On Dyson's Crank Conjecture and the Uniform Asymptotic

ON DYSON’S CRANK CONJECTURE AND THE UNIFORM ASYMPTOTIC BEHAVIOR OF CERTAIN INVERSE THETA FUNCTIONS

KATHRIN BRINGMANN AND JEHANNE DOUSSE

Abstract. In this paper we prove a longstanding conjecture by Freeman Dyson concerning the limiting shape of the crank generating function. We ﬁt this function in a more general family of inverse theta functions which play a key role in physics.

1. Introduction and statement of results Dyson’s crank was introduced to explain Ramanujan’s famous partition congruences with modulus 5, 7, and 11. Denoting for n ∈ N by p(n) the number of integer partitions of n, Ramanujan [22] proved that for n ≥ 0 p (5n + 4) ≡ 0 (mod 5), p (7n + 5) ≡ 0 (mod 7), p (11n + 6) ≡ 0 (mod 11). A key ingredient of his proof is the modularity of the partition generating function ∞ 1 X 1 q 24 P (q) := p(n)qn = = , (q; q) η(τ) n=0 ∞ Qj−1 ` 2πiτ where for j ∈ N0 ∪ {∞} we set (a)j = (a; q)j := `=0(1 − aq ), q := e , and 1 ∞ 1 24 Q n η(τ) := q n=1(1 − q ) is Dedekind’s η-function, a modular form of weight 2 . Ramanujan’s proof however gives little combinatorial insight into why the above congruences hold. In order to provide such an explanation, Dyson [8] famously introduced the rank of a partition, which is deﬁned as its largest part minus the number of its parts. He conjectured that the partitions of 5n + 4 (resp. 7n + 5) form 5 (resp. 7) groups of equal size when sorted by their ranks modulo 5 (resp. 7). This

Date: July 7, 2014. 2010 Mathematics Subject Classification. 05A17, 11F03, 11F30, 11F50, 11P55, 11P82. 1 2 KATHRIN BRINGMANN AND JEHANNE DOUSSE conjecture was proven by Atkin and Swinnerton-Dyer [4]. Ono and the first author [7] showed that partitions with given rank satisfy also Ramanujan-type congruences. Dyson further postulated the existence of another statistic which he called the “crank” and which should explain all Ramanujan congruences. The crank was later found by Andrews and Garvan [1, 12]. If for a partition λ, o(λ) denotes the number of ones in λ, and µ(λ) is the number of parts strictly larger than o(λ), then the crank of λ is defined as largest part of λ if o(λ) = 0, crank(λ) := µ(λ) − o(λ) if o(λ) > 0. Denote by M(m, n) the number of partitions of n with crank m. Mahlburg [19] then proved that partitions with fixed crank also satisfy Ramanujan-type congruences. In this paper, we solve a longstanding conjecture by Dyson [9] concerning the limiting shape of the crank generating function. Conjecture 1.1 (Dyson). As n → ∞ we have 1 1 M (m, n) ∼ βsech2 βm p(n) 4 2 with β := √π . 6n Dyson then asked the question about the precise range of m in which this asymptotic holds and about the error term. In this paper, we answer all of these questions. √ Theorem 1.2. The Dyson-Conjecture is true. To be more precise, if |m| ≤ √1 n log n, π 6 we have as n → ∞ β 2 βm 1 1 M(m, n) = sech p(n) 1 + O β 2 |m| 3 . (1.1) 4 2 Remarks. 1. For fixed m one can directly obtain asymptotic formulas since the generating function is the convolution of a modular form and a partial theta function [6]. However, Dyson’s conjecture is a bivariate asymptotic. Indeed, this fact is the source of the difficulty of this problem. 2. We note that Theorem 1.2 is of a very different nature than known asymptotics in the literature. For example, the partition function can be approximated as p(n) = M(n) + O n−α , ON DYSON’S CRANK CONJECTURE 3

