Ahlgren and Andersen Research in Number Theory (2015) 1:10 DOI 10.1007/s40993-015-0011-9
RESEARCH ARTICLE Open Access Weak harmonic Maass forms of weight 5/2 and a mock modular form for the partition function
Scott Ahlgren* andNickolasAndersen
*Correspondence: [email protected] Abstract Department of Mathematics, We construct a natural basis for the space of weak harmonic Maass forms of weight University of Illinois, Urbana IL, USA 5/2 on the full modular group. The nonholomorphic part of the first element of this basis encodes the values of the ordinary partition function p(n). We obtain a formula for the coefficients of the mock modular forms of weight 5/2 in terms of regularized inner products of weakly holomorphic modular forms of weight −1/2, and we obtain Hecke-type relations among these mock modular forms.
1 Introduction AnumberofrecentworkshaveconsideredbasesforspacesofweakharmonicMaass forms of small weight. Borcherds [4] and Zagier [25] (in their study of infinite product
expansions of modular forms, among many other topics) made use of the basis {fd}d>0 −d defined by f−d = q +O(q) for the space of weakly holomorphic modular forms of weight 1/2 in the Kohnen plus space of level 4. Duke, Imamoglu¯ and Tóth [14] extended this to
abasis{fd}d∈Z for the space of weak harmonic Maass forms of the same weight and level and interpreted the coefficients in terms of cycle integrals of the modular j-function. In subsequent work, [16] they constructed a similar basis in the case of weight 2 for the full modular group, and related the coefficients of these forms to regularized inner products of an infinite family of modular functions. To construct these bases requires various types of Maass-Poincaré series. These have played a fundamental role in the theory of weak harmonic Maass forms (see for example [6, 10] among many other works). Here, we will construct a natural basis for the space of weak harmonic Maass forms of
weight 5/2onSL2(Z) with a certain multiplier. To develop the necessary notation, let the Dedekind eta-function be defined by 1 n 2πiz η(z) := q 24 (1 − q ), q := e , z = x + iy ∈ H, n≥1 where H denotes the complex upper half-plane. We have the generating function −1 n− 1 η (z) = p(n)q 24 , n≥0
where p(n) is the ordinary partition function. The transformation property 1 ab η(γz) = ε(γ )(cz + d) 2 η(z), γ = ∈ SL (Z) (1.1) cd 2
© 2015 Ahlgren and Andersen. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited. Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 2 of 16
defines a multiplier system ε of weight 1/2onSL2(Z) which takes values in the 24-th roots ! (ε) of unity. Let M− 1 denote the space of weakly holomorphic modular forms of weight 2 −1 −1/2 and multiplier ε on SL2(Z).Thenη (z) is the first element of a natural basis {gm} −1 forthisspace.Toconstructthebasis,wesetg1 := η ,andforeachm ≡ 1mod24we define gm := g1P(j),where P(j) is a suitable monic polynomial in the Hauptmodul j(z) − m 23 such that gm = q 24 + O q 24 .Welistthefirstfewexampleshere:
− g = η 1 1 − 1 23 47 71 95 119 143 167 191 = q 24 + q 24 + 2q 24 + 3q 24 + 5q 24 + 7q 24 + 11q 24 + 15q 24 + O q 24 , − g = η 1( j − 745) 25 − 25 23 47 71 95 = q 24 + 196885q 24 + 21690645q 24 + 886187500q 24 + O q 24 , (1.2) − g = η 1( j2 − 1489j + 160511) 49 − 49 23 47 71 95 = q 24 + 42790636q 24 + 40513206272q 24 + 8543738297129q 24 + O q 24 .
! (ε) / Denote by H 5 the space of weak harmonic Maass forms of weight 5 2 and multiplier 2 ε on SL2(Z). These are real analytic functions which transform as 5 ab F(γ z) = ε(γ )(cz + d) 2 F(z), γ = ∈ SL (Z), cd 2
which are annihilated by the weight 5/2 hyperbolic Laplacian ∂2 ∂2 ∂ ∂ 2 5 5 :=−y + + iy + i , 2 ∂x2 ∂y2 2 ∂x ∂y and which have at most linear exponential growth at ∞ (see the next section for details). ! (ε) / If M 5 denotes the space of weakly holomorphic modular forms of weight 5 2and 2 ε ! (ε) ⊆ ! (ε) multiplier ,thenwehaveM 5 H 5 . 2 2 { } ! (ε) ∈ Here we construct a basis hm for the space H 5 . For negative m we have hm 2 3 ! (ε) ξ ( ) = 6 2 ξ ! (ε) → ! (ε) M 5 , while for positive m we have 5 hm πm gm,where 5 : H 5 M− 1 2 2 2 2 2 is the ξ-operator defined in (2.2) below. In particular, the nonholomorphic part of h1 encodes the values of the partition function. To state the first result it will be convenient to introduce the notation β( ) = − 3 πy y 2 , 6 ,
( ) = ∞ −t s−1 where the incomplete gamma function is given by s, y : y e t dt.
