w foow from propertes of poynomas, as n the proof of Dyson s conjecture, f we can show that for bn + 1 dstnct vaues of a, M na, b, k) = M n a, b, k) Lemma 31 reduces the probem to showng that M na, b, k) = M n a, b, k) f k b Frst we show that the equaty hods for a = 0: From 1), we have 0 H 0, 1,, n ; 0, b, k) = 1 j n 1 ) k 33) j Thus by the equa parameter case of Dyson s conjecture, M n0, b, k) = nk)!/k!) n, whch s equa to M n 0, b, k) The remanng vaues are obtaned from the foowng emma, whch competes the proof of Theorem 1 10

11 Lemma 3 For fed nonnegatve ntegers b and k b, H 0, 1,, n ; a, b, k) vanshes when a equas one of the vaues n 3) Proof We prove the contrapostve: Suppose that h [nk] but the constant term of H 0, 1,, n ; h, b, k) = 1) kn )+nb 1 <j n j ) k n n =1 n 1)k+b =1 1 ) h b s not 0 We sha show that t 1)k + b < h tk for some t, e, h s not n 3) We appy Lemma wth u j = k, m = n 1)k + b, and v = h b Then for some subset T [n] we have k h b) T 34) T and,j T <j 1 <j n k k + ) n 1)k + h 35) T,j T <j Let t = T Then 34) may be wrtten as and ths mpes that for t 0, t 1)tk th b) t, But 36) aso hods for t = 0, snce h 1 > b k Smary, 35) gves t 1)k + b < h 36) n 1)nk tt 1)k + n t)n 1)k + h), whch smpfes to tn t)k n t)h, so for t n, h tk 37) But 37) aso hods for t = n, snce h nk Thus by 36) and 37), whch competes the proof t 1)k + b < h tk, We note that t s possbe to hande the case k < b drecty wthout appyng Proposton 4 by usng the foowng fact: z 0 s a root of a poynoma P z) wth mutpcty r f d and ony f dz P z 0) = 0 for = 0, 1,, r 1 For nstance, we can fnd roots of M na, b, k) wth mutpcty at east by consderng the constant term of a H 0, 1,, n ; a, b, k) = n s=1 n 1 s 0 ) n =1 1 ) a 1 ) b 0 1 ) k 0 j 1 j n 11

12 4 The Aomoto constant term dentty In ths secton we w prove Aomoto s dentty usng our eementary approach Frst we note that f m 0 or m n, then 15) reduces to the Morrs dentty, so we assume here that 1 m n 1 The proof s smar to that of the Morrs dentty but s more compcated We provde ony the detas of the key ponts In contrast wth the Morrs dentty, t s not easy to show that 15) hods when a = 0 So nstead of provng equaty at a bn + 1st pont, we show that both sdes of 15) have the same eadng coeffcents as poynomas n a Proposton 41 1 Both sdes of 15) are poynomas n a of degree at most bn The eft sde and the rght sde of 15) have the same eadng coeffcents n a 3 The rght sde of 15) vanshes when a equas one of the vaues n the foowng tabe 1,,, b; k + 1), k + ),, k + b); [n m 1)k + 1], [n m 1)k + ],, [n m 1)k + b]; 41) [n m)k + ], [n m)k + 3],, [n m)k + b + 1]; [n 1)k + ], [n 1)k + 3],, [n 1)k + b + 1] Proof of Proposton 41 sketch) As wth the Morrs dentty, parts 1 and 3 are straghtforward To show part, we rewrte the rght sde of 15) as n 1 =0 [a + k + χ n m) + 1] [a + k + χ n m) + b]k + 1))!, b + k)! k! whose eadng coeffcent s now ceary n 1 =0 k + 1))! b + k)! k! 4) On the other hand, a cacuaton smar to that n Lemma 1 shows that the eadng coeffcent of the eft sde of 15) equas 1 n nb)! n ) nb b 1 ) k, j whch s equa to 4) by Coroary 63 =1 1 j n As n the proof of the Morrs dentty, we may assume k to be suffcenty arge by Proposton 4 Then we can compete the proof of the Aomoto dentty by the foowng emma 1

