Congruences of the Partition Function

Transcription

1 Y Yang (2011) Congruences ohe Partition Function, Internationa Mathematics Research Notices, Vo 2011, No 14, Advance Access ubication October 12, 2010 doi:101093/imrn/rnq194 Congruences ohe Partition Function Yifan Yang Deartment of Aied Mathematics, Nationa Chiao Tung University and Nationa Center for Theoretica Sciences, Hsinchu, Taiwan 300 Corresondence to be sent to: Let (n) denote the artition function In this artice, we wi show that congruences of the form (m k n+ B) 0modm for a n 0 exist for a rimes m and satisfying m 13 and 2, 3, m, where B is a suitaby chosen integer deending on m and Here, the integer k deends on the Hecke eigenvaues of a certain invariant subsace of S m/2 1 (Γ 0 (576), χ 12 ) and can be exicity comuted More generay, we wi show that for each integer i > 0 there exists an integer k such that with a roery chosen B the congruence hods for a integers n not divisibe by 1 Introduction (m i k n+ B) 0modm i Let (n) denote the number of ways to write a ositive integer n as unordered sums of ositive integers For convenience, we aso set (0) = 1, (n) = 0forn< 0, and (α) = 0 if α Z A remarkabe discovery of Ramanujan [14] is that the artition function (n) Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 Received November 6, 2009; Revised Juy 2, 2010; Acceted Setember 6, 2010 c The Author(s) 2010 Pubished by Oxford University Press A rights reserved For ermissions, ease e-mai: journasermissions@oucom

2 3262 Y Yang satisfies the congruences (An + B) 0modm, (1) for a nonnegative integers n for the tries (A, B, m) = (5, 4, 5), (7, 5, 7), (11, 6, 11) Ramanujan aso conjectured that congruences (1) exist for the cases A= 5 j,7 j,or11 j This conjecture was roved by Watson [18] for the cases of owers of 5 and 7 and Atkin [4] for the cases of owers o1 (Aarenty, Ramanujan actuay found a roof ohe congruences moduo owers of 5 himsef The roof was contained in an unubished manuscrit, which was hidden from the ubic unti 1988 It aeared that Ramanujan intended to rove congruences moduo owers of 7 aong the same ine of attack, but his aiing heath revented him from working out the detais See the commentary at the end of [8] for more detais) Since then, the robem of finding more exames of such congruences has attracted a great dea of attention However, Ramanujan-tye congruences aear to be very sarse Prior to the ate twentieth century, there are ony a handfu of such exames [5, 7] In those exames, the integer A is no onger a rime ower It turns out that if we require the integer A to be a rime, then the congruences roved or conjectured by Ramanujan are the ony ones This was roved recenty in a remarkabe aer of Ahgren and Boyan [2] On the other hand, if A is aowed to be a nonrime ower, a surrising resut of Ono [13] shows that for each rime m 5and each ositive integer k, a ositive roortion of rime has the roerty ( m k 3 ) 0modm (2) for a nonnegative integers n reativey rime to This resut was ater extended to comosite m, (m, 6) = 1, by Ahgren [1] The resuts of [1, 13] were further extended by Ahgren and Ono [3] Neither of [1, 13] addressed the agorithmic asect of finding congruences ohe form (2) For the cases m {13, 17, 19, 23, 29, 31}, thiswasdonebyweaver[19] In effect, she found 76,065 new congruences (However, we shoud remark that some congruences isted in [19, Theorem 2] were aready discovered by Atkin [5] Had Atkin had the comuting ower ohe resent day, he woud have undoubtedy discovered many more congruences) For rimes m 37, this was addressed by Chua [9], athough no exicit exames were given therein Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014

3 Congruences ohe Partition Function 3263 Another remarkabe discovery of Ono [13, Theorem 5] is that the artition function ossesses a certain eriodic roerty moduo a rime m Secificay, he showed that for every rime m 5, there exist integers 0 N(m) m 48(m3 2m+1) and 1 P (m) m 48(m3 2m+1) such that ( m i ) ( m P (m)+i ) mod m (3) for a nonnegative integers n and a i N(m) Note that the bound m 48(m3 2m+1) can be imroved greaty using a resut of Garvan [10] See Coroary 33 in Section 3 for detais In this aer, we wi obtain new congruences for the artition function and discuss reated robems In articuar, we wi show that there exist congruences of the form (m k n+ B) 0modm for a rimes m and such that m 13 and not equa to 2, 3, and m, where B is a suitaby chosen integer deending on m and Theorem 11 Let m and be rimes such that m 13 and 2, 3, m Then there exists an exicity comutabe ositive integer k 2 such that for a nonnegative integers n reativey rime to m ( m 2k 1 ) 0modm (4) For instance, in Section 5, we wi find that for m = 37, congruences (4) hodwith k Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 As far as we know, this is the first exame in iterature where a congruence (1) moduo arimem 37 is exicity given Theorem 11 is in fact a simified version of one ohe main resuts (See Theorem 36) In the fu version, we wi see that the integer k in Theorem 11 can be determined quite exicity in terms ohe Hecke oerators on a certain invariant subsace ohe sace S m/2 1 (Γ 0 (576), χ 12 ) of cus forms of eve 576 and weight m/2 1