where M(n) is the main term which is a sum of varying length of Kloosterman 3 sums and Bessel functions and α > 0. Rademacher [20] obtained α = 8 , 1 Lehmer improved this to 2 − ε and Folsom and Masri [11] in their recent −δ 1 work obtained an impressive error of n for some absolute δ > 2 . Our result has a very different flavor due to the nonmodularity of the generating function and the bivariate asymptotics. 1 2 3. In fact we could replace the error by O(β 2 mα (m)) for any α(m) such that log n 1 √ 1 = o (α(m)) for all |m| ≤ √ n log n and βmα(m) → 0 as n → ∞. n 4 π 6 − 1 Here we chose α(m) = |m| 3 to avoid complicated expressions in the proof. A straightforward calculation shows Corollary 1.3. Almost all partitions satisfy Dyson’s conjecture. To be more precise √ n ] λ ` n|crank(λ)| ≤ √ log n ∼ p(n). (1.2) π 6 Remarks. 1. We thank Karl Mahlburg for pointing out Corollary 1.3 to us. 2. We can improve (1.2) and give the size of the error term. Dyson’s conjecture follows from a more general result concerning the coefficients Mk(m, n) defined for k ∈ N by ∞ ∞ X X (q)2−k C (ζ; q) = M (m, n) ζmqn := ∞ . k k (ζq) (ζ−1q) n=0 m=−∞ ∞ ∞

Note that M(m, n) = M1(m, n). Denoting by pk(n) the number of partitions of n allowing k colors, we have. Theorem 1.4. For k ﬁxed and |m| ≤ 1 log n, we have as n → ∞ 6βk

1 βk 2 βkm 1 M (m, n) = sech p (n) 1 + O β 2 |m| 3 , k 4 2 k k

q k with βk := π 6n . Remarks. 4 KATHRIN BRINGMANN AND JEHANNE DOUSSE

1. We note that for k ≥ 3, the functions Ck(ζ; q) are well-known to be generating functions of Betti numbers of moduli spaces of Hilbert schemes on (k − 3)−point blow-ups of the projective plane [13] (see also [6] and references therein). The results of this paper immediately gives the limiting proﬁle of the Betti numbers for large second Chern class of the sheaves. Recently, Hausel and Rodriguez-Villegas [16] also determined proﬁles of Betti numbers for other moduli spaces. 2. Note that our method of proof would allow determining further terms in the asymptotic expansion of Mk(m, n). 1 2 2 3. Again we could replace the error by O(βk mαk(m)) for any αk(m) such that log n 1 1 = o (αk(m)) for all |m| ≤ log n and βkmαk(m) → 0 as n → ∞. (kn) 4 6βk 4. The function Ck can also be represented as a so-called Lerch sum. To be more precise, we have [1]

n n(n+1) X m n 1 − ζ X (−1) q 2 M (m, n) ζ q = n . (1.3) (q)∞ 1 − ζq m∈Z n∈Z n≥0 This representation, which was a key representation in [6], is not used in this paper. 5. The special case k = 2 yields the birank of partitions [14]. 6. We expect that our methods also apply to show an analogue of (1.1) for the rank. The case of ﬁxed m is considered in upcoming work by Byungchan Kim, Eunmi Kim, and Jeehyeon Seo [17]. This paper is organized as follows. In Section 2, we recall basic facts on modular and Jacobi forms which are the base components of Ck and collect properties on Euler polynomials. In Section 3, we determine the asymptotic behavior of Ck. In Section 4, we use Wright’s version of the Circle Method to ﬁnish the proof of Theorem 1.4. In Section 5, we illustrate Theorem 1.2 numerically.

Acknowledgements The research of the ﬁrst author was supported by the Alfried Krupp Prize for Young University Teachers of the Krupp foundation and the research leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP/2007-2013) / ERC Grant agreement n. 335220 - AQSER. Most of this research was conducted while the second author was ON DYSON’S CRANK CONJECTURE 5 visiting the University of Cologne funded by the Krupp foundation. The ﬁrst author thanks Freeman Dyson and Jan Manschot for enlightening conversation. Moreover we thank Michael Mertens, Karl Mahlburg, and Larry Rolen for their comments on an earlier version of this paper. Moreover we thank the referee for valuable comments that improved the exposition of this paper.

