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(m`kn + B) ≡ 0 mod m for all n ≥ 0
exist for all primes m and ` satisfying m ≥ 13 and ` 6= 2, 3, m, where B is a suitably chosen integer depending on m and `. Here the integer k depends 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 with a properly chosen B the congruence
p(mi`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 unordered 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 [14] is that the partition function p(n) satisfies the congruences
(1) p(An + B) ≡ 0 mod m,
for all nonnegative 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 [18] for the cases of powers of 5 and 7 and Atkin [4] for the cases of powers of 11. 1 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 [5, 7]. 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
Date: July 2, 2010.
2000 Mathematics Subject Classification. Primary 11P83; Secondary 11F25, 11F37, 11P82.
1Apparently, Ramanujan actually found a proof of the congruences modulo powers of 5 himself. The proof
was contained in an unpublished manuscript, which was hidden from the public until 1988. It appeared that Ramanujan intended to prove congruences modulo powers of 7 along the same line of attack, but his ailing health prevented him from working out the details. See the commentary at the end of [8] for more details.
power, a surprising result of Ono [13] 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
for all nonnegative integers n relatively prime to `. This result was later extended to com-posite m, (m, 6) = 1, by Ahlgren [1]. The results of [13] and [1] were further extended by Ahlgren and Ono [3].
Neither of [13] 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 [19]. In effect, she found 76,065 new congruences. (However, we should remark that some con-gruences listed in Theorem 2 of [19] were already discovered by Atkin [5]. Had Atkin had the computing power of the present day, he would have undoubtedly discovered many more congruences.) For primes m ≥ 37, this was addressed by Chua [9], although no explicit examples were given therein.
Another remarkable discovery of Ono [13, Theorem 5] is that the partition function possesses a 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 nonnegative integers n and all i ≥ N (m). Note that the bound m48(m3−2m+1)
can be improved greatly using a result of Garvan [10]. See Corollary 3.3 in Section 3 for details.
In this paper, we will obtain new congruences for the partition function and discuss related problems. In particular, we will show that there exist congruences of the form
p(m`kn + B) ≡ 0 mod m
for all primes m and ` such that m ≥ 13 and ` not equal to 2, 3, m, where B is a suitably chosen integer depending on m and `.
Theorem 1.1. 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`
2k−1n + 1
24
≡ 0 mod m for all nonnegative integersn relatively prime to m.
For instance, in Section 5 we will find that for m = 37, 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 1.1 is in fact a simplified version of one of the main results. (See Theorem 3.6). In the full version, we will see that the integer k in Theorem 1.1 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
Garvan [10] and rediscovered by the author of the present paper. 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 [13] and other subsequent papers [9, 19]. 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 6, 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
≡ 0 mod mi
for all positive integers n not divisible by `. One example worked out in Section 6 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 matrix γ = a b
c d ∈ GL(2, Q) and a modular form f(τ) of an even weight k, the
slash operator is defined by
f (τ ) kγ := (det γ)k/2(cτ + d)−kf
aτ + b cτ + d
.
For a power series f (q) =P af(n)qnand a positive integer N , we let UNand 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.
ACKNOWLEDGMENT
The author would like to thank the referees for thorough reading of the manuscript and providing many invaluable comments. In particular, the author is very grateful to one of the referees for giving a more accurate account of the history of the partition congruence problem and to another referee for bringing his attention to a very recent paper of Garvan [10]. Also, the proof of Proposition 2.1 presented here was suggested by the second referee.
The author was partially supported by Grant 98-2115-M-009-001 of the National Sci-ence Council of Taiwan. He was also supported by the National Center for Theoretical Sciences.
2. WORKS OFONO[13], WEAVER[19],ANDCHUA[9] In this section, we will review the ideas in [13, 19, 9].
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 [13] 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= ∆(τ )(m2k−1)/24 mod m,
where ∆(τ ) = η(τ )24is the normalized cusp form of weight 12 on SL(2, Z). From these, Ono [13, 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 χ
12 = 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),
where χmis the Kronecker character attached to Q(
√
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 the general Hecke theory for half-integral weight modular forms states that if f (τ ) = P∞
n=1af(n)q n ∈ S
λ+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 [16] and a result of Serre [15, 6.4].