4 3264 Y Yang with character χ 12 = ( 12 ) This invariant subsace of S m/2 1 (Γ 0 (576), χ 12 ) was first discovered by Garvan [10] and rediscovered by the author ohe resent aer To describe this invariant subsace and to see how it comes into ay with congruences ohe artition function, erhas we shoud first review the work of Ono [13] and other subsequent aers [9, 19] Thus, we wi ostone giving the statements of our main resuts unti Section 3 Our method can be easiy extended to obtain congruences of (n) moduo a rime ower In Section 6, we wi see that for each rime ower m i and a rime 2, 3, m, there aways exists a ositive integer k such that ( m i 2k 1 ) 0modm i for a ositive integers n not divisibe by One exame worked out in Section 6 is ( ) 0 mod 13 2 In the same section, we wi aso discuss congruences oye (5 j k n+ B) 0mod5 j+1 11 Notation Throughout the aer, we et S λ (Γ 0 (N), χ) denote the sace of cus forms of weight λ and eve N with character χ By an invariant subsace of S λ (Γ 0 (N), χ) we mean a subsace that is invariant under the action ohe Hecke agebra on the sace Foramatrixγ = ( ab cd) GL(2, Q) and a moduar form f(τ) of an even weight k, the sash oerator is defined by f(τ) k γ := (det γ) k/2 (cτ + d) k f ( ) aτ + b cτ + d For a ower series f(q) = a f (n)q n and a ositive integer N, weetu N and V N denote the oerators Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 U N : f(q) f(q) U N := a f (Nn)q n, n=0 V N : f(q) f(q) V N := a f (n)q Nn n=0

5 Congruences ohe Partition Function 3265 Moreover, if ψ is a Dirichet character, then f ψ denotes the twist f ψ := af (n)ψ(n)q n Finay,forarimem 5 and a ositive integer j, wewrite F m, j = n 0,m j n 1 mod ( m j ) q n Note that we have 2 Works of Ono [13], Weaver [19], and Chua [9] In this section, we wi review the ideas in [9, 13, 19] F m, j U m = F m, j+1 (5) First of a, by a cassica identity of Euer, we know that the generating function of (n) has an infinite roduct reresentation If we set q = e 2πiτ,thenwehave (n)q n 1 = 1 q n n=0 q 1/ n=0 n=1 (n)q n = η(τ) 1, where η(τ) is the Dedekind eta function Now assume that m is a rime greater than 3 Ono [13] considered the function η(m k τ) mk /η(τ) On the one hand, one has ( η(m k τ) mk η(τ) U m k = (1 q n ) mk (n)q 1)/) n+(m2k U m k n=1 n=0 Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 On the other hand, one has η(m k τ) mk η(τ) m2k 1 = Δ(τ) (m2k 1)/ mod m, η(τ)

6 3266 Y Yang where Δ(τ) = η(τ) is the normaized cus form of weight 12 on SL(2, Z) From these, Ono [13, Theorem 6] deduced that F m,k (Δ(τ)(m2k 1)/ U m k) V η(τ) mk mod m Now it can be verified that for k = 1, the right-hand side ohe above congruence is contained in the sace S (m 2 m 1)/2(Γ 0 (576m), χ 12 ) of cus forms of eve 576m and weight (m 2 m 1)/2 with character χ 12 = ( 12 ) Thenby(5) and the fact that U m defines a inear ma U m : S λ+1/2 (Γ 0 (4Nm), ψ) S λ+1/2 (Γ 0 (4Nm), ψχ m ), where χ m is the Kronecker character attached to Q( m), one sees that F m,k G m,k = a m,k (n)q n mod m for some G m,k S (m 2 m 1)/2(Γ 0 (576m), χ 12 χ k 1 m ) Now the genera Hecke theory for haf-integra weight moduar forms states that if f(τ) = n=1 a f(n)q n S λ+1/2 (Γ 0 (4N), ψ) and is a rime not dividing 4N, then the Hecke oerator defined by T 2 : f(τ) n=1 ( a f ( 2 n) + ψ() ( ( 1) λ n ) λ 1 a f (n) + ψ( 2 ) 2λ 1 a f ( n 2 ) ) q n sends f(τ) to a cus form in the same sace In the situation under consideration, if is a rime not dividing 576m such that G m,k T 2 0modm, Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 then we have 0 (G m,k T 2) U mod m = n=1 ( ( n )) a m,k ( 3 n) + ψ( 2 ) m2 m 3 a m,k q n