≡ ( ) ∈ ! (ε) Theorem 1. For each m 1 mod 24 ,thereisauniquehm H 5 with Fourier 2 expansion of the form ( ) = m + +( ) n < hm z q 24 pm n q 24 if m 0, 0 { } ! (ε) < ∈ ! (ε) > The set hm forms a basis for H 5 .Form 0 we have hm M 5 ,whileform 0 we 2 2 have 3 ξ ( ) = 6 2 · ∈ ! (ε) 5 hm πm gm M− 1 . 2 2 = +( ) Furthermore, for m n, the coefficients pm n are real. When m = 1, we have 3 3 2 2 + 6 −1 6 n 1 n ξ 5 (h1) = η (z) = p q 24 . 2 π π 24 n≥−1 Using Lemma 7 below we obtain the following corollary. Corollary 2. The function P(z) defined by + n + 3 − n ( ) = ( ) 24 − n 1 2 β( ) 24 P z : p1 n q p 24 n ny q 0 5 (Z) ε is a weight 2 weak harmonic Maass form on SL2 with multiplier . The coefficients can be computed using the formula in Proposition 11 below. We have √ 1 25 49 73 ( ) = − ... + 4 π 24 + ... 24 + ... 24 + ... 24 +··· P z 3.96 i 3 q 111.40 q 254.26 q 86.52 q 1 3 − 23 3 − 47 +iβ(−y)q 24 − 1 · 23 2 β(23y)q 24 − 2 · 47 2 β(47y)q 24 +··· . (1.3) We will construct the functions hm(z) in Section 3 using Maass-Poincaré series. Unfor- tunately, the standard construction does not produce any nonholomorphic forms in ! (ε) H 5 , and we must therefore consider derivatives of these series with respect to an aux- 2 iliary parameter. This method was recently used by Duke, Imamoglu¯ and Tóth [16] in the case of weight 2 (see also [5, 20]). The construction provides an exact formula for ±( ) = the coefficients pm n .Inparticular,whenm 1 we obtain the famous exact formula of Rademacher [23] for p(n) as a corollary (see Section 4 for details). −( ) Work of Bruinier and Ono [11] provides an algebraic formula for the coefficients pm n (and in particular the values of p(n)) as the trace of certain weak Maass forms over CM points. Forthcoming work of the second author [3] investigates the analogous arithmetic +( ) and geometric nature of the coefficients pm n . In analogy with [14] and [9], the coef- ficients are interpreted as the real quadratic traces (i.e. sums of cycle integrals) of weak Maass forms. Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 4 of 16 Remark. When m < 0, we can construct the forms hm directly as in (1.2). We list a few =− d examples here. Let j : q dq j.Then h− = η j 23 − 23 1 25 49 73 97 = q 24 − q 24 − 196885q 24 − 42790636q 24 − 2549715506q 24 + O q 24 , h− = h− ( j − 743) 47 23 − 47 1 25 49 73 = q 24 − 2q 24 − 21690645q 24 − 40513206272q 24 + O q 24 , h− = h− ( j2 − 1487j + 355910) 71 23 − 71 1 25 49 73 = q 24 − 3q 24 − 886187500q 24 − 8543738297129q 24 + O q 24 . Together with the family (1.2), these form a “grid” in the sense of Guerzhoy [17] or Zagier [25] (note that the integers appearing as coefficients are the same up to sign as those in (1.2)). > +( ) For m 0, the formula for pm n which results from the construction is an infinite series whose terms are Kloosterman sums multiplied by a derivative of the J-Bessel func- tion in its index. Here we give an alternate interpretation of these coefficients