YIFAN YANG
Dedicated to Professor B. C. Berndt on the occasion of his 70th birthday
ABSTRACT. Let p(n) denote the partition function. In this article, we will show that congruences of the form
p(mj`kn + B) ≡ 0 mod m for all n ≥ 0
exist for all primes m and ` satisfying m ≥ 13 and ` 6= 2, 3, m. Here the integer k de-pends on the Hecke eigenvalues of a certain invariant subspace of Sm/2−1(Γ0(576), χ12)
and can be explicitly computed.
More generally, we will show that for each integer i > 0 there exists an integer k such that for every non-negative integers j ≥ i with a properly chosen B the congruence
p(mj`kn + B) ≡ 0 mod mi holds for all integers n not divisible by `.
1. INTRODUCTION
Let p(n) denote the number of ways to write a positive integer n as sums of positive integers. For convenience, we also set p(0) = 1, p(n) = 0 for n < 0, and p(α) = 0 if α 6∈ Z. A remarkable discovery of Ramanujan [13] is that the partition function p(n) satisfies the congruences
(1) p(An + B) ≡ 0 mod m,
for all non-negative integers n for the triples
(A, B, m) = (5, 4, 5), (7, 5, 7), (11, 6, 11).
Ramanujan also conjectured that congruences (1) exist for the cases A = 5j, 7j, or 11j. This conjecture was proved by Watson [17] for the cases of powers of 5 and 7 and Atkin [3] for the cases of powers of 11. Since then, the problem of finding more examples of such congruences has attracted a great deal of attention. However, Ramanujan-type congruences appear to be very sparse. Prior to the late twentieth century, there are only a handful of such examples [4, 6]. In those examples, the integers A are no longer prime powers.
It turns out that if we require the integer A to be a prime, then the congruences proved or conjectured by Ramanujan are the only ones. This was proved recently in a remarkable paper of Ahlgren and Boylan [2]. On the other hand, if A is allowed to be a non-prime power, a surprising result of Ono [12] shows that for each prime m ≥ 5 and each positive integer k, a positive proportion of primes ` have the property
(2) p m k`3n + 1 24 ≡ 0 mod m Date: September 15, 2009.
2000 Mathematics Subject Classification. Primary 11P83; Secondary 11F25, 11F37, 11P82. This work is partially supported by the National Center for Theoretical Sciences in Taiwan.
for all non-negative integers n relatively prime to `. This result was later extended to com-posite m, (m, 6) = 1, by Ahlgren [1]. Neither of [12] and [1] addressed the algorithmic aspect of finding congruences of the form (2). For the cases m ∈ {13, 17, 19, 23, 29, 31} this was done by Weaver [18]. In effect, she found 76,065 new congruences. For primes m ≥ 37, this was addressed by Chua [8]. Although no explicit examples of congruences (2) for m ≥ 37 were given in [8], in principle, if one is patient enough, one will eventually find such congruences.
Another remarkable discovery of Ono [12, Theorem 5] is that the partition function possesses certain periodic property modulo a prime m. Specifically, he showed that for every prime m ≥ 5, there exist integers 0 ≤ N (m) ≤ m48(m3−2m+1)and 1 ≤ P (m) ≤ m48(m3−2m+1)such that (3) p m in + 1 24 ≡ p m P (m)+in + 1 24 mod m
for all non-negative integers n and all i ≥ N (m). In [8], Chua raised a conjecture (Con-jecture 2.1 in Section 3 below), which, if is true, will greatly improve Ono’s bound. (See Corollary 5 below.)
In this note, we will obtain new congruences for the partition function and discuss re-lated problems. In particular, we will show that there exist congruences of the form
p(mj`kn + B) ≡ 0 mod m for all primes m and ` such that m ≥ 13 and ` not equal to 2, 3, m.
Theorem 1. Let m and ` be primes such that m ≥ 13 and ` 6= 2, 3, m. Then there exists an explicitly computable positive integerk ≥ 2 such that
(4) p m
j`2k−1n + 1
24
≡ 0 mod m
for all non-negative integersn relatively prime to m and all positive integers j.
For instance, in Section 6 we will find that for m = 37 and arbitrary j, congruences (4) hold with
` 5 7 11 13 17 19 23 29 31 41 43 47 53 59 61
k 228 57 18 684 38 38 684 684 228 171 18 333 18 12 684 As far as we know, this is the first example in literature where a congruence (1) modulo a prime m ≥ 37 is explicitly given.
Theorem 1 is in fact a simplified version of one of the main results. (See Theorem 7). In the full version, we will see that the integer k in Theorem 1 can be determined quite explicitly in terms of the Hecke operators on a certain invariant subspace of the space Sm/2−1(Γ0(576), χ12) of cusp forms of level 576 and weight m/2 − 1 with character
χ12 = 12· . To describe this invariant subspace and to see how it comes into play with
congruences of the partition function, perhaps we should first review the work of Ono [12] and other subsequent papers [8, 18]. Thus, we will postpone giving the statements of our main results until Section 3.
Our method can be easily extended to obtain congruences of p(n) modulo a prime power. In Section 7, we will see that for each prime power miand a prime ` 6= 2, 3, m, there always exists a positive integer k such that
p m
i`2k−1n + 1
24
for all positive integers n not divisible by `. One example worked out in Section 7 is p 13 2· 556783n + 1 24 ≡ 0 mod 132.