7 Congruences ohe Partition Function 3267 since ( n ) = 0 In articuar, if n is not divisibe by, then a m,k ( 3 n) 0modm, which imies ( m k 3 ) 0modm Finay, to show that there is a ositive roortion of rimes such that G m,k T 2 0modm, Ono invoked the Shimura corresondence between haf-integra weight moduar forms and integra weight moduar forms [16] and a resut of Serre [15, 64] As mentioned earier, Ono [13] did not address the issue of finding exicit congruences ohe form (2) However, [13, Section 4] did give us some hints on how one might roceed to discover new congruences, at east for sma rimes m The key observation is the foowing The moduar form G m,k itsef is in a vector sace of big dimension, so to determine whether G m,k T 2 vanishes moduo m, one needs to comute the Fourier coefficients of G m,k for a huge number oerms However, it turns out that F m,k is congruent to another haf-integra weight moduar form of a much smaer weight For exame, using Sturm s theorem [17] Ono verified that F 13,2k+1 G 13,2k k η(τ) 11 mod 13, (6) F 13,2k+2 G 13,2k k η(τ) 23 mod 13 for a nonnegative integers k The moduar form η(τ) 11 is in fact a Hecke eigenform (The moduar form η(τ) 23 is aso a Hecke eigenform It has been known since Morris Newman s work in the 1950s that for odd r with 0 < r <, the function η(τ) r is a Hecke eigenform) More generay, for m {13, 17, 19, 23, 29, 31}, itisshownin[13, Section 4], [11, Proosition 6] and [19, Proosition 5] that G m,1 is congruent to a Hecke eigenform of weight m/2 1 Using this observation, Weaver [19] then devised an agorithm to find exicit congruences ohe form (2) form {13, 17, 19, 23, 29, 31} The roof of congruences (6) givenin[11, 19] is essentiay verification in the sense that they a used Sturm s criterion [17] That is, by Sturm s theorem to show that two moduar forms on a congruence subgrou Γ are congruent to each other moduo a rime m, it suffices to comare sufficienty many coefficients, deending on the weight Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014

8 3268 Y Yang and index (SL(2, Z) : Γ) Naturay, this kind of argument wi not be very usefu in roving genera resuts In [9], instead ohe congruence η(mτ) m η(τ) m2 1 mod m η(τ) used by Ono, Chua considered the congruence η(mτ) m η(mτ) m 1 η(τ) m 1 mod m η(τ) as the starting oint The function on the right is a moduar form of weight m 1on Γ 0 (m) Thus, by the eve reduction emma of Atkin and Lehner [6, Lemma 7], one has η(mτ) m 1 η(τ) m 1 (U m + m (m 1)/2 1 W m ) S m 1 (SL(2, Z)), where for a moduar form f(τ) of an even integra weight k on Γ 0 (m), the Atkin Lehner oerator W m is defined by It foows that ( ) 0 1 W m : f(τ) f(τ) k = ( mτ) k f m 0 1 F m,1 = η(τ) U m f m(τ) η(τ) mod m m ( 1 ) (7) mτ for some cus form f m (τ) S m 1 (SL(2, Z)) (Incidenty, this aso roves Ramanujan s congruences for m = 5, 7, and 11, since there are no nontrivia cus forms of weight 4, 6, and 10 on SL(2, Z)) By examining the order of vanishing of f m (τ) at, Chua [9, Theorem 11] then concuded that if we et r m denote the integer in the range 0 < r m < such that m r m mod, then Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 F m,1 η(τ) r m φ m,1 (τ)mod m for some moduar form φ m,1 on SL(2, Z) of weight (m r m 2)/2 More generay, one has the foowing roosition

9 Congruences ohe Partition Function 3269 Proosition 21 Let m 13 be a rime and r m be the integer in the range 0 < r m < such that m r m mod Set r m, j = r m if j is odd, 23 if j is even Then F m, j η(τ) r m, j φ m, j (τ)mod m for some moduar form φ m, j (τ) on SL(2, Z), where the weight of φ m, j is (m r m 2)/2 if j is odd and is m 13 if j is even Proof Consider the function f m, j (τ) = η(m j τ) m j /η(τ) It is a moduar form of weight (m j 1)/2 onγ 0 (m j ) with character ( m ) j Consider aso the auxiiary function η(τ) m if j is odd, η(mτ) h m, j (τ) = ( η(τ) m ) 2 if j is even η(mτ) It is a moduar form on Γ 0 (m) with character ( /m) j and satisfies h m, j (τ) 1modm (8) By the eve reduction emma of Atkin and Lehner [6, Lemma 7], if we ay U m to f m,i j 1 times and then mutiy the resuting function by h m, j, we get a moduar form on Γ 0 (m) with trivia character That is, f m, j (τ) U j 1 m h m, j(τ) Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 is a moduar form on Γ 0 (m) with trivia character The weight is λ m, j = (m j 1) 2 (m j 1) 2 + (m 1) 2 if j is odd, + m 1 if j is even

10 3270 Y Yang Then by the eve reduction emma again ( f m, j (τ) U j 1 m h m, j(τ)) (U m + m λ m, j/2 1 W m ) is a moduar form on SL(2, Z), where W m is the Atkin Lehner oerator defined in (7) Considering the order of vanishing at, we see that this moduar form on SL(2, Z) is where Δ(τ) = η(τ), Δ(τ) μ m, j φ m, j (τ), μ m, j = m2 j + ν m, j 1 m j with ν m, j being the unique integer satisfying 0 <ν m, j < m j and ν m, j 1modm j,and φ m, j is a moduar form of weight λ m, j 12μ m, j on SL(2, Z) Now observe that h m, j m λ m, j/2 1 W m is congruent to 0 moduo a high ower of m Then, by (8), we have In other words, we have Δ(τ) μ m, j φ m, j (τ) f m, j (τ) Um j V = η(τ) m j F m, j mod m F m, j η(τ) μ m, j m j φ m, j (τ)= η(τ) (ν m, j 1)/m j φ m, j (τ)mod m The integer (ν m, j 1)/m j is in the range between 0 and Aso, it is congruent to 1/m j moduo Thus, we have ν m, j 1 r m if j is odd, = r m j m, j = 23 if j is even Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 From this, we get (m r m 2) λ m, j 12μ m, j = 2 m 13 if j is odd, if j is even This roves the roosition