2. Preliminaries 2.1. Modularity of the generating functions. A key ingredient of our asymptotic results is to employ the modularity of the functions Ck. To be more precise, we write (throughout q := e2πiτ , ζ := e2πiw with τ ∈ H, w ∈ C)

1 − 1 k 3−k i ζ 2 − ζ 2 q 24 η (τ) C (ζ; q) = , (2.1) k ϑ (w; τ) where ∞ 1 Y n η(τ) := q 24 (1 − q ) , n=1 ∞ 1 1 Y n n −1 n−1 ϑ (w; τ) := iζ 2 q 8 (1 − q ) (1 − ζq ) 1 − ζ q . n=1 The function η is a modular from, whereas ϑ is a Jacobi form. To be more precise, we have the following transformation laws (see e.g. [20]). Lemma 2.1. We have 1 √ η − = −iτη(τ), τ w 1 √ πiw2 ϑ ; − = −i −iτe τ ϑ (w; τ) . τ τ 2.2. Euler polynomials. Recall that the Euler polynomials may be deﬁned by their generating function ∞ 2ext X tr =: E (x) . (2.2) et + 1 r r! r=0 The following lemma may easily be concluded by diﬀerentiating the generating function (2.2). For the readers convenience we give a proof. 6 KATHRIN BRINGMANN AND JEHANNE DOUSSE

Lemma 2.2. We have ∞ 1 t X t2r − sech2 = E (0) . 2 2 2r+1 (2r)! r=0 Proof. We have ∞ ∞ X t2r d X t2r+1 E (0) = E (0) . 2r+1 (2r)! dt 2r+1 (2r + 1)! r=0 r=0 Now ∞ 2 X tr = E (0) , et + 1 r r! r=0 ∞ 2 X (−t)r = E (0) . e−t + 1 r r! r=0 Taking the diﬀerence gives the claim of the lemma since d 1 1 1 t − = − sech2 . dt et + 1 e−t + 1 2 2 We also require an integral representation of Euler polynomials. To be more precise, setting for j ∈ N0 Z ∞ w2j+1 Ej := dw, (2.3) 0 sinh(πw) we obtain Lemma 2.3. We have (−1)j+1E (0) E = 2j+1 . j 2 i Proof. We make the change of variables w → w+ 2 and then use the Residue Theorem to shift the path of integration back to the real line. Using the Binomial Theorem, we may thus write

i 2j+1 2j+1 2j+1−` i Z w + i X 2j + 1 i Z w` E = − 2 dw = − dw. j 2 cosh(πw) 2 ` 2 cosh(πw) R `=0 R ON DYSON’S CRANK CONJECTURE 7

−` ` 1 The last integral is known to equal (−2i) E`, where E` := 2 E`( 2 ) denotes the `th Euler number (see page 41 of [10]). The claim now follows using the well-known identity (see page 41 of [10]) j j−` X j 1 E` E (x) = x − . j ` 2 2` `=0

3. Asymptotic behavior of the function Ck.

Since Mk(−m, n) = Mk(m, n) we from now on assume that m ≥ 0. The goal of this section is to study the asymptotic behavior of the generating function of Mk(m, n). We deﬁne ∞ 1 Z 2 X n 2πiw −2πimw Cm,k(q) := Mk (m, n) q = Ck e ; q e dw 1 n=0 − 2 k 1 q 24 Z 2 = 2 k g (w; τ) cos(2πmw)dw, η (τ) 0 where 1 − 1 3 i ζ 2 − ζ 2 η (τ) g (w; τ) := . ϑ (w; τ) Here we used that g(−w; τ) = g(w; τ). In this section we determine the asymptotic behavior of Cm,k(q), when q is near an essential singularity on the unit circle. It turns out that the dominant pole lies at q = 1. Throughout the rest of the paper let 1 iz − 1 πm 3 τ = , z = βk(1 + ixm 3 ) with x ∈ satisfying |x| ≤ . 2π R βk 3.1. Bounds near the dominant pole. In this section we consider the range |x| ≤ 1. We start by determining the asymptotic main term of g. Lemma 2.1 and the deﬁnition of ϑ and η immediately imply. Lemma 3.1. For 0 ≤ w ≤ 1 we have for |x| ≤ 1 as n → ∞

2 2 iz 2π sin(πw) 2π w −4π2(1−w)Re 1 g w; = e z 1 + O e ( z ) . 2π 2π2w z sinh z 8 KATHRIN BRINGMANN AND JEHANNE DOUSSE

In view of Lemma 3.1 it is therefore natural to deﬁne

1 4π Z 2 sin(πw) 2π2w2 G (z) := e z cos(2πmw)dw, m,1 z 2π2w 0 sinh z 1 ! Z 2 iz 2π sin(πw) 2π2w2 G (z) := 2 g w; − e z cos(2πmw)dw. m,2 2π 2π2w 0 z sinh z Thus k q 24 C (q) = (G (z) + G (z)) . (3.1) m,k ηk(τ) m,1 m,2