In the same section, we will also discuss congruences of type p(5j`kn+B) ≡ 0 mod 5j+1.
Notations. Throughout the paper, we let Sλ(Γ0(N ), χ) denote the space of cusp forms
of weight λ and level N with character χ. By an invariant subspace of Sλ(Γ0(N ), χ) we
mean a subspace that is invariant under the action of the Hecke algebra on the space. For a power series f (q) = P af(n)qn and a positive integer N , we let UN and VN
denote the operators
UN : f (q) 7−→ f (q) UN := ∞ X n=0 af(N n)qn, VN : f (q) 7−→ f (q) VN := ∞ X n=0 af(n)qN n.
Moreover, if ψ is a Dirichlet character, then f ⊗ψ denotes the twist f ⊗ψ :=P af(n)ψ(n)qn.
Finally, for a prime m ≥ 5 and a positive integer j, we write Fm,j= X n≥0,mjn≡−1 mod 24 p m jn + 1 24 qn.
Note that we have
(5) Fm,j
Um= Fm,j+1.
2. WORKS OFONO[12], WEAVER[18],ANDCHUA[8] In this section, we will review the ideas in [12, 18, 8].
First of all, by a classical identity of Euler, we know that the generating function of p(n) has an infinite product representation
∞ X n=0 p(n)qn = ∞ Y n=1 1 1 − qn.
If we set q = e2πiτ, then we have q−1/24
∞
X
n=0
p(n)qn= η(τ )−1,
where η(τ ) is the Dedekind eta function. Now assume that m is a prime greater than 3. Ono [12] considered the function η(mkτ )mk/η(τ ). On the one hand, one has
η(mkτ )mk η(τ ) Umk = ∞ Y n=1 (1 − qn)mk· ∞ X n=0 p(n)qn+(m2k−1)/24 ! Umk.
On the other hand, one has η(mkτ )mk
η(τ ) ≡ η(τ )
m2k−1
where ∆(τ ) = η(τ )24is the normalized cusp from of weight 12 on SL(2, Z). From these, Ono [12, Theorem 6] deduced that
Fm,k≡ (∆(τ )(m2k−1)/24 Umk) V24 η(24τ )mk mod m.
Now it can be verified that for k = 1, the right-hand side of the above congruence is contained in the space S(m2−m−1)/2(Γ0(576m), χ12) of cusp forms of level 576m and
weight (m2− m − 1)/2 with character χ12 = 12·. Then by (5) and the fact that Um
defines a linear map
Um: Sλ+1/2(Γ0(4N m), ψ) → Sλ+1/2(Γ0(4N m), ψχm), χm=
m ·
, one sees that
Fm,k≡ Gm,k=
X
am,k(n)qn mod m
for some Gm,k∈ S(m2−m−1)/2(Γ0(576m), χ12χk−1m ).
Now recall the general Hecke theory for half-integral weight modular forms states that if f (τ ) = P∞
n=1af(n)qn ∈ Sλ+1/2(Γ0(4N ), ψ) and ` is a prime not dividing 4N , then
the Hecke operator defined by T`2 : f (τ ) 7→ ∞ X n=1 af(`2n) + ψ(`) (−1)λn ` `λ−1af(n) + ψ(`2)`2λ−1af(n/`2) qn sends f (τ ) to a cusp form in the same space. In the situation under consideration, if ` is a prime not dividing 576m such that
Gm,k T`2 ≡ 0 mod m, then we have 0 ≡ (Gm,k T`2) U` mod m = ∞ X n=1 am,k(`3n) + ψ(`2)`m 2−m−3 am,k(n/`) qn
since `n` = 0. In particular, if n is not divisible by `, then am,k(`3n) ≡ 0 mod m, which implies p m k`3n + 1 24 ≡ 0 mod m.
Finally, to show that there is a positive proportion of primes ` such that Gm,k
T`2 ≡ 0
mod m, Ono invoked the Shimura correspondence between half-integral weight modular forms and integral weight modular forms [15] and a result of Serre [14, 6.4].
As mentioned earlier, Ono [12] did not address the issue of finding explicit congruences of the form (2). However, Section 4 of [12] did give us some hints on how one might proceed to discover new congruences, at least for small primes m. The key observation is the following.
The modular form Gm,kitself is in a vector space of big dimension, so to determine
whether Gm,k
T`2 vanishes modulo m, one needs to compute the Fourier coefficients of
half-integral weight modular form of a much smaller weight. For example, using Sturm’s theorem [16] Ono verified that
F13,2k+1≡ G13,2k+1≡ 11 · 6kη(24τ )11 mod 13,
F13,2k+2≡ G13,2k+2≡ 10 · 6kη(24τ )23 mod 13
(6)
for all non-negative integers k. The modular form η(24τ )11is in fact a Hecke eigenform.
(The modular form η(24τ )23is also a Hecke eigenform as we shall see in Section 3.) More
generally, for m ∈ {13, 17, 19, 23, 29, 31}, it is shown in [12, Section 4], [9, Proposition 6] and [18, Proposition 5] that Gm,1is congruent to a Hecke eigenform of weight m/2 − 1.
Using this observation, Weaver [18] then devised an algorithm to find explicit congruences of the form (2) for m ∈ {13, 17, 19, 23, 29, 31}.