11 Congruences ohe Partition Function 3271 Remark 22 Proosition 21 was stated as [9, Conjecture 1] The roof sketched here was suggested by one ohe referees and was adated from the roof of [2, Theorem 3] Aternativey, one can combine Proosition 31 with an induction ste roved in [9] to get the same concusion See the arxiv version arxiv: ohe resent aer for more detais 3 Main resuts In this section, we wi state our main resuts Before doing so, et us first reca a roerty about the subsace {η(τ) r f(τ): f M s (SL(2, Z))} of S s+r/2 (Γ 0 (576), χ 12 ), in which the function η(τ) r m, j φ m, j (τ) in Proosition 21 ies Proosition 31 ([10, Proosition 31]) Let r be an odd integer with 0 < r < Let s be a nonnegative even integer Then the sace S r,s :={η(τ) r f(τ): f(τ) M s (SL(2, Z))} (9) is an invariant subsace of S s+r/2 (Γ 0 (576), χ 12 ) under the action ohe Hecke agebra That is, for a rimes 2, 3anda f S r,s,wehave f T 2 S r,s Remark 32 This roerty of S r,s was first discovered by Garvan [10], and ater rediscovered by the author ohe resent aer (See the arxiv version arxiv: of the resent aer) Garvan stated the roosition under the assumtion that (r, 6) = 1 instead of 2 r, but it can be easiy checked that his roof aso works for the cases r = 3, 9, 15, and 21 as we The author s roof is more comicated, but can be aied in other simiar situations However, at the hindsight, the invariance of S r,s under the action of Hecke agebra is best exained (and roved) as foows The usua definition of moduar forms of haf-integra weights, as er Shimura [16], is given in terms ohe theta function θ(τ)= n Z qn2 Secificay, we say a hoomorhic function f : H C is a moduar form of haf-integra weight λ on Γ 0(4N) with character χ, where χ is a Dirichet character moduo 4N, if f(τ) is hoomorhic at each cus and satisfies f(γ τ) χ(d)θ(γτ)2λ+1 f(τ) = θ(τ) 2λ+1 Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014

12 3272 Y Yang for a γ = ( ab cd) Γ0 (4N) It is in this sense we say η(τ) is a moduar form of weight 1/2 onγ 0 (576) with character χ 12 Now the choice of θ in the definition of haf-integra moduar forms is erhas the most natura and simest from the view oint of Wei reresentations, but one drawback ohis choice is that the eves ohe moduar forms have to be a mutie of 4 On the other hand, if we define moduar forms of haf-integra weights in terms of η(τ), then the eves can be taken a the way down to 1 Exicity, et Γ be a congruence subgrou of SL(2, Z) for an odd integer r with 0 < r < and a nonnegative even integer s, wesayafunction f : H C is a moduar form of (η r, s)-tye on Γ if it is hoomorhic in H andateachcussuchthat f(γ τ) f(τ) = (cτ + d)s η(γ τ)r η(τ) r for a γ = ( ab cd) Γ LetSr,s (Γ ) be the sace of a such moduar forms on Γ Consider the case Γ = SL(2, Z) On the sace S r,s (SL(2, Z)), we can aso define Hecke oerators T 2 for rimes 2, 3 and show that their actions on f(τ) = af (n)q n/ S r,s (SL(2, Z)) is T 2 : f(τ) n=1 ( a f ( 2 n) + ( )( 12 ( 1) λ n ) λ 1 a f (n) + 2λ 1 a f ( n 2 ) ) q n/ with λ = (r + 2s 1)/2 Now observe that if g(τ) S r,s (SL(2, Z)), then g(τ + 1) = e 2πir/ g(τ), which imies that g(τ) = q r/ (c 0 + c 1 q + ), c i C Therefore, f(τ) = g(τ)/η(τ) r is a function hoomorhic on H and at each cus and satisfies f(γ τ) = (cτ + d) s f(τ) for a γ = ( ab cd) SL(2, Z) In other words, and S r,s (SL(2, Z)) ={η(τ) r f(τ) : f M s (SL(2, Z))} Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 S r,s ={g(τ): g S r,s (SL(2, Z))} This exains why S r,s is an invariant subsace of S r/2+s (Γ 0 (576), χ 12 ) Using the igeonhoe rincie, one can see that Proositions 21 and 31 yied Ono s eriodicity resut (3), with an imroved bound