The dominant contribution comes from Gm,1. Lemma 3.2. Assume that |x| ≤ 1 and m ≤ 1 log n. Then we have as n → ∞ 6βk z 2 βkm 2 2 2 βkm G (z) = sech + O β m 3 sech . m,1 4 2 k 2 Proof. Inserting the Taylor expansion of sin, exp, and cos, we get

r 2 2 j+ν 2 2π w X (−1) 2j+1 2ν 2π 2j+2ν+2r+1 sin(πw)e z cos(2πmw) = π (2πm) w . (2j + 1)!(2ν)!r! z j,ν,r≥0 This yields that

r 4π X (−1)j+ν 2π2 G (z) = π2j+1(2πm)2ν I m,1 z (2j + 1)!(2ν)!r! z j+ν+r j,ν,r≥0 where for ` ∈ N0 we deﬁne

1 2`+1 Z 2 w I := dw. ` 2π2w 0 sinh z

We next relate I` to E` defined in (2.3). For this, we note that Z ∞ w2`+1 I = dw − I0 (3.2) ` 2π2w ` 0 sinh z ON DYSON’S CRANK CONJECTURE 9 with Z ∞ 2`+1 Z ∞ 0 w 2`+1 −2π2wRe 1 I := dw w e ( z )dw ` 2π2w 1 sinh 1 2 z 2 1−2`−2 1 Re Γ 2` + 2; π2Re . z z R ∞ −w α−1 Here Γ(α; x) := x e w dw denotes the incomplete gamma function and throughout g(x) f(x) means that g(x) = O (f(x)). Using that as x → ∞ Γ(`; x) ∼ x`−1e−x (3.3) thus yields that −1 0 1 −π2Re 1 −π2Re 1 I Re e ( z ) ≤ e ( z ). ` z zw In the first summand in (3.2) we make the change of variables w → 2π and then shift the path of integration back to the real line by the Residue Theorem. Thus we obtain that Z ∞ w2`+1 z 2`+2 z 2`+2 (−1)`+1E (0) dw = E = 2`+1 , 2π2w 2π ` 2π 2 0 sinh z where for the last equality we used Lemma 2.3. Thus X (−1)r+1 G (z) = m2νz2j+2ν+r+1 m,1 22j+r+1(2j + 1)!(2ν)!r! j,ν,r≥0 −2j−2ν−2r−2 −π2Re( 1 ) × E2j+2ν+2r+1(0) + O |z| e z ∞ X (mz)2ν z z mz = − E (0) + O |z|2 = sech2 + O |z|2 cosh(mz) , (2ν)! 2 2ν+1 4 2 ν=0 where for the last equality we used Lemma 2.2. To finish the proof we have to 2 mz approximate sech 2 and cosh(mz). We have 2 cosh(mz) = cosh βkm + iβkm 3 x

2 2 = cosh(βkm) cos βkm 3 x + i sinh(βkm) sin βkm 3 x

2 = cosh(βkm) 1 + O βkm 3 . 10 KATHRIN BRINGMANN AND JEHANNE DOUSSE

This implies that mz 1 1 sech = = , mz β m 2 2 cosh k 3 2 cosh 2 1 + O βkm yielding 2 mz 2 βkm 2 sech = sech 1 + O β m 3 . 2 2 k Thus we obtain

z βkm 2 1 2 3 2 Gm,1(z) = sech 1 + O βkm + O βk 1 + 2 cosh(βkm) 4 2 m 3 z 2 βkm 2 2 2 βkm 2 = sech + O β m 3 sech + O β cosh(β m) . 4 2 k 2 k k

We may now easily ﬁnish the proof distinguishing the cases on whether βkm is bounded or goes to ∞.

We next turn to bounding Gm,2. Lemma 3.3. Assume that |x| ≤ 1. Then we have as n → ∞ 1 5π2 − 4β Gm,2(q) e k . βk Proof. By Lemma 3.1 we obtain that 1 1 Z 2 sin(πw) 2 1 2 2π Re( z )(w +w−2) Gm,2(z) 2 e dw. − 4π w |z| 0 1 − e z It is not hard to see that sin(πw) 2 1. − 4π w 1 − e z Moreover, q − 2 2 |z| = βk 1 + m 3 x βk, 1 1 Re ≥ . z 2βk ON DYSON’S CRANK CONJECTURE 11