The proof of congruences (6) given in [9] and [18] is essentially “verification” in the sense that they all used Sturm’s criterion [16]. That is, by Sturm’s theorem to show that two modular forms on a congruence subgroup Γ are congruent to each other modulo a prime m, it suffices to compare sufficiently many coefficients, depending on the weight and index (SL(2, Z) : Γ). Naturally, this kind of argument will not be very useful in proving general results. In [8], Chua found a more direct way to prove congruences (6) for Fm,1. In particular, he [8, Theorem 1.1] was able to show that for each prime m ≥ 5, Fm,1
is congruent to a modular form of weight m/2 − 1 modulo m. Instead of the congruence
η(mτ )m η(τ ) ≡ η(τ )
m2−1 mod m
used by Ono, Chua considered the congruence η(mτ )m
η(τ ) ≡ η(mτ )
m−1η(τ )m−1 mod m
as the starting point. The function on the right is a modular form of weight m − 1 on Γ0(m). Thus, by the level reduction lemma of Atkin and Lehner [5, Lemma 7], one has
η(mτ )m−1η(τ )m−1(Um+ m(m−1)/2−1Wm) ∈ Sm−1(SL(2, Z)),
where Wmdenotes the Atkin-Lehner involution. It follows that
Fm,1= 1 η(24τ ) Um≡ fm(24τ ) η(24τ )m mod m
for some cusp form fm(τ ) ∈ Sm−1(SL(2, Z)). (Incidently, this also proves Ramanujan’s
congruences for m = 5, 7, 11, since there are no non-trivial cusp forms of weight 4, 6, 10.) By examining the order of vanishing of fm(τ ) at ∞, Chua [8, Theorem 1.1] then concluded
that if we let rm denote the integer in the range 0 < rm < 24 such that m ≡ −rm
mod 24, then
Fm,1≡ η(24τ )rmφm(24τ )
for some modular form φmon SL(2, Z) of weight (m − rm− 2)/2. Furthermore, based
on an extensive numerical computation, Chua made the following conjecture.
Conjecture 2.1 (Chua [8, Conjecture 1]). Let m ≥ 13 be a prime and rmbe the integer
in the range0 < rm< 24 such that m ≡ −rm mod 24. Set
rm,j=
(
rm, ifj is odd,
Then
Fm,j ≡ η(24τ )rm,jφm,j(24τ ) mod m
for some modular formφm,j(τ ) on SL(2, Z), where the weight of φm,jis(m − rm− 2)/2
ifj is odd and is m − 13 if j is even.
In [8, Section 4], Chua established the induction step for the case of even j assuming the conjecture holds for odd j − 1. However, as remarked by Chua, it appears difficult to prove the induction step from cases of even j − 1 to cases of odd j. In the next section, we will see that this conjecture is a simple consequence of our Theorem 2.
Remark 2.2. Professor H. H. Chan has kindly informed us that Serre has indicated to him an argument to establish Conjecture 2.1. The argument will be given in a forthcoming article [7].
3. STATEMENTS OF MAIN RESULTS
The functions η(24τ )rm,kφ
m,k(24τ ) appearing in Chua’s conjecture (Conjecture 2.1)
are all half-integral weight modular forms of level 576 and character χ12. Thus, our first
main result is concerned with the space Sλ+1/2(Γ0(576), χ12).
Theorem 2. Let r be an odd integer with 0 < r < 24. Let s be a non-negative even integer. Then the space
(7) Sr,s := {η(24τ )rf (24τ ) : f (τ ) ∈ Ms(SL(2, Z))}
is an invariant subspace ofSs+r/2(Γ0(576), χ12) under the action of the Hecke algebra.
That is, for all primes` 6= 2, 3 and all f ∈ Sr,s, we havef
T`2∈ Sr,s.
The following corollary is immediate.
Corollary 3. Let r be an odd integer with 0 < r < 24. Let E4(τ ) and E6(τ ) be the
Eisenstein series of weights4 and 6 on SL(2, Z) and f (τ ) be one of the function 1, E4(τ ),
E6(τ ), E4(τ )2,E4(τ )E6(τ ), and E4(τ )2E6(τ ). Then the function η(24τ )rf (24τ ) is a
Hecke eigenform. In particular, form ∈ {13, 17, 19, 23, 29, 31}, the function η(24τ )rmφ
m,1(24τ )
in Conjecture 2.1 is a Hecke eigenform.
Note that the assertion about gm:= η(24τ )rmφm,1was already proved in Proposition 6
of [9]. In the same proposition, it was also proved that the image of gmunder the Shimura
correspondence is Gm⊗χ12, where Gmis the unique normalized newform of weight m−3
on Γ0(6) whose eigenvalues for the Atkin-Lehner involutions W2and W3are − m2 and
− 3
m, respectively.
We now apply Theorem 2 to study congruences of the partition function. We first con-sider Conjecture 2.1. Observe that the Hecke operator Tm2 is the same as the operator
Um2 modulo m. Also, the case j = 1 and the induction step from j = 1 to j = 2 have
already been proved in [8]. Thus, from Theorem 2 we immediately conclude that Chua’s conjecture indeed holds in general.