As mentioned earlier, Ono [13] did not address the issue of finding explicit congruences of the form (2). However, Section 4 of [13] 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
Gm,kfor a huge number of terms. However, it turns out that Fm,kis congruent to another
half-integral weight modular form of a much smaller weight. For example, using Sturm’s theorem [17] 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 nonnegative integers k. The modular form η(24τ )11is in fact a Hecke eigenform. (The modular form η(24τ )23is also a Hecke eigenform. It has been known since Morris Newman’s work in the 50’s that for odd r with 0 < r < 24, the function η(24τ )ris a Hecke eigenform.) More generally, for m ∈ {13, 17, 19, 23, 29, 31}, it is shown in [13, Section 4], [11, Proposition 6] and [19, Proposition 5] that Gm,1is congruent to a Hecke eigenform
of weight m/2 − 1. Using this observation, Weaver [19] 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 [11] and [19] is essentially “verification” in the sense that they all used Sturm’s criterion [17]. 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 [9], instead of the congruence η(mτ )m
η(τ ) ≡ η(τ )
m2−1
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 [6, Lemma 7], one has
η(mτ )m−1η(τ )m−1 (Um+ m(m−1)/2−1Wm) ∈ Sm−1(SL(2, Z)),
where, for a modular form f (τ ) of an even integral weight k on Γ0(m), the Atkin-Lehner
operator Wmis defined by (7) Wm: f (τ ) 7−→ f (τ ) k 0 −1 m 0 = (√mτ )−kf − 1 mτ . 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 nontrivial cusp forms of weight 4, 6, 10 on SL(2, Z).) By examining the order of vanishing of fm(τ ) at ∞, Chua [9, Theorem 1.1]
then concluded that if we let rmdenote the integer in the range 0 < rm < 24 such that
m ≡ −rm mod 24, then
Fm,1≡ η(24τ )rmφm,1(24τ ) mod m
for some modular form φm,1on SL(2, Z) of weight (m − rm− 2)/2. More generally, one
has the following proposition.
Proposition 2.1. Let m ≥ 13 be a prime and rmbe the integer in the range0 < rm< 24
such thatm ≡ −rm mod 24. Set
rm,j= ( rm, ifj is odd, 23, ifj is even. 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.
Proof. Consider the function fm,j(τ ) = η(mjτ )m j
/η(τ ). It is a modular form of weight (mj− 1)/2 on Γ
0(mj) with character m· j
. Consider also the auxiliary function hm,j(τ ) =
(
η(τ )m/η(mτ ), if j is odd, (η(τ )m/η(mτ ))2
, if j is even. It is a modular form on Γ0(m) with character m·
j
and satisfies
(8) hm,j(τ ) ≡ 1 mod m.
By the level reduction lemma of Atkin and Lehner [6, Lemma 7], if we apply Umto fm,i
j − 1 times and then multiply the resulting function by hm,j, we get a modular form on
Γ0(m) with trivial character. That is,
fm,j(τ )
is a modular form on Γ0(m) with trivial character. The weight is
λm,j =
(
(mj− 1)/2 + (m − 1)/2, if j is odd,
(mj− 1)/2 + m − 1, if j is even. Then by the level reduction lemma again
fm,j(τ ) Umj−1· hm,j(τ ) (Um+ m λm,j/2−1W m)
is a modular form on SL(2, Z), where Wm is the Atkin-Lehner operator defined in (7).
Considering the order of vanishing at ∞, we see that this modular form on SL(2, Z) is ∆(τ )µm,jφ m,j(τ ), where ∆(τ ) = η(τ )24, µm,j = m2j+ 24ν m,j− 1 24mj
with νm,jbeing the unique integer satisfying 0 < νm,j < mj and 24νm,j ≡ 1 mod mj,
and φm,jis a modular form of weight λm,j− 12µm,jon SL(2, Z).
Now observe that hm,j
mλm,j/2−1W
m is congruent to 0 modulo a high power of m.
Then, by (8), we have ∆(24τ )µm,jφ m,j(24τ ) ≡ fm,j(τ ) Umj V24= η(24τ )m j Fm,j mod m.
In other words, we have Fm,j ≡ η(24τ )24µm,j−m
j
φm,j(24τ ) = η(24τ )(24νm,j−1)/m j
φm,j(24τ ) mod m.