13 Lemma 4 For fed nonnegatve ntegers b, k b, and m [n], f a equas one of the vaues n 41), then A m 0, 1,, n ; a, b, k) vanshes Proof We prove the contrapostve: Suppose that h [nk + 1] but the constant term of A m 0, 1,, n ; h, b, k) = 1) kn )+nb 1 <j n j ) k n n =1 n 1)k+b =1 1 ) h b χ m) s not equa to 0 We sha show that t 1)k + b + 1 h tk for some t wth 1 t < n m, or t 1)k + b + 1 h tk + 1 for t = n m, or t 1)k + b + h tk + 1 for some t wth n m t n That s, h s not n 41) We appy Lemma wth u j = k, m = n 1)k + b, and v = h b χ m) Then for some subset T [n] we have k h b χ m)) T 43) T and,j T <j 1 <j n k k + ) n 1)k + h χ m) 44) T,j T <j Let t = T Then 43) may be wrtten as t 1)tk th b) t T χ m), and ths mpes that for t 0, t 1)k + b + χt [m] = ) h 45) But 45) aso hods for t = 0, snce h 1 b k + 1 Smary, 44) gves n 1)nk tt 1)k + n t)n 1)k + h) χ m) T Takng a terms n the k to the eft gves tn t)k n t)h T χ m), so for t n, h tk + χt [m]) 46) But 46) aso hods for t = n, snce h nk + 1 Thus by 45) and 46), t 1)k + b + χt [m] = ) h tk + χt [m]) Now accordng to the three cases t < n m, t = n m, and t > n m, the mnmum vaues of χt [m] = ) are 1, 1, and 0, respectvey, and the mamum vaues of χt [m]) are 0, 1, and 1, respectvey Ths competes the proof 13

14 5 On the Forrester conjecture We can appy our method to Forrester s constant term to obtan some parta resuts It s routne to obtan the foowng Proposton 51 1 Both sdes of 16) are poynomas n a of degree at most bn If a = 0, then the eft sde of 16) s equa to the rght sde of 16) 3 The rght sde of 16) vanshes when a equas one of the vaues n the foowng tabe 1,,, b; k + 1), k + ),, k + b); [n 0 1)k + 1], [n 0 1)k + ],, [n 0 1)k + b]; n 0 k + 1), n 0 k + ),, n 0 k + b); [n 0 + 1)k + ], [n 0 + 1)k + 3],, [n 0 + 1)k + b + 1]; [n 1)k + n 1 ], [n 1)k + n 1 + 1],, [n 1)k + n 1 + b 1] 51) Note that the vaues n 51) are dstnct f k b Therefore, by appyng Proposton 4, Forrester s conjecture woud be estabshed f we coud show that for suffcenty arge k, the eft sde of 16) vanshes when a equas any vaue n 51) However, we are ony abe to show that t vanshes for some of these vaues Denote by F n0 ; a, b, k) the eft sde of 16) We obtan the foowng Lemma 5 Assume k s suffcenty arge For t wth 0 t n 1, et M := mn{n 1, t} If a = h wth h satsfyng the condtons tk + C h tk + b, f 0 t n 0, 5) tk + C + 1 h tk + b + C 3, f n t n 1, 53) where C 1 = n 1 4n t) Mn1 M) n t Mn1 M) n t n 1 4n t), f n 1 M,, f n 1 > M,, f n 1 M, C =, f t n 0 < n 1 < M, t n 0, f n 1 t n 0, t n0 + 1)t n 0 ) C 3 =, t + 1 then F n0 ; a, b, k) vanshes 14