Corollary 4 (Conjecture of Chua). Let m ≥ 13 be a prime and rmbe the integer in the
range0 < rm< 24 such that m ≡ −rm mod 24. Set
rm,j=
(
rm, ifj is odd,
Then
Fm,j ≡ η(24τ )rm,jφm,j(24τ ) mod m
for some modular formφm,j(τ ) on SL(2, Z), where the weight of φm,jis(m − rm− 2)/2
ifj is odd and is m − 13 if j is even. Remark 3.1. Note that for odd j, we have
(8) dim Srm,(m−rm−2)/2= jm 12 k −jm 24 k .
To see this, we observe that dim Mλ(SL(2, Z)) − bλ/12c is periodic of period 12. Thus,
to show (8), we only need to verify case by case according the residue of m modulo 24. Using the pigeonhole principle, one can see that Theorem 2 also yields Ono’s periodicity result (3), with an improved bound.
Corollary 5. Let m ≥ 5 be a prime. Then there exist integers 0 ≤ N (m) ≤ mA(m)and 0 ≤ P (m) ≤ mA(m)such that
p m in + 1 24 ≡ p m P (m)+in + 1 24 mod m for all non-negative integersn, where
(9) A(m) = 2 dim M(m−rm−2)/2(SL(2, Z))
andrmis the integer satisfying0 < rm< 24 and m ≡ −rm mod 24.
Corollary 6. Let r be an odd integer satisfying 0 < r < 24 and s be a non-negative even integer. LetSr,s be defined as(7) and {f1, . . . , ft} be a Z-basis for the Z-module
Z[[q]] ∩ Sr,s. Given a prime` ≥ 5, assume that A is the t × t matrix such that
f1 .. . ft T`2 = A f1 .. . ft . Then we have f1 .. . ft U`k2 = Ak f1 .. . ft + Bk g1 .. . gt + Ck f1 .. . ft V`2,
wheregj= fj⊗ ·`, and Ak,Bk, andCkaret × t matrices satisfying
Ak Ak−1 =A −` r+2s−2I t It 0 kI t 0 , and Bk= −`s+(r−3)/2 (−1)(r−1)/212 ` Ak−1, Ck= −`r+2s−2Ak−1.
Theorem 7. Let m ≥ 13 be a prime and j be a positive integer. Set rmto be the integer
satisfying0 < rm< 24 and m ≡ −rm mod 24. Let
t =jm 12 k −jm 24 k
be the dimension ofSrm,(m−rm−2)/2and assume that{f1, . . . , ft} is a basis for the
Z-module Z[[q]] ∩ Srm,(m−rm−2)/2. Let` be a prime different from 2, 3, and m, and assume
thatA is the t × t matrix such that f1 .. . ft T`2 = A f1 .. . ft . Assume that the order of the square matrix
(10) A −` m−4I t It 0 mod m inPGL(2t, Fm) is K, then we have (11) p m j`2uK−1+ 1 24 ≡ 0 mod m
for all positive integersj and u and all positive integers n not divisible by `. Also, if the order of the matrix(10) in GL(2t, Fm) is M , then we have
(12) p m j`in + 1 24 ≡ p m j`2M +in + 1 24 mod m for all non-negative integeri and all positive integers j and n.
Remark 3.2. Note that if the matrix A in the above theorem vanishing modulo m, then the matrix in (10) has order 2 in PGL(2t, Fm), and the conclusion of the theorem asserts
that p m j`3n + 1 24 ≡ 0 mod m. This is the congruence appearing in Ono’s theorem.
Remark 3.3. In general, the integer K in Theorem 7 may not be the smallest positive integer such that congruence (4) holds. We choose to state the theorem in the current form because of its simplicity. See the remark following the proof of Theorem 7.
4. PROOF OFTHEOREM2 We first recall the following lemma of Atkin and Lehner.
Lemma 4.1 ([5, Lemma 7]). Let f be a cusp form of weight s on Γ0(N ) and ` be a prime.
Then
(a) If `|N , then fU` is a cusp form onΓ0(N ). Furthermore, if `2|N , then f
U`is
modular onΓ0(N/`).
(b) If `|N but `2- N , then f(U`+ `s/2−1W`) is a cusp form on Γ0(N/`).
The following transformation formula for η(τ ) is frequently used. Lemma 4.2 ([19, pp.125–127]). For
γ =a b c d
∈ SL2(Z),
the transformation formula forη(τ ) is given by, for c = 0, η(τ + b) = eπib/12η(τ ),
and, forc > 0, η(γτ ) = (a, b, c, d) r cτ + d i η(τ ) with (a, b, c, d) = d c
i(1−c)/2eπi(bd(1−c2)+c(a+d))/12, ifc is odd, c
d
eπi(ac(1−d2)+d(b−c+3))/12, ifd is odd, where d
c
is the Legendre-Jacobi symbol.
We now prove Theorem 2. Assume that g(τ ) ∈ Sr,s, say g(τ ) = η(24τ )rf (24τ ) for
some f (τ ) ∈ Ms(SL(2, Z)). Assume g(τ ) = qrP∞n=0a(n)q24n. Then by the definition
of T`2, we have gT`2 = X n≥0,n≡−r/24 mod `2 a(n)q(24n+r)/`2 + `s+(r−3)/2 (−1) (r−1)/212 ` ∞ X n=0 24n + r ` a(n)q24n+r + `r+2s−2 ∞ X n=0 a(n)q`2(24n+r). (13)
We now consider the function
h(τ ) = η(`2τ )24−rg(τ /24) = η(`2τ )24−rη(τ )rf (τ ).