13 Coroary 33 Congruences ohe Partition Function 3273 Let m 5 be a rime Then there exist integers N(m) and P (m) with 0 (N(m) 1)/2 m A(m) and 0 P (m) m A(m) such that ( m i ) ( m 2P (m)+i ) mod m for a nonnegative integers n and a i N(m), where m m A(m) = dim M (m rm 2)/2(SL(2, Z)) = 12 and r m is the integer satisfying 0 < r m < and m r m mod From Proosition 31, we can deduce the foowing coroary, which wi be roved in the next section Coroary 34 Let r be an odd integer satisfying 0 < r < and s be a nonnegative even integer Let S r,s be defined as (9)and{,, } be a Z-basis for the Z-modue Z[[q]] S r,s Given a rime 5, assume that A is the t t matrix such that Then we have T 2 = A U k = A 2 k + B k g 1 g t + C k V 2, where g j = f j ( ), and for nonnegative integers k, A k, B k,andc k are t t matrices satisfying ( ) ( ) k ( ) Ak A r+2s 2 I t It = I t 0 0 A k 1 (10) Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 with I t being the t t identity matrix, and ( ( 1) B k = s+(r 3)/2 (r 1)/2 ) 12 A k 1, C k = r+2s 2 A k 1

14 3274 Y Yang Remark 35 It is we known that for nonnegative even integer s, the sace M s (SL(2, Z)) has a basis consisting of g 1,,g d satisfying g i Z[[q]] and g i = q i 1 +, where d= dim M s (SL(2, Z)) (Usuay,g i are chosen to be roducts of Δ(τ) and Eisenstein series) Then it can be easiy verified that the functions f i (τ) = η(τ) r g i (τ) form a Z-basis of the Z-modue Z[[q]] S r,s In articuar, the rank ohe Z-modue Z[[q]] S r,s is the same as the dimension of S r,s Note aso that if r + 2s 3, then the Hecke oerator T 2 mas Z[[q]] S r,s into Z[[q]] S r,s Therefore, the matrix A in the above coroary has entries in Z This roerty is crucia in our subsequent discussion when we need to take A moduo a rime Now we can state our main resuts The first one is a more recise version of Theorem 11 The roof utiizes the coroary above and wi be given in the next section Theorem 36 m r m mod Let Let m 13 be a rime Set r m to be the integer satisfying 0 < r m < and m m t = 12 be the dimension of S rm,(m r m 2)/2 and assume that {,, } is a Z-basis for the Z- modue Z[[q]] S rm,(m r m 2)/2 Let be a rime different from 2, 3, and m, and assume that A is the t t matrix such that T 2 = A Assume that the order ohe square matrix ( ) A m 4 I t mod m (11) I t 0 Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 in PGL(2t, F m ) is K Thenwehave ( m 2uK 1 ) 0modm (12) for a ositive integers u and a ositive integers n not divisibe by

15 Congruences ohe Partition Function 3275 Aso, ihe order ohe matrix (11) ingl(2t, F m ) is M, thenwehave ( m i ) ( m 2M+i ) mod m (13) for a nonnegative integer i and a ositive integers n Remark 37 Note that ihe matrix A in the above theorem vanishes moduo m, then the matrix in (11) has order 2 in PGL(2t, F m ), and the concusion ohe theorem asserts that ( m j 3 ) 0modm This is the congruence aearing in Ono s theorem Remark 38 In genera, the integer K in Theorem 36 may not be the smaest ositive integer such that congruence (4) hods We choose to state the theorem in the current form because of its simicity See the remark foowing the roof of Theorem 36 4 Proof of Coroary 34 and Theorem 36 Proof of Coroary 34 where g j = f j ( ) and By the definition of T 2,wehave U 2 = A 1 + B 1 g 1 g t + C 1 V 2, ( ( 1) A 1 = A, B 1 = s+(r 3)/2 (r 1)/2 ) 12 I t, C 1 = r+2s 2 I t Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 Now we make the key observation g j U 2 = 0, f j V 2 U 2 = f j,

16 3276 Y Yang from which we obtain U 2 = (A C 1) + A 1 B 1 g 1 g t + A 1C 1 V 2 Iterating, we see that in genera if U k = A 2 k + B k then the coefficients satisfy the recursive reation g 1 g t + C k V 2, A k+1 = A k A 1 + C k, B k+1 = A k B 1, C k+1 = A k C 1 (Note that B 1 and C 1 are scaar matrices Thus, a coefficients are oynomias in A) Finay, we note that the reation A k+1 = A k A 1 + C k = A k A 1 + C 1 A k 1 can be written as which yieds ( Ak+1 This roves the coroary A k ( Ak+1 A k ) = ( A C1 I t 0 )( Ak A k 1 ) ( ) k ( ) ( ) k+1 ( ) A C1 A A C1 It = = I t 0 I t 0 0 I t ), Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 Proof of Theorem 36 Let m 13 be a rime Let r be the integer satisfying 0 < r < and m r mod and set s = (m r 2)/2 By Proosition 21, F m,1 is congruent to a moduar form in S r,s, where S r,s is defined by (9) Now et {,, } be a Z-basis for Z[[q]] S r,s and A be given as in the statement ohe theorem (Note that A has entries in