2 1 The claim now follows, using that the maximum of w + w − 2 on [0, 2 ] is obtained 1 for w = 2 . Combining the above yields. Proposition 3.4. Assume that |x| ≤ 1. Then we have as n → ∞

k +1 √ z 2 βkm kπ2 k +2 2 βkm kn 2 6z 2 3 2 π 6 Cm,k (q) = k sech e + O βk m sech e . 4(2π) 2 2 2 Proof. Recall from (3.1) that

k q 24 C (q) = (G (z) + G (z)) . m,k ηk(τ) m,1 m,2 Lemma 2.1 easily gives that

k k 2 q 24 z 2 kπ = e 6z (1 + O(β )) . ηk(τ) 2π k

The functions Gm,1 and Gm,2 are now approximated using Lemma 3.2 and Lemma 3.3, respectively. It is not hard to see that the main error term arises from approximation Gm,1. We thus obtain

k +1 2 2 z 2 kπ βkm k 2 βkm π k Re 1 6z 2 2 2 3 2 6 ( z ) Cm,k(q) = k e sech + O |z| βkm sech e . 4(2π) 2 2 2 The claim follows now using that

|z| βk, √ 1 1 6n Re ≤ = √ . z βk π k

3.2. Bounds away from the dominant pole. We next investigate the behavior of Cm,k away from the dominant cusp q = 1. To be more precise, we consider the range 1 1 ≤ x ≤ πm 3 . Let us start with the following lemma, which proof uses the same idea βk as in [23]. 12 KATHRIN BRINGMANN AND JEHANNE DOUSSE

1 Lemma 3.5. Assume that τ = u + iv ∈ H with Mv ≤ |u| ≤ 2 for u > 0 and v → 0, we have that √ 1 π 1 1 |P (q)| v exp − 1 − √ . v 12 2π 1 + M 2 Proof. We rewrite ∞ ∞ ∞ ∞ X X X qnm X qm log(P (q)) = − log(1 − qn) = = . m m(1 − qm) n=1 n=1 m=1 m=1 Therefore we may estimate ∞ ∞ X |q|m |q| |q| X |q|m | log(P (q))| ≤ ≤ − + m|1 − qm| |1 − q| 1 − |q| m(1 − |q|m) m=1 m=1 1 1 = log(P (|q|)) − |q| − . 1 − |q| |1 − q| 1 We now split u into 2 ranges. If Mv ≤ |u| ≤ 4 , then we have cos(2πu) ≤ cos(2πMv). Therefore |1 − q|2 = 1 − 2e−2πv cos(2πu) + e−4πv ≥ 1 − 2e−2πv cos(2πMv) + e−4πv. Taylor expanding around v = 0 we ﬁnd that √ |1 − q| ≥ 2πv 1 + M 2 + O v2 . (3.4)

1 1 For 4 ≤ |u| ≤ 2 we have cos(2πu) ≤ 0. Therefore √ |1 − q| ≥ 1 > 2πv 1 + M 2. 1 Hence, for all Mv ≤ |u| ≤ 2 , √ |1 − q| ≥ 2πv 1 + M 2 + O v2 . (3.5) Furthermore we have 1 − |q| = 1 − e−2πv = 2πv + O v2 . (3.6) By Lemma 2.1, we have

− 2πv e 24 √ π P (|q|) = = ve 12v (1 + O(v)) . η(iv) ON DYSON’S CRANK CONJECTURE 13

Thus π 1 log(P (|q|)) = + log(v) + O(v). (3.7) 12v 2 Combining (3.5), (3.6), and (3.7), we obtain π 1 1 1 | log(P (q))| ≤ + log(v) + O(v) − 1 − √ + O(1) 12v 2 2πv 1 + M 2 1 π 1 1 1 = − 1 − √ + log(v) + O(1). v 12 2π 1 + M 2 2

Exponentiating yields the desired result.

We are now able to bound |Cm,k(q)| away from q = 1.