Using Newman’s criterion [11, Theorem 1], we see that h(τ ) is a cusp form of weight s + 12 on Γ0(`2). Then by Lemma 4.1, h
(U`2+ `s/2+5U`W`) is a cusp form on SL(2, Z),
that is,
(14) h(U`2+ `s/2+5U`W`) = η(τ )24h(τ )˜
for some modular form ˜h(τ ) of weight s on SL(2, Z). We claim that (15) h(U`2+ `s/2+5U`W`) = η(τ )24−r(gT`2)(τ /24).
Once this is proved, by comparing (14) and (15), we immediately get Theorem 2. We now verify (15).
By the definition of U`2we have
hU`2 = ∞ Y n=1 (1 − q`2n)24−r ∞ X n=0 a(n)q`2−r(`2−1)/24+n ! U`2 = q ∞ Y n=1 (1 − qn)24−r X n≥0,n≡−r/24 mod `2 a(n)q(24n−r(`2−1))/24`2 = η(τ )24−r X n≥0,n≡−r/24 mod `2 a(n)q(24n+r)/24`2. (16)
The term involving U`W`is more complicated. We have
hU`W`= 1 ` `−1 X k=0 h 1 k 0 ` ! W`= `−s/2−7τ−s−12 `−1 X k=0 h 1 k 0 ` 0 −1 ` 0 .
The term k = 0 gives us
`−s/2−7τ−s−12h(−1/`2τ ) = `−s/2−7τ−s−12η(−1/τ )24−rη(−1/`2τ )rf (−1/`2τ ). Using the formula η(−1/τ ) =p(τ /i)η(τ ) and the assumption that f(τ ) ∈ Ms(SL(2, Z)),
this reduces to (17) `3s/2+r−7η(τ )24−rη(`2τ )rf (`2τ ) = `3s/2+r−7η(τ )24−r ∞ X n=0 a(n)q`2(24n+r)/24. We now consider the contribution from the cases k 6= 0.
We have η(`2τ ) k` −1 `2 0 = η k`τ − 1 τ . By Lemma 4.2, this is equal to
(18) η(`2τ ) k` −1 `2 0 = e2πik`/24r τ iη(τ ). For η(τ ) and f (τ ), we observe that
k`τ − 1 `2τ = k u ` k0 (τ − k0/`),
where k0denotes the multiplicative inverse of k modulo ` and u = (kk0− 1)/`. Thus, by Lemma 4.2, (19) η(τ ) k` −1 `2 0 = k 0 ` i(1−`)/2e2πi`(k+k0)/24 r `τ i η τ −k 0 ` . Also, (20) f (τ ) k` −1 `2 0 = (`τ )sf τ −k 0 ` . Combining (18), (19), and (20), we obtain
`−s/2−7τ−s−12h k` −1 `2 0 = `(s+r)/2−7 k 0 ` ir(1−`)/2e2πir`k0/24η(τ )24−rη τ −k 0 ` r f τ −k 0 ` , and `−s/2−7τ−s−12 `−1 X k=1 h k` −1 `2 0 = `(s+r)/2−7ir(1−`)/2η(τ )24−r `−1 X k=1 k ` e2πir`k/24g τ − k/` 24 . (21)
The sum in the last expression is equal to
`−1 X k=1 k ` e2πirk(`2−1)/24` ∞ X n=0 e−2πikn/`a(n)qn+r/24. (22)
With the well-known evaluation
`−1 X k=1 k ` e2πikn/`=n ` i(`−1)2/4√`
of the Gaussian sum, (22) can be simplified to i(`−1)2/4 √ ` ∞ X n=0 r(`2− 1)/24 − n ` a(n)qn+r/24 = i(`−1)2/4√` −24 ` ∞ X n=0 24n + r ` a(n)qn+r/24. Substituting this into (21) and using
−1 ` = (−1)(`−1)/2, 2 ` = (−1)(`2−1)/8, we arrive at `−s/2−7τ−s−12 `−1 X k=1 h k` −1 `2 0 = `(s+r+1)/2−7ir(1−`)/2+(`−1)2/4 −24 ` η(τ )24−r ∞ X n=0 24n + r ` a(n)qn+r/24 = `(s+r+1)/2−7 −1 ` (r−1)/2 12 ` η(τ )24−r ∞ X n=0 24n + r ` a(n)qn+r/24. (23)
Together with (17), (23) implies that `s/2+5hU`W`= `2s+r−2η(τ )24−r ∞ X n=0 a(n)q`2(24n+r)/24 + `s+(r−3)/2η(τ )24−r (−1) (r−1)/212 ` ∞ X n=0 24n + r ` a(n)qn+r/24. (24)
Comparing (16) and (24) with (13), we see that (15) indeed holds. The proof of Theorem 2 is now complete.
5. PROOF OFCOROLLARY6ANDTHEOREM7 Proof of Corollary 6. By the definition of T`2, we have
f1 .. . ft U`2 = A1 f1 .. . ft + B1 g1 .. . gt + C1 f1 .. . ft V`2, where gt= ft⊗ `· and A1= A, B1= −`s+(r−3)/2 (−1)(r−1)/212 ` It, C1= −`r+2s−2It.