17 Z See Remark 35) Then by Coroary 34, we know that U k = A 2 k + B k Congruences ohe Partition Function 3277 g 1 g t + C k V 2, where g j = f j ( ),anda k, B k,andc k are t t matrices satisfying with ( Ak A k 1 for a k 1 Now we have ) ( ) = X k It, (14) 0 ( ( 1) B k = (m 5)/2 (r 1)/2 ) 12 A k 1, C k = m 4 A k 1 (15) X = ( ) A m 4 I t I t 0 X 1 = (m 4) ( 0 I t ) m 4 I t A Therefore, ihe order of X mod m in PGL(2t, F m ) is K, then we have, for a ositive integers u, ( AuK 1 A uk 2 ) ( ) ( ) = X uk 1 It 0 mod m 0 U for some t t matrix U, that is, A uk 1 0modm The rest of roof foows Ono s argument We have Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 U uk 1 2 B uk 1 g 1 g t + C uk 1 V 2 mod m

18 3278 Y Yang and U uk 1 2 U C uk 1 V mod m This imies that the 2uK 1 nth Fourier coefficients of f j vanishes moduo m for a j and a nnot divisibe by SinceF m,1 is a inear combination of f j moduo m, the same thing is true for the ( 2uK 1 n)th Fourier coefficients of F m,1 This transates to for a n not divisibe by This roves (12) ( m 2uK 1 ) 0modm Finay, ihe matrix X has order M in GL(2t, F m ), then from the recursive reations (14) and(15), it is obvious that (13) hods This cometes the roof Remark 41 In genera, the integer K in Theorem 36 may not be the smaest ositive integer such that congruence (4) hods For exame, consider the case where S r,s has dimension t 2 and the reduction of Z[[q]] S r,s moduo m has a basis consisting of Hecke eigenforms,, defined over F m Suose that the eigenvaues of T 2 ( ) for f i moduo m are a (1),,a (t) F m Letk i denote the order of a (i) m 4 in PGL(2, F m )Letk 1 0 be the east common mutie of k i Then we can show that f i U 2k 1 c i f i V mod m for some c i F m and consequenty congruence (4) hods Of course, the east common mutie of k i may be smaer than the integer K in Theorem 36 in genera 5 Exames Exame 51 Let m = 13 According to Proosition 21, we have Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 F 13,1 cη(τ) 11 mod 13 for some c F 13 (In fact, c = 11 See [13, age 303]) The eigenvaues a moduo 13 of T 2 for the first few rimes are

19 Congruences ohe Partition Function a For = 5, the matrix ( ) a 9 ( ) 10 8 X = mod 13 has eigenvaues 5 ± 7 over F 13 Now the order of (5 + 7)/(5 7) in F 169 is 14 This imies that 14 is the order of X in PGL(2, F 13 ) and we have ( u 1 ) 0 mod 13 for a ositive integers u and a ositive integers n not divisibe by 5 Likewise, we find that congruence (4) hods with k Exame 52 Let m = 37 By Proosition 2,1, we know that F 37,1 is congruent to a cus form in S 11,12 moduo 37 In fact, according to [9, Tabe 31], F 37,1 η(τ) 11 (E 4 (τ) Δ(τ))mod 37 The two eigenforms of S 11,12 are defined over a certain rea quadratic number fied, but the reduction of S 11,12 Z[[q]] moduo 37 has eigenforms defined over F 37 They are = η(τ) 11 (E 4 (τ) 3 + Δ(τ)), f 2 = η(τ) 11 Δ(τ) Let a (i) denote the eigenvaue of T 2 associated to f i Wehavethefoowingdata Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, a (1) a (2)

20 3280 Y Yang Let ( ) a (i) 33 X i = 1 0 For = 5, we find the orders of X 1 and X 2 in PGL(2, F 37 ) are 38 and 12, resectivey The east common mutie ohe orders is 228 Thus, we have ( u 1 ) 0 mod 37 for a ositive integers u and a ositive integers n not divisibe by 5 Note that this is an exame showing that the integer K in the statement of Theorem 36 is not otima (Here we have K = 456) For other sma rimes, we find that the congruence hods for a n not divisibe by with ( 37 2uk 1 ) 0 mod k Generaizations There are severa directions in which one may generaize Theorem 36 Here, we ony consider congruences ohe artition function moduo rime owers The case m = 5 wi be deat with searatey because in this case we have a very recise congruence resut In his roof of Ramanujan s conjecture for the cases m = 5 and 7, Watson [18, age 111] estabished a formua Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 i 1 F 5, j = i 1 c j,i η(120τ) 6i 1 η(τ) 6i c j,i η(120τ) 6i η(τ) 6i+1 if j is odd, if j is even,

21 Congruences ohe Partition Function 3281 where 3 j 1 5 j mod 5 j+1 if i = 1, c j,i 0mod5 j+1 if i 2 From the identity, one deduces that η(τ) 19 mod 5 j+1 F 5, j 3 j 1 5 j η(τ) 23 mod 5 j+1 if j is odd, if j is even Then Lovejoy and Ono [12] used this formua to study congruences ohe artition function moduo higher owers of 5 One distinct feature of [12] is the foowing emma Lemma 61 (Lovejoy and Ono [12, Theorem 22]) Let 5bearimeLetaand b be the eigenvaues of η(τ) 19 and η(τ) 23 for the Hecke oerator T 2, resectivey Then we have ( ) 15 a, b (1 + ) mod 5 With this emma, Lovejoy and Ono obtained congruences ohe form ( 5 j k ) 0mod5 j+1 for rimes congruent to 3 or 4 moduo 5 Here, we sha deduce new congruences using our method Theorem 62 Let 7 be a rime Set 5 if 1mod5, K = 4 if 2, 3mod5, 2 if 4mod5 (16) Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 Then we have ( 5 j 2uK 1 ) 0mod5 j+1 for a ositive integers j and u and a integers n not divisibe by