The integer (24νm,j− 1)/mj is in the range between 0 and 24. Also, it is congruent to
−1/mjmodulo 24. Thus, we have
24νm,j− 1
mj = rm,j =
(
rm, if j is odd,
23, if j is even. From this, we get
λm,j− 12µm,j=
(
(m − rm− 2)/2, if j is odd,
m − 13, if j is even.
This proves the proposition.
Remark 2.2. Proposition 2.1 was stated as Conjecture 1 in [9]. The proof sketched here was suggested by one of the referees and was adapted from the proof of Theorem 3 in [2]. Alternatively, one can combine Proposition 3.1 below with an induction step proved in [9] to get the same conclusion. See the arxiv version arXiv:0904.2530 of the present paper for more details.
3. MAIN RESULTS
In this section, we will state our main results. Before doing so, let us first recall a property about the subspace
{η(24τ )rf (24τ ) : f ∈ Ms(SL(2, Z))}
of Ss+r/2(Γ0(576), χ12), in which the function η(24τ )rm,jφm,j(24τ ) in Proposition 2.1
Proposition 3.1 ([10, Proposition 3.1]). Let r be an odd integer with 0 < r < 24. Let s be a nonnegative even integer. Then the space
(9) 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.
Remark 3.2. This property of Sr,swas first discovered by Garvan [10], and later
rediscov-ered by the author of the present paper. (See the arxiv version arXiv:0904.2530 of the present paper.) Garvan stated the proposition under the assumption that (r, 6) = 1 instead of 2 - r, but it can be easily checked that his proof also works for the cases r = 3, 9, 15, 21 as well. The author’s proof is more complicated, but can be applied in other similar situa-tions. However, at the hindsight, the invariance of Sr,sunder the action of Hecke algebra
is best explained (and proved) as follows.
The usual definition of modular forms of half-integral weights, as per Shimura [16], is given in terms of the theta function θ(τ ) =P
n∈Zq n2
. Specifically, we say a holomorphic function f : H → C is a modular form of half-integral weight λ + 1/2 on Γ0(4N ) with
character χ, where χ is a Dirichlet character modulo 4N , if f (τ ) is holomorphic at each cusp and satisfies
f (γτ )
f (τ ) = χ(d)
θ(γτ )2λ+1
θ(τ )2λ+1
for all γ = a b
c d ∈ Γ0(4N ). It is in this sense we say η(24τ ) is a modular form of weight
1/2 on Γ0(576) with character χ12.
Now the choice of θ in the definition of half-integral modular forms is perhaps the most natural and simplest from the view point of Weil representations, but one drawback of this choice is that the levels of the modular forms have to be a multiple of 4. On the other hand, if we define modular forms of half-integral weights in terms of η(τ ), then the levels can be taken all the way down to 1. Explicitly, let Γ be a congruence subgroup of SL(2, Z) for an odd integer r with 0 < r < 24 and a nonnegative even integer s, we say a function f : H → C is a modular form of (ηr, s)-type on Γ if it is holomorphic in H and at each cusp such that
f (γτ )
f (τ ) = (cτ + d)
sη(γτ )r
η(τ )r
for all γ = a b
c d ∈ Γ. Let Sr,s(Γ) be the space of all such modular forms on Γ.
Consider the case Γ = SL(2, Z). On the space Sr,s(SL(2, Z)), we can also define Hecke
operators T`2 for primes ` 6= 2, 3 and show that their actions on f (τ ) =P af(n)qn/24 ∈
Sr,s(SL(2, Z)) is T`2 : f (τ ) 7→ ∞ X n=1 af(`2n) + 12 ` (−1)λn ` `λ−1af(n) + `2λ−1af(n/`2) qn/24 with λ = (r + 2s − 1)/2. Now observe that if g(τ ) ∈ Sr,s(SL(2, Z)), then g(τ + 1) =
e2πir/24g(τ ), which implies that g(τ ) = qr/24(c0+ c1q + · · · ), ci ∈ C. Therefore,
f (τ ) = g(τ )/η(τ )ris a function holomorphic on H and at each cusp and satisfies f (γτ ) = (cτ + d)sf (τ ) for all γ = a b
c d ∈ SL(2, Z). In other words,
Sr,s(SL(2, Z)) = {η(τ )rf (τ ) : f ∈ Ms(SL(2, Z))}
and
This explains why Sr,sis an invariant subspace of Sr/2+s(Γ0(576), χ12).