involving the regularized Petersson inner product, in analogy with [16] and [15]. For modular forms f and g of weight k,define f , g = lim f (z)g(z)yk dx dy , (1.4) reg →∞ y2 Y F(Y) provided that this limit exists. Here F(Y) denotes the usual fundamental domain for SL2(Z) truncated at height Y. Theorem 3. For positive m, n ≡ 1 (mod 24) with m = n we have + 3 ( ) = 6 2 · pm n : πm gm, gn reg. We note that when n = m the integral defining this inner product does not converge. The following is an immediate corollary of Theorem 3. Corollary 4. For positive m, n ≡ 1 (mod 24) we have 3 · +( ) = 3 · +( ) n 2 pn m m 2 pm n . 2 There are also Hecke relations among the forms hm.LetT 5 ( ) denote the Hecke 2 2 ! (ε) operator of index on H 5 (see Section 6 for definitions). 2 Theorem 5. For any m ≡ 1 (mod 24) and for any prime ≥ 5 we have ( 2) = 3 + 3m hm T 5 h 2m hm. 2 +( ) Using Theorem 5 it is possible to deduce many relations among the coefficients pm n . We record a typical example as a corollary. Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 5 of 16 m n Corollary 6. If m, n ≡ 1 (mod 24) and ≥ 5 is a prime with = ,then +( 2 ) = 3 + ( ) pm n p 2m n . Remark. It is possible to derive results analogous to Theorem 5 and Corollary 6 involv- 2k ing the operators T 5 for any k (see, for example, [1, 2]). For brevity, we will not 2 record these statements here. In the next section we provide some background material. In Section 3 we adapt the method of [16] to construct the basis described in Theorem 1. The last three sections contain the proofs of the remaining results. 2 Weak harmonic Maass forms We require some preliminaries on weak harmonic Maass forms (see for example [8, 22, = (Z) ∈ 1 Z 24] for further details). For convenience, we set : SL2 .Ifk 2 then we say that f has weight k and multiplier ν for if γ = ν(γ) f k f (2.1) for every γ ∈ .Here denotes the slash operator defined for γ = ab ∈ GL (Q) k cd 2 with det(γ ) > 0by + k −k az b f γ := (det γ)2 (cz + d) f . k cz + d We choose the argument of each nonzero τ ∈ C in (−π, π], and we define τ k using ∈ 1 Z ξ the principal branch of the logarithm. For any k 2 ,let k denote the Maass-type differential operator which acts on differentiable functions f on H by ∂f ξ (f ) = 2iyk . (2.2) k ∂z This operator satisfies ξ γ = (ξ ) γ k f k kf 2−k (2.3) for any γ ∈ .Soiff ismodularofweightk,thenξk(f ) ismodularofweight2− k. Furthermore, ξk(f ) = 0ifandonlyiff is holomorphic. We define the weight k hyperbolic Laplacian k by ∂2 ∂2 ∂ ∂ 2 :=−ξ − ◦ ξ =−y + + iky + i . k 2 k k ∂x2 ∂y2 ∂x ∂y In this paper, we are interested in the multiplier system ε which is attached to the Dedekind eta function. An explicit description ofε is given, for instance, in ([19], Section 2.8). Defining e(α) := e2πiα,wehaveε 11 = e 1 and ε(−I) = e − 1 .For 01 24 4 γ = ab with c > 0, we have cd a + d − 3c 1 ε(γ ) = e − s(d, c) , (2.4) 24c 2 Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 6 of 16 where s(d, c) is the