15 Proof We prove the contrapostve: Suppose that h [nk + n 1 ] but the constant term of F n0 ; h, b, k) = ± 1 <j n j ) k+χn 0 j ) n =1 n 1)k+b+n 1 1)χ>n 0 ) n =1 1 ) h b, where χ n 0 j = χ > n 0 )χj > n 0 ), s not equa to 0 We sha obtan condtons on h from whch the emma foows We appy Lemma wth u j = k + χ n 0 j ), m = n 1)k + b + n 1 1)χ > n 0 ), and v = h b Then for some subset T [n] we have k + χ n 0 j ) h b) T 54) T and,j T <j 1 <j n k + χ n 0 j ),j T <j k + χ n 0 j ) + T n 1)k + h + n1 1)χ > n 0 ) ) 55) Let t = T and assume that n T there are t 0 eements ess than or equa to n 0 and t 1 eements greater than n 0 Then 54) may be wrtten as t 1)tk + t 1 t 1 1) th b) t, where ma{0, t n 0 } t 1 mn{n 1, t} by ts defnton, and the above equaton mpes that for T, t 1)k + b t 1t 1 1) h 56) t But 56) aso hods for T = f t 1 t 1 1)/t s taken as 1 when t = 0 hence t 1 = 0), snce h 1 > b k Smary, 55) gves n 1)nk + n 1 1)n 1 tt 1)k + t 1 t 1 1) + n t) n 1)k + h ) + n 1 1)n 1 t 1 ) Takng a terms n the k to the eft gves so for T [n], tn t)k n t)h t 1 n 1 t 1 ), h tk + t 1n 1 t 1 ) 57) n t But 57) aso hods for T = [n] f t 1 n 1 t 1 )/n t) s taken as n 1 when t = n hence t 1 = n 1 ), snce h nk + n 1 Thus by 56) and 57), [ h It, t 1 ) := It foows that F n0 ; h, b, k) vanshes f t 1)k + b t 1t 1 1), tk + t 1n 1 t 1 ) t n t h [nk + n 1 ] \ It, t 1 ), 15 t 1 t ]

16 where t ranges from 0 to n and t 1 ranges from ma{0, t n 0 } to mn{n 1, t} The above condton can be smpfed further: For 0 t n 1 f r1 n 1 r 1 ) tk + n t + 1 h tk + b + r r 1) t ) hods for every r 1 wth ma{0, t n 0 } r 1 mn{n 1, t} and r wth ma{0, t n 0 +1} r mn{n 1, t + 1}, then F n0 ; h, b, k) vanshes Ths s because when k s suffcenty arge, the eft and rght endponts of It, r 1 ) are aways to the eft of the correspondng endponts of It+1, r ) for any r 1 and r n ther range Therefore, after removng the ntervas r 1 It, r 1 ) and r It + 1, r ), each remanng vaue of h beongs to an open nterva possby empty), from the rght endpont of It, r 1 ) to the eft endpont of It + 1, r ) for some r 1 and r By anayzng the etreme vaues of 58) among the range of r 1 and r, t s straghtforward to obtan 5) and 53) Remark 53 The frst two nes of 51) are aways mped by Lemma 5 to be roots for n 0 1 Ths foows easy by checkng the cases t = 0 and t = 1 Coroary 54 Conjecture 14 hods n the etreme cases n 1 = and n 1 = n 1 Proof We verfy ths drecty by Lemma 5 If n 1 =, then n 0 = n The frst two nes of 51) are roots by Remark 53 If t n 0 = n, then M = mn{, t} =, and C 1 = 0 Thus we obtan the range tk + 1 h tk + b, whch s consstent wth the t + 1)st ne of 51) If t = n 1, then M =, and C = C 3 = 1 Thus we obtan the range n 1)k + h n 1)k + b + 1, whch s consstent wth the nth ne of 51) Therefore, Lemma 5 mpes that a vaues of a n 51) are roots, and Forrester s conjecture hods n ths case If n 1 = n 1, then n 0 = 1 The cases t n 0 = 1 of 51) are deat wth n Remark 53 If t n 1, then M = t, C 3 = tt 1) t+1 = t+1)t 1) t 1) t+1 = t 1, and a three cases of C are equa to t 1 Thus we obtan the range tk + t h tk + b + t 1, whch s consstent wth that of 51) As n the case n 1 =, Forrester s conjecture hods Proposton 55 Forrester s conjecture hods when n 5 Proof The cases n 4 are consequences of Coroary 54; the case n = 5 can be verfed by Lemma 5 Further routne cacuatons by Lemma 5 gves us the foowng tabe: n 1 = n 3 n n 1 M r = n 1 + n 1 5 n 1 + n ) where M r s an upper bound for the number of mssng roots n 51), e, roots that are not mped by Lemma 5 For brevty, we verfy n deta ony the case n 1 = 3; the other cases are smar For n 1 = 3, we have n 0 = n 3 The cases t = 0, 1 are guaranteed by Remark 53, so henceforth we w aways assume t and by Proposton 55) n 6 If t =, then M = and n 1 = 3/ < M Therefore C 1 = = 0, for n 5 If 3 t n 3, then M = 3, 3 4n ) 16