22 3282 Y Yang Proof In view of (16), We need to study when a Fourier coefficient of η(τ) 19 or η(τ) 23 vanishes moduo 5 Let f = η(τ) 19 Let 7bearimeanda be the eigenvaue of T 2 associated to f By Coroary 34 we have f U k 2 = a k f + b k f ( ) + c k f V 2, (17) where a 1 = a, b 1 = 8 ( 12/), c 1 = 17,anda k = a k 1 a 1 + c k 1, b k = a k 1 b 1, c k = a k 1 c 1 According to the roof of Theorem 36, ihe order of in PGL(F 5 ) is k, then ( ) a 17 mod 5 (18) 1 0 f U 2uk 1 f V mod 5 (19) for a ositive integers u Now by Lemma 61 the characteristic oynomia of (18) has a factorization ( x ( 15 )) ( x ( 15 ) ) moduo 5 From this we see that the order of (18) inpgl(f 5 ) is 5 if 1mod5, K = 4 if 2, 3mod5, 2 if 4mod5 Thus, (19) hodswithk = K This yieds the congruence ( 5 j 2uK 1 ) 0mod5 j+1 Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 for odd j, ositive integer u, and a ositive integers n not divisibe by The roof ohe case j even is exacty the same because mod 5 Remark 63 Watson [18] aso had an identity for F 7, j, with which one can study congruences moduo higher owers of 7 However, because there does not seem to exist an anaog of Lemma 61 in this case, we do not have a resut as recise as Theorem 62

23 Congruences ohe Partition Function 3283 The next congruence resut is an anaog of [19, Theorem 2], which in turn originates from the argument outined in [13, age 301] Theorem 64 Let 7 be a rime Assuming one ohe three situations beow occurs, we set k and m to be (1) k = 2andm = 5if 1mod 5, ( n/) = 1, (2) k = 2andm = 4if 2mod 5, ( n/) = 1, and Then (3) k = 1andm = 4if 3 mod 5, ( n/) = 1 ( 5 i 2(um +k ) ) 0mod5 i+1 for a nonnegative integers u and a ositive integers i Proof Assume first that i is odd Again, in view of (16), we need to study when the Fourier coefficients of f(τ) = η(τ) 19 vanish moduo 5 have Let 7 be a rime and a be the eigenvaue of T 2 where a k, b k,andc k satisfy ( ak a k 1 f U k 2 = a k f + b k f associated to By(17), we ( ) + c k f V 2, (20) ) ( ) k ( ) a 17 ( ) 1 12 =, b k a k 1, c k a k 1 mod From Lemma 61, we know that for 1 mod 5, we have a 1 2ɛ and thus the vaues of a k moduo 5 are a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 2ɛ 3 4ɛ 0 ɛ 2 3ɛ ɛ 3 Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 where ɛ = (15/) Now assume that f(τ) = c(n)q n Comaring the nth Fourier coefficients ohe two sides of (20) for integers n reativey rime to, weobtain ( ( n )) ( ( )) 12n c( 2k n) = a k + b k c(n) a k a k 1 c(n) mod 5

24 3284 Y Yang When k = 5u+ 2 for a nonnegative integer u, wehave ( 15 c( 2(5u+2) n) 3 ( 15 = 3 ) u ( 1 + ) u ( 1 + ( )( 15 12n ( n )) c(n) )) c(n) mod 5 (21) Thus, if ( n/) = 1, then c( 2(5u+2) n) 0 mod 5 This transates to the congruence ( 5 i 2(5u+2) ) 0mod5 i+1 This roves the first case ohe theorem The roof ohe other cases is simiar Remark 65 Note that the case 4 mod 5 is missing in Theorem 64 This is because in this case, by Lemma 61, the Hecke eigenvaues of T 2 both muties of 5 Then the numbers a k in (20) satisfy ( ak a k 1 ) ( ) k ( ) = From this, we see that a k ± a k 1 can never vanish moduo 5 for η(τ) 19 and η(τ) 23 are Exame 66 We now give some exames of congruences redicted in Theorem 64 (1) Let = 11, i = 1, and n= 67 Then the first situation occurs We find ( ) = (204364) = , Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 whichisamutieof25 (2) Let = 11, i = 1, and n= 19 The condition in the theorem is not fufied, but (21) imies that ( ) ( ) mod 25