1 πm 3 Proposition 3.6. Assume that 1 ≤ |x| ≤ β . Then we have, as n → ∞, r √ ! 3−k kn 6kn − 2 |C (q)| n 4 exp π − m 3 . m,k 6 8π

Proof. We have by deﬁnition

1 Z 2 k Cm,k(q) = 2P (q) g(w; τ) cos(2πmw)dw. 0 Note that by (1.3)

2 2 n n +n n n +n X (−1) q 2 X (−1) q 2 g (w; τ) = 1 + (1 − ζ) + 1 − ζ−1 . 1 − ζqn 1 − ζ−1qn n≥1 n≥1 We thus may bound

n2+n β n2 3 X |q| 2 1 X − k − 3 g (w; τ) e 2 β 2 n 4 . 1 − |q|n 1 − |q| k n≥1 n≥1 Thus k 3 |Cm,k(q)| P (q) n 4 . 14 KATHRIN BRINGMANN AND JEHANNE DOUSSE

1 β β m− 3 x 1 k k − 3 Using Lemma 3.5 with v = 2π , u = 2π , and M = m , yields for 1 ≤ |x| ≤ 1 πm 3 , βk " !!# − 1 2π π 1 1 |P (q)| n 4 exp − 1 − . p − 2 βk 12 2π 1 + m 3 Therefore " !!# 3−k 2πk π 1 1 |Cm,k(q)| n 4 exp − 1 − p − 2 βk 12 2π 1 + m 3 " r √ !# 3−k kn 6kn 1 n 4 exp π − 1 − p − 2 6 π 1 + m 3 r √ ! 3−k kn 6kn − 2 n 4 exp π − m 3 . 6 8π

4. The Circle Method In this section we use Wright’s variant of the Circle Method and complete the proof of Theorem 1.4 and thus the proof of Dyson’s conjecture. We start by using Cauchy’s Theorem to express Mk as an integral of its generating function Cm,k: Z 1 Cm,k(q) Mk (m, n) = n+1 dq, (4.1) 2πi C q where the contour is the counterclockwise transversal of the circle C := {q ∈ C ; |q| = 1 −βk − e }. Recall that z = βk(1 + ixm 3 ). Changing variables we may write Z βk −z nz Mk(m, n) = 1 Cm,k(e )e dx. 2πm 3 1 |x|≤ πm 3 βk We split this integral into two pieces

Mk(m, n) = M + E ON DYSON’S CRANK CONJECTURE 15 with Z βk −z nz M := 1 Cm,k e e dx, 2πm 3 |x|≤1 Z βk −z nz E := 1 1 Cm,k e e dx. 2πm 3 1≤|x|≤ πm 3 βk In the following we show that M contributes to the asymptotic main term whereas E is part of the error term.

4.1. Approximating the main term. The goal of this section is to determine the asymptotic behavior of M. We show Proposition 4.1. We have 1 !! βk 2 βkm m 3 M = sech pk(n) 1 + O 1 . 4 2 n 4 A key step for proving this proposition is the investigation of

1 1+im− 3 Z √ 1 s π kn v+ 1 P := v e 6 ( v )dv s,k 2πi 1 1−im− 3 for s > 0. These integrals may be related to Bessel functions. Denoting by Is the usual I-Bessel function of order s, we have. Lemma 4.2. As n → ∞ r ! r !! 2kn kn 1 Ps,k = I−s−1 π + O exp π 1 + . − 2 3 6 1 + m 3 Proof. We use the following loop integral representation for the I-Bessel function [4] (x > 0) Z 1 −`−1 x(t+ 1 ) I`(2x) = t e t dt, (4.2) 2πi Γ where the contour Γ starts in the lower half plane at −∞, surrounds the origin counterclockwise and then returns to −∞ in the upper half-plane. We choose for Γ 16 KATHRIN BRINGMANN AND JEHANNE DOUSSE the piecewise linear path that consists of the line segments i i i i γ4 : −∞ − 1 , −1 − 1 , γ3 : −1 − 1 , −1 − 1 , 2m 3 2m 3 2m 3 m 3 i i i i γ2 : −1 − 1 , 1 − 1 , γ1 : 1 − 1 , 1 + 1 , m 3 m 3 m 3 m 3 which are then followed by the corresponding mirror images γ0 , γ0 , and γ0 . Note that R 2 3 4 Ps,k = . Thus, to ﬁnish the proof, we have to bound the integrals along γ4, γ3, and γ1 γ2− the corresponding mirror images follow in the same way. First    s 1 r − 1 − 1 Z Z kn im 3 1 im 3 exp π t − + t − dt   − 1  −∞ 6 2 t − im 3 2 γ4 2 ∞ s Z √ − 1 −π kn t im 3 6 e t + dt 2 1 Z ∞ √ r ! √ s −π kn t − s+1 kn − 1 −π kn t e 6 dt n 2 Γ s + 1; π n 2 e 6 , 1 6 using (3.3). Next 1 ! Z Z r s √ − 1 kn 1 − 1 −π kn m 3 exp −π 1 + 1 + im 3 t dt e 6 . − 2 6 1 + m 3 t2 γ3 1 2 Finally 1 ! Z Z r s kn t − 1 exp π t + t − im 3 dt − 2 6 t2 + m 3 γ2 −1 r ! kn 1 exp π 1 + , − 2 6 1 + m 3 t where we used that t + 2 obtains its maximum at t = 1. This ﬁnishes the t2rm− 3 proof. ON DYSON’S CRANK CONJECTURE 17