Now we make the key observation gj U`2 = 0, fj V`2 U`2 = fj,
from which we obtain f1 .. . ft U`22 = (A 2 1+ C1) f1 .. . ft + A1B1 g1 .. . gt + A1C1 f1 .. . ft V`2.
Iterating, we see that in general if f1 .. . ft U`k2 = Ak f1 .. . ft + Bk g1 .. . gt + Ck f1 .. . ft V`2,
then the coefficients satisfy the recursive relation
Ak+1= AkA1+ Ck, Bk+1= AkB1, Ck+1= AkC1.
(Note that B1 and C1 are scalar matrices. Thus, all coefficients are polynomials in A.)
Finally, we note that the relation Ak+1= AkA1+ Ck = AkA1+ C1Ak−1can be written
as Ak+1 Ak =A C1 It 0 Ak Ak−1 . which yields Ak+1 Ak =A C1 It 0 kA It =A C1 It 0 k+1I t 0 .
This proves the theorem.
Proof of Theorem 7. Let m ≥ 13 be a prime. Let r be the integer satisfying 0 < r < 24 and m ≡ −r mod 24 and set s = (m − r − 2)/2. By Corollary 4, Fm,1 congruent to
a modular form in Sr,s, where Sr,sis defined by (7). Now let {f1, . . . , ft} be a basis for
Sr,sand A be given as in the statement of the theorem. Then by Corollary 6, we know that
f1 .. . ft U`k2 = Ak f1 .. . ft + Bk g1 .. . gt + Ck f1 .. . ft V`2,
where gj = fj⊗ `·, and Ak, Bk, and Ckare t × t matrices satisfying
(25) Ak Ak−1 = XkIt 0 , (26) Bk= −`(m−5)/2 (−1)(r−1)/212 ` Ak−1, Ck= −`m−4Ak−1 with X =A −` m−4I t It 0
for all k ≥ 1. Now we have
X−1= `−(m−4)
0 `m−4It
−It A
Therefore, if the order of X mod m in PGL(2t, Fm) is K, then we have
AuK−1 AuK−2 = XuK−1It 0 ≡ 0 U mod m.
for some t × t matrix U , that is, AuK−1 ≡ 0 mod m. The rest of proof follows Ono’s
We have f1 .. . ft U`uK−12 ≡ BuK−1 g1 .. . gt + CuK−1 f1 .. . ft V`2 mod m and f1 .. . ft U`uK−12 U`≡ CuK−1 f1 .. . ft V` mod m.
This implies that the `2uK−1nth Fourier coefficients of f
jvanishes modulo m for all j and
all n not divisible by `. Since Fm,1is a linear combination of fjmodulo m, the same thing
is true for the `2uK−1nth Fourier coefficients of Fm,1. This translates to
p m`
2uK−1n + 1
24
≡ 0 mod m for all n not divisible by `. This proves (11) for the case j = 1.
For the case j > 1, we note that the operators U`and Umcommutes. Thus,
f1 .. . ft UmjU`k2 = Ak f1 .. . ft Umj + Bk g1 .. . gt Umj + Ck f1 .. . ft V`2 Umj,
where Ak, Bk, and Ck satisfy the same relations (25) and (26). Taking the fact (5) into
account, we see that the same argument in the case j = 1 gives us the general congruence. Finally, if the matrix X has order M in GL(2t, Fm), then from the recursive relations
(25) and (26), it is obvious that (12) holds. This completes the proof. Remark 5.1. In general, the integer K in Theorem 7 may not be the smallest positive integer such that congruence (4) hold. To see this, for simplicity, we assume that Sr,s
has dimension t and its reduction modulo m has a basis consisting of Hecke eigenforms f1, . . . , ftdefined over Fm. If the eigenvalues of T`2for fimodulo m are a(1)` , . . . , a(t)` ∈
Fm. Let kidenote the order of
a(i)` −`m−4
1 0
in PGL(2, Fm). Let k be the least common
multiple of ki. Then we can show that
fi
U`2k−1≡ cifi
V` mod m
for some ci ∈ Fmand consequently congruence (4) holds. Of course, the least common
multiple of kimay be smaller than the integer K in Theorem 7 in general.
6. EXAMPLES
Example 6.1. Let m = 13. According to Corollary 4, we have F13,1≡ cη(24τ )11 mod 13
for some c ∈ F13. (In fact, c = 11. See [12, page 303].) The eigenvalues a`modulo 13 of
T`2for the first few primes ` are
` 5 7 11 17 19 23 29 31 37 41 43 47 53 59 61 67 73
a` 10 8 5 1 8 8 4 4 5 9 12 6 10 0 2 4 0
For ` = 5, the matrix X =a` −` 9 1 0 ≡10 8 1 0 mod 13 has eigenvalues 5 ±√7 over F13. Now the order of (5 +
√
7)/(5 −√7) in F169is 14. This
implies that 14 is the order of X in PGL(2, F13) and we have
p 13 · 5
28u−1n + 1
24
≡ 0 mod 13
for all positive integers u and all positive integers n not divisible by 5. Likewise, we find congruence (4) holds with
` 5 7 11 17 19 23 29 31 37 41 43 47 53 59 61 67 73
k 14 14 14 7 14 3 6 12 14 12 7 12 7 2 13 12 2
Example 6.2. Let m = 37. By Corollary 4, we know that F37,1 is congruent to a cusp
form in S11,12modulo 37. In fact, according to [8, Table 3.1],
F37,1≡ η(24τ )11(E4(24τ )3+ 17∆(24τ )) mod 37.