Using the pigeonhole principle, one can see that Propositions 2.1 and 3.1 yield Ono’s periodicity result (3), with an improved bound.
Corollary 3.3. Let m ≥ 5 be a prime. Then there exist integers N (m) and P (m) with 0 ≤ (N (m) − 1)/2 ≤ mA(m)and0 ≤ P (m) ≤ mA(m)such that
p m in + 1 24 ≡ p m 2P (m)+in + 1 24 mod m for all nonnegative integersn and all i ≥ N (m), where
(10) A(m) = dim M(m−rm−2)/2(SL(2, Z)) = jm 12 k −jm 24 k andrmis the integer satisfying0 < rm< 24 and m ≡ −rm mod 24.
From Proposition 3.1, we can deduce the following corollary, which will be proved in the next section.
Corollary 3.4. Let r be an odd integer satisfying 0 < r < 24 and s be a nonnegative even integer. LetSr,s be defined as(9) 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 for nonnegative integers k, Ak,Bk, andCk aret × t matrices
satisfying Ak Ak−1 =A −` r+2s−2I t It 0 kI t 0 withItbeing thet × t identity matrix, and
Bk= −`s+(r−3)/2
(−1)(r−1)/212
`
Ak−1, Ck= −`r+2s−2Ak−1.
Remark 3.5. It is well-known that for nonnegative even integer s, the space Ms(SL(2, Z))
has a basis consisting of g1, . . . , gd satisfying gi ∈ Z[[q]] and gi = qi−1+ · · · , where
d = dim Ms(SL(2, Z)). (Usually, gi are chosen to be products of ∆(τ ) and Eisenstein
series.) Then it can be easily verified that the functions fi(τ ) = η(24τ )rgi(24τ ) form a
Z-basis of the Z-module Z[[q]] ∩ Sr,s. In particular, the rank of the Z-module Z[[q]] ∩ Sr,s
is the same as the dimension of Sr,s.
Note also that if r + 2s ≥ 3, then the Hecke operator T`2 maps Z[[q]] ∩ Sr,s into
Z[[q]] ∩ Sr,s. Therefore, the matrix A in the above corollary has entries in Z. This property
is crucial in our subsequent discussion when we need to take A modulo a prime.
Now we can state our main results. The first one is a more precise version of Theorem 1.1. The proof utilizes the corollary above and will be given in the next section.
Theorem 3.6. Let m ≥ 13 be a prime. Set rmto be the integer satisfying0 < rm< 24
andm ≡ −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
(11) A −` m−4I t It 0 mod m inPGL(2t, Fm) is K. Then we have (12) p m` 2uK−1n + 1 24 ≡ 0 mod m for all positive integersu and all positive integers n not divisible by `.
Also, if the order of the matrix(11) in GL(2t, Fm) is M , then we have
(13) p m` in + 1 24 ≡ p m` 2M +in + 1 24 mod m for all nonnegative integeri and all positive integers n.
Remark 3.7. Note that if the matrix A in the above theorem vanishes modulo m, then the matrix in (11) 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 3.8. In general, the integer K in Theorem 3.6 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 3.6.
4. PROOF OFCOROLLARY3.4ANDTHEOREM3.6 Proof of Corollary 3.4. By the definition of T`2, we have
f1 .. . ft U`2 = A1 f1 .. . ft + B1 g1 .. . gt + C1 f1 .. . ft V`2, where gj = fj⊗ `· and A1= A, B1= −`s+(r−3)/2 (−1)(r−1)/212 ` It, C1= −`r+2s−2It.
Now we make the key observation gj
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 corollary.
Proof of Theorem 3.6. 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 Proposition 2.1, Fm,1is congruent
to a modular form in Sr,s, where Sr,sis defined by (9). Now let {f1, . . . , ft} be a Z-basis
for Z[[q]] ∩ Sr,sand A be given as in the statement of the theorem. (Note that A has entries
in Z. See Remark 3.5.) Then by Corollary 3.4, 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
(14) Ak Ak−1 = XkIt 0 , (15) 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−4I t −It A
Therefore, if the order of X mod m in PGL(2t, Fm) is K, then we have, for all positive integers u, 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
argument. 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 fjvanishes 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−1n)th 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 (12).