25 Congruences ohe Partition Function 3285 Indeed, we have (4) = 5, (57954) = , and they are congruent to each other moduo 25 (3) Let = 7, i = 2, and n= 23 Then the second situation occurs We have Theorem 67 ( ) = (575) = , which is indeed a mutie of 5 3 Let m 13 be a rime and be a rime different from 2, 3, and m For each ositive integer i, there exists a ositive integer K such that for a u 1anda ositive integers n not divisibe by, the congruence ( m i 2uK 1 ) 0modm i hods There is aso another ositive integer M such that hods for a nonnegative integers n and r Proof ( m i r ) ( m i M+r ) mod m i Let β m,i be the integer satisfying 1 β m,i m i 1andβ m,i 1modm i Define (m i 1 + 1)(m 1) k m,i = 2 m i 1 (m 1) 12 m if i is odd, if i is even Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 By [2, Theorem 3], for a i 1, there is a moduar form f M km,i (SL(2, Z)) such that F m,i η(τ) (β m,i 1)/m i f(τ)mod m i The rest of roof is arae to that of Theorem 36

26 3286 Y Yang Exame 68 Consider the case m = 13 and i = 2 of Theorem 67 and assume that is a rime different from 2, 3, and 13 By [2, Theorem 3], F 13,2 is congruent to a moduar form in the sace S 23,144 of dimension 13 Choose a Z-basis f i = η(τ) 23 E 4 (τ) 3(13 i) Δ(τ) i 1, i = 1,,13, for Z[[q]] S 23,144 and et A be the matrix of T 2 with resect to this basis Ihe order of the matrix ( A 309 I 13 in PGL(26, Z/169) is K, then we have I 13 0 ) mod 169 ( 169 2K 1 ) 0 mod 169 for a integers n not divisibe by For instance, for = 5, we find A= Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014 moduo 169, and the order K is 28, 392, which yieds ( ,783 ) 0 mod 13 2 for a n not divisibe by 5

27 Congruences ohe Partition Function 3287 Acknowedgements The author woud ike to thank the referees for thorough reading ohe manuscrit and roviding many invauabe comments In articuar, the author is very gratefu to one ohe referees for giving a more accurate account ohe history ohe artition congruence robem and to another referee for bringing his attention to a very recent aer of Garvan [10] Aso, the roof of Proosition 21 resented here was suggested by the second referee This aer was dedicated to Prof B C Berndt on the occasion of his 70th birthday Funding This work was artiay suorted by the Nationa Science Counci of Taiwan, grant no M (to YY) This work was aso suorted by the Nationa Center for Theoretica Sciences References [1] Ahgren, S Distribution ohe artition function moduo comosite integers M Mathematische Annaen 318, no 4 (2000): [2] Ahgren, S and M Boyan Arithmetic roerties ohe artition function Inventiones Mathematicae 153, no 3 (2003): [3] Ahgren, S and K Ono Congruence roerties for the artition function Proceedings of the Nationa Academy of Sciences USA 98, no 23 (2001): (eectronic) [4] Atkin, A O L Proof of a conjecture of Ramanujan Gasgow Mathematica Journa 8 (1967): [5] Atkin, A O L Mutiicative congruence roerties and density robems for (n) Proceedings ohe London Mathematica Society (3) 18 (1968): [6] Atkin, A O L and J Lehner Hecke oerators on Γ 0 (m) Mathematische Annaen 185 (1970): [7] Atkin, A O L and J N O Brien Some roerties of (n) and c(n) moduo owers o3 Transactions ohe American Mathematica Society 126 (1967): [8] Berndt, B C and K Ono Ramanujan s unubished manuscrit on the artition and tau functions with roofs and commentary Seminaire Lotharingien de Combinatoire 42 (1999): Art B42c, 1 63 (eectronic) (The Andrews Festschrift, Maratea, 1998) [9] Chua, K S Exicit congruences for the artition function moduo every rime Archiv der Mathematik 81, no 1 (2003): [10] Garvan, F G Congruences for Andrews smaest arts artition function and new congruences for Dyson s rank Internationa Journa of Number Theory 6, no 2 (2010): [11] Guo, L and K Ono The artition function and the arithmetic of certain moduar L- functions Internationa Mathematics Research Notices no 21 (1999): [12] Lovejoy, J and K Ono Extension of Ramanujan s congruences for the artition function moduo owers of 5 Journa für die Reine und Angewandte Mathematik 542 (2002): Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014

28 3288 Y Yang [13] Ono, K Distribution ohe artition function moduo m Annas of Mathematics (2) 151, no 1 (2000): [14] Ramanujan, S Coected Paers of Srinivasa Ramanujan, edited by G H Hardy, P V Seshu Aiyar, and B M Wison Providence, RI: American Mathematica Society, Chesea Pubishing, 2000 (Third rinting ohe 1927 origina, With a new reface and commentary by Bruce C Berndt) [15] Serre, J-P Divisibiité de certaines fonctions arithmétiques Enseignement des Mathematiques (2), 22, no 3 4 (1976): [16] Shimura, G On moduar forms of haf integra weight Annas of Mathematics (2) 97 (1973): [17] Sturm, J On the congruence of moduar forms Number theory (New York, ), Lecture Notes in Mathematics 10 Berin: Sringer, 1987 [18] Watson, G N Ramanujans Vermutung über Zerfäungsanzahen Journa für die Reine und Angewandte Mathematik 179, no 2 (1938): [19] Weaver, R L New congruences for the artition function Ramanujan Journa 5, no 1 (2001): Downoaded from htt://imrnoxfordjournasorg/ at Nationa Chiao Tung University Library on Ari, 2014