The two eigenforms of S11,12 are defined over a certain real quadratic number field, but
the reduction of S11,12∩ Z[[q]] modulo 37 has eigenforms defined over F37. They are
f1= η(24τ )11(E4(24τ )3+ 24∆(24τ )), f2= η(24τ )11∆(24τ ).
Let a(i)` denote the eigenvalue of T`2associated to fi. We have the following data.
` 5 7 11 13 17 19 23 29 31 41 43 47 53 59 61 a(1)` 1 33 22 7 11 0 1 9 35 11 28 14 30 24 12 a(2)` 32 10 0 6 7 8 31 36 9 10 1 35 9 3 16 `33 8 26 36 8 23 8 6 31 31 11 6 1 10 23 29 Let Xi= a(i)` −`33 1 0 .
For ` = 5, we find the orders of X1and X2in PGL(2, F37) are 38 and 12, respectively.
The least common multiple of the orders is 228. Thus, we have p 37 · 5
456u−1n + 1
24
≡ 0 mod 37
for all positive integers u and all positive integers n not divisible by 5. Note that this is an example showing that the integer K in the statement of Theorem 7 is not optimal. (Here we have K = 456.)
For other small primes `, we find congruence p 37`
2uk−1n + 1
24
≡ 0 mod 37 holds for all n not divisible by ` with
` 5 7 11 13 17 19 23 29 31 41 43 47 53 59 61
7. GENERALIZATIONS
There are several directions one may generalize Theorem 7. Here we only consider congruences of the partition function modulo prime powers. The case m = 5 will be dealt with separately because in this case we have a very precise congruence result.
In his proof of Ramanujan’s conjecture for the cases m = 5, 7, Watson [17, page 111] established a formula F5,j = X i≥1 cj,i η(120τ )6i−1 η(24τ )6i , if j is odd, X i≥1 cj,i η(120τ )6i η(24τ )6i+1, if j is even, where cj,i≡ ( 3j−15j mod 5j+1, if i = 1, 0 mod 5j+1, if i ≥ 2.
From the identity, one deduces that (27) F5,j ≡ 3j−15j
(
η(24τ )19 mod 5j+1, if j is odd,
η(24τ )23 mod 5j+1, if j is even.
Then Lovejoy and Ono [10] used this formula to study congruences of the partition func-tion modulo higher powers of 5. One distinct feature of [10] is the following lemma. Lemma 7.1 (Lovejoy and Ono [10, Theorem 2.2]). Let ` ≥ 5 be a prime. Let a and b be the eigenvalues ofη(24τ )19andη(24τ )23for the Hecke operatorT`2, respectively. Then
we have
a, b ≡ 15 `
(1 + `) mod 5.
With this lemma, Lovejoy and Ono obtained congruences of the form p 5
j`kn + 1
24
≡ 0 mod 5j+1
for primes ` congruent to 3 or 4 modulo 5. Here we shall deduce new congruences using our method.
Theorem 8. Let ` ≥ 7 be a prime. Set
K`= 5, if` ≡ 1 mod 5, 4, if` ≡ 2, 3 mod 5, 2, if` ≡ 4 mod 5. Then we have p 5 j`2uK`−1n + 1 24 ≡ 0 mod 5j+1 for all positive integersj and u and all integers n not divisible by `.
Proof. In view of (27), We need to study when a Fourier coefficient of η(24τ )19or η(24τ )23
vanishes modulo 5.
Let f = η(24τ )19. Let ` ≥ 7 be a prime and a be the eigenvalue of T
`2associated to f . By Corollary 6 we have (28) fU`k2= akf + bkf ⊗ · ` + ckf V`2,
where a1 = a, b1 = −`8 −12` , c1 = −`17, and ak = ak−1a1+ ck−1, bk = ak−1b1,
ck = ak−1c1. According to the proof of Theorem 7, if the order of
(29) a −` 17 1 0 mod 5 in PGL(F5) is k, then (30) fU`2uk−1 ≡ f V` mod 5
for all positive integers u. Now by Lemma 7.1 the characteristic polynomial of (29) has a factorization x − 15 ` x − 15 ` `
modulo 5. From this we see that the order of (29) in PGL(F5) is
K`= 5, if ` ≡ 1 mod 5, 4, if ` ≡ 2, 3 mod 5, 2, if ` ≡ 4 mod 5. Thus, (30) holds with k = K`. This yields the congruence
p 5
j`2uK`−1n + 1
24
≡ 0 mod 5j+1 for odd j, positive integer u, and all positive integers n not divisible by `.
The proof of the case j even is similar to the above and is omitted. Remark 7.2. In [17], Watson also had an identity for F7,j, with which one can study
congruences modulo higher powers of 7. However, because there does not seem to exist an analog of Lemma 7.1 in this case, we do not have a result as precise as Theorem 8
The next congruence result is an analog of Theorem 2 of [18], which in turn is originated from the argument outlined in [12, page 301].