Dedekind sum c− 1 r 1 dr dr 1 s(d, c) := − − − . (2.5) c 2 c c 2 r=1 If f = 0 satisfies (2.1) with ν = ε then we must have 2k ≡ 1 (mod 4),since (− )−k = (− ) = ε(− ) 1 f f k I I f . (2.6) In what follows, we assume this consistency condition so that the forms in question are not identically zero. Suppose that f : H → C is real analytic and satisfies γ = ε(γ ) f k f (2.7) for all γ ∈ .Thenf has a Fourier expansion at ∞ which is supported on exponents of the form n/24 with n ≡ 1 (mod 24). If, in addition, f satisfies kf = 0, (2.8) then by the discussion in Section 3.2 below we have the Fourier expansion + n − πny − n ( ) = ( ) 24 + ( ) ( − ) 24 f z a n q a n 1 k, 6 q . (2.9) n≡1 (24) n≡−1 (24) ! (ε) Let Hk denote the space of functions satisfying (2.7) and (2.8) with the additional prop- erty that only finitely many of the a±(n) with n ≤ 0 in (2.9) are nonzero. We call elements ! (ε) ε of Hk weak harmonic Maass forms of weight k and multiplier . Note that these forms (ε) ⊆ (ε) ⊆ ! (ε) ⊆ are allowed to have poles in the nonholomorphic part. Let Sk Mk Mk ! (ε) Hk denote the subspaces of cusp forms, modular forms, and weakly holomorphic modular forms, respectively. The next lemma follows from a computation. Care must be taken with the two cases m > 0andm < 0. Lemma 7. For any m and any constant c, we have m 3 m − 6 2 ξ 5 c · β(my)q 24 =−c · π q 24 . 2 m ∈ ! (ε) Suppose that f H 5 has Fourier expansion (2.9). By Lemma 7 and (2.3) we find that 2 3 n ξ ( ) =− −( ) 6 2 24 ∈ ! (ε) 5 f a n πn q M− 1 . 2 2 n≡−1 (24) ± Since S− 1 (ε) ={0} and S 5 (ε) ={0},weseethatf = 0ifandonlyifa (n) = 0 for all 2 2 n < 0. Because of this, each form hm in Theorem 1 is uniquely determined by a single m m term. If m < 0thistermisq 24 and if m > 0thistermisβ(−my)q 24 . 3 Construction of harmonic Maass forms and proof of Theorem 1 In order to construct the basis described in Theorem 1, we first construct Poincaré series attached to the usual Whittaker functions (similar constructions can be found in [5, 7, 10, 14, 16, 20] among others). It turns out that for positive m these series are identically zero. So we must differentiate with respect to an auxiliary parameter s inordertoobtain nontrivial forms in this case. The construction is carried out in several subsections and is summarized in Proposition 11 below. Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 7 of 16 3.1 Poincaré series ∈ 1 Z ≡ ( ) ϕ( ) In this section, fix k with 2k 1 mod 4 .Supposethat z is a smooth function 2 satisfying ϕ(z + 1) = e 1 ϕ(z).Sinceε 1 n γ = e n ε (γ ), the expression 24 01 24 − ε (γ ) (cz + d) kϕ (γ z) 1 n only depends on the coset ∞γ ,where ∞ := : n ∈ Z . So the series 01 − P(z) := ε(γ )(cz + d) kϕ(γz) γ ∈ ∞\ is well-defined if ϕ is chosen so that it converges absolutely. Each coset of ∞ \ corre- sponds to a pair (c, d) with c ∈ Z and (d, c) = 1. Because of (2.6) the terms corresponding to c and −c are equal. Let m ≡ 1 (mod 24) and define 0 y mx ϕm(z) := ϕ e , (3.1) 