We now turn to the proof of Proposition 4.1. Proof of Proposition 4.1. Using Proposition 3.4 and making a change of variables, we obtain by Lemma 4.2

k +2 2 √ β βkm k +3 1 βkm 2kn k 2 2 3 2 π 3 M = k sech P k +1,k + O βk m sech e 4(2π) 2 2 2 2 k +2 2 r ! βk 2 βkm 2kn = k sech I− k −2 π 4(2π) 2 2 2 3 !! ! r k √ kn 1 +3 1 2 βkm π 2kn + O exp π 1 + + O β 2 m 3 sech e 3 . − 2 k 6 1 + m 3 2 Using the Bessel function asymptotic (see (4.12.7) in [3]) ex ex I`(x) = √ + O 3 2πx x 2 yields √ √ k +2  2kn 2kn ! 2 π 3 π 3 βk 2 βkm e e M = k sech  √ 1 + O 3 4(2π) 2 2 2kn 4 n 4 π 2 3 !!! r k √ kn 1 +3 1 2 βkm π 2kn + O exp π 1 + + O β 2 m 3 sech e 3 . − 2 k 6 1 + m 3 2 It is not hard to see that the last error term is the dominant one. Thus √ k +2 2kn 2 π β βkm e 3 1 1 k 2 3 − 4 M = k sech √ 1 1 + O m n . 4(2π) 2 2 2kn 4 π 2 3 Using that [15, 21]

1+k 4 √ k − 3+k π 2kn 1 p (n) = 2 (8n) 4 e 3 1 + O √ k 3 n now easily gives the claim. 18 KATHRIN BRINGMANN AND JEHANNE DOUSSE

4.2. The error arc. We ﬁnally bound E and show that it is exponentially smaller than M. The following proposition then immediately implies Theorem 1.4. Proposition 4.3. As n → ∞ r √ ! 3−k 2kn 6kn − 2 E n 4 exp π − m 3 . 3 8π Proof. Using Proposition 3.6, we may bound √ Z r ! βk 3−k kn 6kn 2 4 − 3 βkn E 1 1 n exp π − m e dx m 3 1≤x≤ πm 3 6 8π βk r √ ! 3−k 2kn 6kn − 2 n 4 exp π − m 3 . 3 8π

5. Numerical data We illustrate our results in 2 tables. M(0,n) n M (0, n) Mf(0, n) Mf(0,n) 20 41 ∼ 45 ∼ 0.912 50 8626 ∼ 9261 ∼ 0.931 500 3.228743492 · 1019 ∼ 3.298285542 · 1019 ∼ 0.979 1000 2.403603986 · 1029 ∼ 2.439699707 · 1029 ∼ 0.985 M(1,n) n M (1, n) Mf(1, n) Mf(1,n) 20 38 ∼ 44 ∼ 0.863 50 8541 ∼ 9185 ∼ 0.930 500 3.226300403 · 1019 ∼ 3.295574297 · 1019 ∼ 0.979 1000 2.402671309 · 1029 ∼ 2.438696696 · 1029 ∼ 0.985 β 2 βm where we set Mf(m, n) := 4 sech ( 2 )p(n). References [1] G. Andrews and F. Garvan, Dyson’s crank of a partition, Bull. Amer. Math. Soc. 18 (1988), 167–171. [2] A. Atkin and H. Swinnerton-Dyer, Some properties of partitions, Proc. Lond. Math. Soc. 4 (1954), 84–106. ON DYSON’S CRANK CONJECTURE 19

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Mathematical Institute, University of Cologne, Weyertal 86-90, 50931 Cologne, Germany E-mail address: [email protected]

LIAFA, Universite Denis Diderot - Paris 7, 75205 Paris Cedex 13, FRANCE E-mail address: [email protected]