Finally, if the matrix X has order M in GL(2t, Fm), then from the recursive relations
(14) and (15), it is obvious that (13) holds. This completes the proof. Remark 4.1. In general, the integer K in Theorem 3.6 may not be the smallest positive integer such that congruence (4) holds. For example, consider the case where Sr,s has
dimension t ≥ 2 and the reduction of Z[[q]]∩Sr,smodulo m has a basis consisting of Hecke
eigenforms f1, . . . , ftdefined over Fm. Suppose that 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 3.6 in general.
5. EXAMPLES
Example 5.1. Let m = 13. According to Proposition 2.1, we have F13,1≡ cη(24τ )11 mod 13
for some c ∈ F13. (In fact, c = 11. See [13, 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 that 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 5.2. Let m = 37. By Proposition 2.1, we know that F37,1is congruent to a cusp
form in S11,12modulo 37. In fact, according to [9, 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 3.6 is not optimal. (Here we have K = 456.)
For other small primes `, we find that the 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
6. GENERALIZATIONS
There are several directions in which one may generalize Theorem 3.6. 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 [18, 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 (16) 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 [12] used this formula to study congruences of the partition func-tion modulo higher powers of 5. One distinct feature of [12] is the following lemma. Lemma 6.1 (Lovejoy and Ono [12, 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 6.2. 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 (16), 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 3.4 we have (17) f U`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 3.6, if the order of
(18) a −` 17 1 0 mod 5 in PGL(F5) is k, then (19) f U`2uk−1 ≡ f V` mod 5
for all positive integers u. Now by Lemma 6.1 the characteristic polynomial of (18) has a factorization x − 15 ` x − 15 ` `
modulo 5. From this we see that the order of (18) in PGL(F5) is
K`= 5, if ` ≡ 1 mod 5, 4, if ` ≡ 2, 3 mod 5, 2, if ` ≡ 4 mod 5. Thus, (19) 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 exactly the same because `21≡ `17 mod 5.
Remark 6.3. In [18], 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 6.1 in this case, we do not have a result as precise as Theorem 6.2.
The next congruence result is an analog of Theorem 2 of [19], which in turn originates from the argument outlined in [13, page 301].
Theorem 6.4. Let ` ≥ 7 be a prime. Assuming one of the three situations below occurs, we setk`andm`to be
(1) k`= 2 and m`= 5 if ` ≡ 1 mod 5, −n` = −1,
(2) k`= 2 and m`= 4 if ` ≡ 2 mod 5, −n` = −1, and
(3) k`= 1 and m`= 4 if ` ≡ 3 mod 5, −n` = −1. Then p 5 i`2(um`+k`)n + 1 24 ≡ 0 mod 5i+1 for all nonnegative integersu and all positive 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(24τ )19vanish modulo 5.
Let ` ≥ 7 be a prime and a be the eigenvalue of T`2associated to `. By (17), we have
(20) f U`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 6.1, 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 (20) 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 nonnegative 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. (21)
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. Remark 6.5. Notice that the case ` ≡ 4 mod 5 is missing in Theorem 6.4. This is be-cause in this case, by Lemma 6.1, the Hecke eigenvalues of T`2for η(24τ )19and η(24τ )23
are both multiples of 5. Then the numbers akin (20) satisfy
ak ak−1 =0 1 1 0 k1 0 . From this we see that ak± ak−1can never vanish modulo 5.
Example 6.6. We now give some examples of congruences predicted in Theorem 6.4. (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 (21) 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 6.7. 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 u ≥ 1 and all positive integersn not divisible by `, the congruence
p m
i`2uK−1n + 1
24
≡ 0 mod mi holds. There is also another positive integerM such that
p m i`rn + 1 24 ≡ p m i`M +rn + 1 24 mod mi holds for all nonnegative integersn and r.
Proof. Let βm,ibe 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 3.6.
Example 6.8. Consider the case m = 13 and i = 2 of Theorem 6.7 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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