24 24 with ϕ0(y) = O(yα) as y → 0forsomeα ∈ R. As in the proof of the convergence of the classical Eisenstein series, we find that the series 1 −k Pm(z) := ε(γ )(cz + d) ϕm(γ z) 2 γ ∈ ∞\ converges absolutely and locally uniformly for k > 2 − 2α. ( ) − ϕ ( ) →∞ The function Pm z m z has polynomial growth as y ,andthefunction x e − (Pm(z) − ϕm(z)) is periodic with period 1. So we have the Fourier expansion 24 − x ( ( ) − ϕ ( )) = ( ) ( ) e 24 Pm z m z a n, y e nx n∈Z with 1 ( ) = − x ( ( ) − ϕ ( )) (− ) a n, y e 24 Pm z m z e nx dx. 0 Using a standard argument (see, for example, [10], Proposition 3.1) we compute the Fourier coefficients a(n, y) as 1 ( ) = − x ε(γ )( + )−kϕ(γ ) (− ) a n, y e 24 cz d z e nx dx 0 γ ∈ ∞\ c>0 1+ d ε(γ ) c −x + d/c − a 1 nd = e z kϕ − e −nx + dx ck d 24 c c2z c (c,d)=1 c c>0 ε(γ ) dm + d(24n + 1) = e k > c 24c c 0 ∗ d mod c ∞ − y mx 1 × z kϕ0 e − − n + x dx. −∞ 24c2|z|2 24c2|z|2 24 Here c∗ indicates that the sum is restricted to residue classes coprime to c,andd denotes γ = a − 1 the inverse of d modulo c. The second equality comes from writing z c c2(z+d/c) and from making the change of variable x → x − d/c. The last equality comes from writing d = d + c with 0 ≤ d < c and gluing together the integrals for each . Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 8 of 16 If we write m = 24m + 1, then by (2.4) we obtain √ dm + d(24n + 1) ε(γ )e = iK(m , n, c), 24c d mod c∗ where K(m, n, c) denotes the Kloosterman sum π ( ) dm + dn K(m, n; c) := e is d,c e . (3.2) c d mod c∗ Therefore we have ( ) = ϕ ( ) + ( ) + 1 Pm z m z c n, y e n 24 x (3.3) n∈Z with √ ∞ K(m , n; c) − y mx c(n, y) = i z kϕ0 e − − n + 1 x dx. ck −∞ 24c2|z|2 24c2|z|2 24 c>0 (3.4) 3.2 Whittaker functions and nonholomorphic Maass-Poincaré series The Poincaré series Pm(z) clearly has the desired transformation behavior; in order to construct harmonic forms, we specialize ϕ0(y) to be a function which has the desired behavior under k. The Whittaker functions Mμ,ν(y) and Wμ,ν(y) are linearly indepen- dent solutions of Whittaker’s differential equation 2 1 μ 1 − 4ν w + − + + w = 0, (3.5) 4 y 4y2 and are defined in terms of confluent hypergeometric functions (see ([21], §13.14) for definitions and properties). Using (3.5) we see after a computation that − k − k y 2 M ( ) k − 1 (4π|n|y)e(nx) and y 2 W ( ) k − 1 (4π|n|y)e(nx) sgn n 2 ,s 2 sgn n 2 ,s 2 are linearly independent solutions of the differential equation ( ) = − k − k − ( ) kF z s 2 1 2 s F z . (3.6) = k At the special value s 2 , we have (by [21], (13.18.2), (13.18.5)) − k −sgn(n) y y 2 M ( ) k k−1 (y) = e 2 (3.7) sgn n 2 , 2 and − y − k e 2 if n > 0, 2 − ( ) = y Wsgn(n) k , k 1 y y (3.8) 2 2 (1 − k, y)e 2 if n < 0. For m ≡ 1 (mod 24) and s ∈ C,define − k ϕ ( ) = π| | y 2 π| | y mx m,s z : 4 m 24 Msgn(m) k ,s− 1 4 m 24 e 24 . (3.9) 2 2 Re(s)− k Then ϕm,s(z) = O y 2 as y → 0, so the series 1 −k Pm(z, s) := ε(γ )(cz + d) ϕm s(γ z) 2 , γ ∈ ∞\ ( )> = k = − k converges for Re s 1. Thus if one of the special values s 2 or s 1 2 is larger than 1, then Pm(z, s) is harmonic at this value. The following proposition describes the Fourier expansion of Pm(z, s). Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 9 of 16 Proposition 8. Let k and Pm(z, s) be as above, and suppose that m ≡ 1 (mod 24) and Re(s)>1. Then we have − k y 2 y nx Pm(z, s) = ϕm,s(z) + gm,n(s)Lm,n(s) 4π|m| W ( ) k − 1 4π|n| e 24 sgn n 2 ,s 2 24 24 n≡1 (24) where √ π −k+ 1 ( ) | / | 2 i 2 2s m n gm,n(s) := + ( ) k s sgn n 2 and ⎧ √ K(m ,n ;c) π |mn| ⎨⎪ − > c J2s 1 6c if mn 0, c>0 √ L (s) := m,n ⎪ K(m ,n ;c) π |mn| ⎩ − < c I2s 1 6c if mn 0. c>0 Here Jα(x) and Iα(x) denote the usual Bessel functions and K(m , n ; c) is defined in (3.2). Proof. In view of (3.9) and (3.1) we take 0 − k ϕ (y) := (4π|m|y) 2 M ( ) k − 1 (4π|m|y). sgn m 2 ,s 2 We write n = 24n + 1. Then (3.3) becomes ( ) = ϕ ( ) + ( ) n Pm z, s m z, s c n , y, s e 24 x . n≡1 (24) Using (3.4) we find that √ K(m , n ; c) c(n , y, s) = i ck c>0 ∞ − k π| | 2 π| | − × −k 4 m y 4 m y mx − nx z Msgn(m) k ,s− 1 e dx. −∞ 24c2|z|2 2 2 24c2|z|2 24c2|z|2 24 (3.10) The integral in (3.10) can be written as − k − 4π|m| y 2 cki k 24 ∞ − k π| | − × y−ix 2 4 m y mx − nx y+ix Msgn(m) k ,s− 1 e dx −∞ 2 2 24c2(x2 + y2) 24c2(x2 + y2) 24 − k − = π| | y 2 k k : 4 m 24 c i I. (3.11) The integral I is computed in ([7], p. 33) in the case m < 0, and the case m > 0is similar. Write x =−yu.Then ∞ k − 2 π = 1 iu 4 m mu + nyu I y M k ,s− 1 e du. −∞ 1 + iu 2 2 24c2y(u2 + 1) 24c2y(u2 + 1) 24 Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 10 of 16 = ny = m Now let A 24 and B 24c2y . By ([18], p. 357) we have ⎧ √ ⎪ 2π |B/A| √ ⎪ ( π| |) − π | | > ⎪ W k ,s− 1 4 A J2s 1 4 AB if A 0, ⎨ s + k 2 2 I = y (2s) · √ 2 ⎪ 2π |B/A| √ ⎪ ( π| |) − π | | < ⎩⎪ W− k ,s− 1 4 A I2s 1 4 AB if A 0, − k 2 2 s 2 √ π √ 2π (2s) |m/n| − | | > = ( π| | y ) · J2s 1 6c √ mn if n 0, Wsgn(n) k ,s− 1 4 n 24 π ( + ( ) k ) 2 2 I − |mn| if n < 0. c s sgn n 2 2s 1 6c Combining this with the expression for m < 0 from ([7], p. 33), we obtain √ π √ 2π (2s) |m/n| − | | > = π| | y · J2s 1 6c √ mn if mn 0, I Wsgn(n) k ,s− 1 4 n 24 π ( + ( ) k ) 2 2 I − |mn| if mn < 0. c s sgn n 2 2s 1 6c Using this with (3.10) and (3.11) we find that √ −k+ 1 2πi 2 (2s) |m/n| − k ( ) = π| | y 2 π| | y c n , y, s 4 m 24 Wsgn(n) k ,s− 1 4 n 24 (s + sgn(n) k ) 2 2 2 √ π ( ) − | | > K m , n ; c J2s 1 6c mn if mn 0, × · π √ c I − |mn| if mn < 0. c>0 2s 1 6c 5 3.3 Derivatives of nonholomorphic Maass-Poincaré series in weight 2 . = 5 = k = 5 We specialize Proposition 8 to the situation k 2 and s 2 4 . Using (3.7) and (3.8) and noting that gm,n(s) = 0forn < 0, we obtain m 3 n 5 = 24 − π | n | 4 5 24 Pm z, 4 q 2 m Lm,n 4 q . (3.12) 0 Since S 5 (ε) ={0},weseethat 2 5 = > Pm z, 4 0form 0. (3.13) In order to construct nontrivial forms when m > 0, we apply the method of [16] and consider the derivative Qm(z) := ∂s| = 5 Pm(z, s). (3.14) s 4 By (3.6) we have 5 ∂sPm(z, s) = (1 − 2s)Pm(z, s). 