17 n and C 1 = 1 9 4n t) = 4n t) = 0 If t = n, then M = 3 and t n 0 = 1 It foows that t n 0 < n 1 n < M, C = 1 4n t) = 1, C 3 = 1 If t = n 1, then M = 3 Ths mpes that n 1 t n 0, C = t n 0 =, and C 3 = 1 In concuson, n the case n 1 = 3, ony one root [n 1)k + b + ] s not mped by Lemma 5 Coroary 56 Conjecture 14 hods n the case n 1 = 3 Proof As we have just seen, for n 1 = 3 we are mssng ony one root But by [3], we know that the q-generazaton of 16) hods when a = k Thus f we et q 1 n ths resut, we get a = k as our bn + 1)st pont We concude ths paper by the foowng observaton Let us take the Forrester constant term as an eampe In the proof of Lemma, we made the epanson y y j ) k y j ) 1 <j n n 0 +1 <j ny = c γ y γ 1 1 yγn n, γ where y = 1, and try to show that the constant term assocated wth y γ 1 1 yγn n s equa to 0 for each γ However, t woud be suffcent to show that for each γ, ether the assocated constant term s 0 or c γ = 0 after canceaton) We conjecture that n our approach to Forrester s conjecture, c γ = 0 when Lemma does not appy We have checked our conjecture for n 6 and k 3 Acknowedgments The authors woud ke to thank the referees for hepfu suggestons to mprove the presentaton Lun Lv and Yue Zhou woud ke to acknowedge the hepfu gudance of ther supervsor Wam YC Chen The thrd author was supported by the 973 Project, the PCSIRT project of the Mnstry of Educaton, the Mnstry of Scence and Technoogy and the Natona Scence Foundaton of Chna 6 Append: Consequences of the poynoma approach From Lemma 1 and ts proof, we can deduce the foowng resut Coroary 61 Let d and Q be as n Lemma 1 wth k 0 = 0 Then the eadng coeffcent of Qa 0,, a n ) n a 0 s 1 d! n ) d and the second eadng coeffcent of Qa 0,, a n ) n a 0 s n =1 a d 1)! n =1 ) d 1 1 n =1 n =1 n d )! a L 1,, n ), 61) =1 ) d ) n =1 a L 1,, n ) 6) 17

18 Proof Takng the eadng coeffcent of ) n a 0 gves 1 1,, 1!! n! n n =1 a = 1 d! L 1,, n ) d! 1 1 n n n a 1,, 1!! n! L 1,, n ) n = 1 d! n ) d n =1 =1 a L 1,, n ) Takng the second eadng coeffcent of ) gves n 1,, n =1 a + 1) 1 1 n n 1!! n! n =1 a L 1,, n ), whch can be rewrtten as 6) Appyng Coroary 61 to the Dyson conjecture gves the foowng dentty, whch appeared n [1, Coroary 54] Coroary n ) a 1+ +a n n =1 a 1 j n 1 ) aj = a a n )! 63) j a 1! a! a n! We omt the formua for the second eadng coeffcent, whch s more compcated Appyng Coroary 61 to Morrs s dentty gves the foowng resut, whch s needed for the proof of Aomoto s dentty We remark that 64) was aso obtaned n [11, Proposton ] through a compcated cacuaton Coroary n ) nb n =1 b 1 j n 1 ) k n 1 k + 1))! = nb)! j b + k)! k! 64) =0 1 n ) nb =1 n =1 b 1 j n The proofs of 63), 64) and 65) are straghtforward 1 ) k = b nk k b + 1 ) n 1 k + 1))! nb )! j b + k)! k! 65) =0 References [1] G E Andrews, Probems and prospects for basc hypergeometrc functons, n Theory and Appcaton of Speca Functons, ed R A Askey, Academc Press, New York, 1975, pp

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