Theorem 9. Let ` ≥ 7 be a prime. Assume that one of the following situations occurs. (1) ` ≡ 1 mod 5, −n` = −1 with k`= 2 and m`= 5, or
(2) ` ≡ 2 mod 5, −n` = (−1)i−1withk`= 2 and m`= 4, or
(3) ` ≡ 3 mod 5, −n` = (−1)i−1withk
`= 1 and m`= 4. Then p 5 i`2(um`+k`)n + 1 24 ≡ 0 mod 5i+1 for all non-negative integersu.
Proof. Assume first that i is odd. Again, in view of (27), we need to study when the Fourier coefficients of f (τ ) = η(24τ )19vanish modulo 5.
Let ` ≥ 7 be a prime and a be the eigenvalue of T`2associated to `. By (28), we have
(31) fU`k2= akf + bkf ⊗ · ` + ckf V`2, where ak, bk, cksatisfy ak ak−1 =a −` 17 1 0 k1 0 , bk≡ − −12 ` ak−1, ck ≡ −`ak−1 mod 5.
From Lemma 7.1, we know that for ` ≡ 1 mod 5, we have a1 ≡ 2 and thus the values
of akmodulo 5 are
a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 . . .
2 3 4 0 2 3 4 0 1 2 3 . . .
where = 15`. Now assume that f (τ ) = P c(n)qn. Comparing the nth Fourier coeffi-cients of the two sides of (31) for integers n relatively prime to `, we obtain
c(`2kn) =ak+ bk n ` c(n) ≡ ak− ak−1 −12n ` c(n) mod 5. When k = 5u + 2 for a non-negative integer u, we have
c(`2(5u+2)n) ≡ 3 15 ` u 1 + 15 ` −12n ` c(n) = 3 15 ` u 1 + −n ` c(n) mod 5. (32)
Thus, if −n` = −1, then c(`2(5u+2)n) ≡ 0 mod 5. This translates to the congruence p 5
i`2(5u+2)n + 1
24
≡ 0 mod 5i+1.
This proves the first case of the theorem. The proof of the other cases is similar. Example 7.3. (1) Let ` = 11, i = 1, and n = 67. Then the first situation occurs. We
find p 5 · 11 4· 67 + 1 24 = p(204364) = 28469 . . . 24450, which is a multiple of 25.
(2) Let ` = 11, i = 1, and n = 19. The condition in the theorem is not fulfilled, but (32) implies that p 5 · 11 4· 19 + 1 24 ≡ p 5 · 19 + 1 24 mod 25. Indeed, we have p(4) = 5, p(57954) = 37834 . . . 45055, and they are congruent to each other modulo 25.
(3) Let ` = 7, i = 2, and n = 23. Then the second situation occurs. We have p 5
2· 74· 23 + 1
24
= p(57524) = 38402 . . . 43875, which is indeed a multiple of 53.
Theorem 10. Let m ≥ 13 be a prime and ` be a prime different from 2, 3, m. For each positive integeri, there exists a positive integer K such that for all j ≥ i, all u ≥ 1 and all positive integersn not divisible by `, the congruence
p m
j`2uK−1n + 1
24
holds. There is also another positive integerM such that p m j`rn + 1 24 ≡ p m j`M +rn + 1 24 mod mi holds for alln.
Proof. Let βm,jbe the integer satisfying 1 ≤ βm,i ≤ mi− 1 and 24βm,i ≡ 1 mod mi.
Define
km,i=
(
(mi−1+ 1)(m − 1)/2 − 12bm/24c − 12, if i is odd,
mi−1(m − 1) − 12, if i is even.
By Theorem 3 of [2], for all i ≥ 1, there is a modular form f ∈ Mkm,i(SL(2, Z)) such that
Fm,i≡ η(24τ )(24βm,i−1)/m i
f (24τ ) mod mi.
The rest of proof is parallel to that of Theorem 7.
Example 7.4. Consider the case m = 13 and i = 2 of Theorem 10 and assume that ` is a prime different from 2, 3, 13. By [2, Theorem 3], F13,2is congruent to a modular form in
the space S23,144of dimension 13. Choose a Z-basis
fi= η(24τ )23E4(24τ )3(13−i)∆(24τ )i−1, i = 1, . . . , 13,
for Z[[q]] ∩ S23,144and let A be the matrix of T`2 with respect to this basis. If the order of
the matrix A −`309I 13 I13 0 mod 169 in PGL(26, Z/169) is K, then we have p 169` 2K−1n + 1 24 ≡ 0 mod 169 for all integers n not divisible by `. For instance, for ` = 5, we find
A = 20 101 52 52 166 148 46 135 96 51 73 49 128 166 164 159 66 123 50 144 85 29 116 22 93 10 158 152 90 65 20 167 27 96 109 154 127 164 76 120 154 132 110 22 113 115 51 25 104 108 82 33 43 148 131 45 81 2 164 145 117 157 4 108 61 134 23 151 120 151 44 30 1 76 32 60 132 165 121 40 83 4 56 88 3 134 100 85 88 18 3 23 20 20 31 66 24 41 126 47 137 33 112 49 143 18 44 26 89 109 118 148 35 16 35 122 150 144 51 47 143 109 164 52 38 92 50 98 60 104 70 165 89 80 28 75 19 110 101 41 155 78 67 123 147 54 4 60 133 49 151 30 32 157 108 82 95 139 50 70 124 168 87 63 13 104 58 107 113
modulo 169, and the order K is 28392, which yields p 13
2· 556783n + 1
24
≡ 0 mod 132 for all n not divisible by 5.
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