2 By (3.13) we find that 5 Qm(z) = 0; it follows that 2 ( ) ∈ ! (ε) Qm z H 5 . 2 The following proposition gives the Fourier expansion of Qm(z), and is an analogue of Proposition 4 of [16]. Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 11 of 16 Proposition 9. For m > 0 let Qm(z) bedefinedasin(3.14).Thenwehave − 5 ( ) = π y 4 ∂ | π y − π y mx Qm z 4 m s s= 5 M 5 ,s− 1 4 m W 5 ,s− 1 4 m e 24 4 4 2 24 4 2 24 24 5 m 3 n n 4 − q 24 − 2π Lm n · q 24 2 m , 0 Proof. We compute ∂ | ( )=∂ | ϕ ( ) s s= 5 Pm z, s s s= 5 [ m z, s ] 4 4 − 5 y 4 y nx + ∂s| = 5 gm,n(s)Lm,n(s) · 4πm W ( ) 5 3 4π|n| e s 4 24 sgn n 4 , 4 24 24 n=0 5 5 − 5 + π y 4 ∂ | π| | y nx gm,n Lm,n 4 m s s= 5 Wsgn(n) 5 ,s− 1 4 n e . 4 4 24 4 4 2 24 24 n=0 By (3.12) and (3.13) we see that if m, n > 0, then 5 0ifm = n, L = (3.15) m,n 1 = 4 2π if m n. Therefore the second sum reduces to − 5 y 4 y mx − 4πm ∂s| = 5 W ( ) 5 − 1 4πm e . 24 s 4 sgn n 4 ,s 2 24 24 If n > 0, then by (3.8), the n-th term in the first sum is − 5 m 4 n ∂ | ( ) ( ) 24 s s= 5 gm,n s Lm,n s q . 4 n Using (3.15) we find that 5 1 ∂ | ( ) ( ) =− δ − π| m | 2 L s s= 5 gm,n s Lm,n s m,n 2 m,n 4 2 n δ (where m,n denotes the Kronecker delta function). 5 If n < 0, then gm,n = 0, so the n-thterminthefirstsumis 4 1 5 5 − 5 − π m 2 π y 4 π| | y nx 2 n Lm,n 4 m 24 W− 5 , 3 4 n 24 e 24 2 4 4 4 n 3 5 5 3 y n =−2π| | 4 Lm n − ,4π|n| q 24 . m 2 , 4 2 24 We must evaluate the first two terms which appear in the expansion of Qm(z). Lemma 10. Suppose that k ∈/ Z and that m > 0.Then y − k y y m π 2 ∂ | π − π mx − ( ) 24 4 m 24 s s= k M k ,s− 1 4 m 24 W k ,s− 1 4 m 24 e 24 k q 2 2 2 2 2 m y m = π π − (− )kπ π 24 + (− )k ( ) − − π 24 cot k 1 csc k q 1 k 1 k, 4 m 24 q . (3.16) Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 12 of 16 Proof. By (13.14.33) of [21], we have ∂s| = k M k − 1 (y) − W k − 1 (y) s 2 2 ,s 2 2 ,s 2 ( − ) ( − ) = ∂ | ( ) − 1 2s ( ) − 2s 1 ( ) s s= k M k ,s− 1 y M k ,s− 1 y M k , 1 −s y 2 2 2 (1 − s − k/2) 2 2 (s − k/2) 2 2 (1 − 2s) (2s − 1) = ∂ | − − ( ) − ∂ | − ( ) s s= k 1 M k , k 1 y s s= k M k , 1 k y 2 (1 − s − k/2) 2 2 2 (s − k/2) 2 2 = (1 − k)M k k−1 (y) − (k − 1)M k 1−k (y). 2 , 2 2 , 2 For the first term, we have (by (3.7)) − k − y y 2 M k k−1 (y) = e 2 . 2 , 2 For the second term, we use (13.14.2) and (13.6.5) of [21] to obtain − k k−1 − y y 2 M k 1−k (y) = (−1) (1 − k)e 2 ( (1 − k) − (1 − k, −y)) . 2 , 2 Therefore, the quantity on the left in (3.16) reduces to k m k y m ( −k) − (− ) (k) ( − k) − (k) q 24 +(− ) (k) − k − πm q 24 1 1 1 1 1 , 4 24 . The lemma follows after using the reflection formula (z) (1 − z) = π csc πz, and its logarithmic derivative ( − ) − ( ) = π π 1 z z cot z. 3.4 Proof of Theorem 1 The next Proposition, which follows directly from Proposition 9 and Lemma 10, summa- rizes the two constructions. 5 ( ) Proposition 11. Let Pm z, 4 and Qm z be as defined in (3.12) and (3.14). For negative m ≡ 1 (mod 24),define m 3 n ( ) = 5 = 24 − π | n | 4 5 24 hm z : Pm z, 4 q 2 m Lm,n 4 q , (3.17) 0 √ √ 3 1 4 m 8 n n h (z) := Q (z) =− i πq 24 − π 4 L q 24 m (5 ) m 3 3 m m,n 2 0 ∈ ! (ε) ∈ ! (ε) Then hm H 5 , and for negative m we have hm M 5 . 2 2 Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 13 of 16 Theorem 1 now follows quickly. Proof of Theorem 1. By the discussion following Lemma 7, there can be at most one ! (ε) element of H 5 having a Fourier expansion of the form (3.17) and at most one having an 2 { } ∈ ! (ε) expansion of the form (3.18). By the same discussion we find that hm spans f H 5 . 2 For m > 0wehave 3 3 m 3 5 n 6 2 − n 4 6 2 − ξ 5 hm(z) = π q 24 + 2π Lm,n π| | q 24 . 2 m m 4 n 0>n≡1 (24) 3 6 2 By comparing principal parts, we conclude that ξ 5 hm = π gm. 2 m We have K(m, n; c) = K(m, n; c), which results from a computation using the relation s(−d, c) =−s(d, c). It follows from +( ) = +( ) > (3.18) that the coefficients pm n are real for m n,andthatpm m isnotrealform 0 (cf. (1.3)). 4 Rademacher’s formula for p(n) Writing n = 24n − 1, we obtain √ 3 1 3 3 ( − ) π − 1 6 2 − − 6 2 K 0, n ; c 24n 1 n − ξk(h1(z)) = π q 24 + 2πn 4 π I 3 q 24 . c 2 6c n >0 c>0 The I-Bessel function satisfies − ( ) = 2 x cosh x sinh x = 2 1/2 d sinh x I 3 x / x , 2 π x3 2 π dx x so that ⎛ ⎞ π 2 1 3 − ⎜sinh c 3 n 24 ⎟ π 1 c 2 3 d 2 − 1 = √ ( − ) 4 ⎜ ⎟ I 3 6 3 n 24 24n 1 ⎝ ⎠ . 2 2 2 π 2 dn − 1 n 24 Together with Theorem 1 this gives 3 −1 π 2 η (z) = ξk(h1(z)) 6 ⎛ ⎞ π 2 − 1 √ ⎜sinh c 3 n 24 ⎟ −1/24 1 d ⎜ ⎟ n− 1 = q + √ K(0, −n; c) c q 24 . π ⎝ ⎠ 2 > > dn − 1 n 0 c 0 n 24 n− 1 Equating coefficients of q 24 gives Rademacher’s exact formula ⎛ ⎞ π sinh 2 n − 1 1 √ d ⎜ c 3 24 ⎟ p(n) = √ K(0, −n; c) c ⎜ ⎟ . π ⎝ ⎠ 2 > dn − 1 c 0 n 24 5 Proof of Theorem 3 We proceed as in [16] and [15]. Let F(Y) denote the usual fundamental domain for SL2(Z), truncated at height Y. We have the following (see, for example, [4], Section 9). Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 14 of 16 ∈ 1 Z ∈ ! (ε) ∈ ! (ε) Lemma 12. Suppose that k 2 ,thath H2−k ,andthatg Mk . Then we have 1 +iY k dxdy 2 g(z)ξ2−kh(z) · y = g(z)h(z) dz. F( ) y2 − 1 + Y 2 iY Proof. We have ∂ k dxdy h g(z)ξ − h(z) · y = g(z) dz ∧ dz =−d g(z)h(z)dz . 2 k y2 ∂z The result follows from Stokes’ theorem. For positive n ≡ 1 (mod 24) we write the forms gn from (1.2) as − n j gn(z) = q 24 + cn(j)q 24 . 0 To prove Theorem 3, we compute 1 + 2 iY gn(z)hm(z) dz − 1 +iY 2 ⎛ ⎞ 1 +iY √ √ 3 2 − n j m 8 π 4 k = ⎝ 24 + ( ) 24 ⎠ − 4 π 24 − k L 24 q cn j q 3 i q 3 m m,kq − 1 +iY 2 0 By Lemma 3.4 of [8] we have (for some constant C) √ √ √ C j C k 5 C |k| |cn(j)|e , L e ,and L e . (5.2) m,k m,k 4 From ([21], (8.11.2)) we have − 5 3 π|k|Y π|k|Y 2 − π|k|Y β (|k|Y) = − , ∼ e 6 as |k|Y →∞. (5.3) 2 6 6 From these estimates we conclude that the expression in (5.1) can be integrated term byterm.Alsonotethatforeachpairj, k with j ≡ 23 (mod 24), k ≡ 1 (mod 24),and j =−k,wehave 1 +iY 2 j k q 24 q 24 dz = 0. − 1 + 2 iY Since n = m we obtain 1 + 2 iY gn(z)hm(z) dz − 1 +iY 2 √ 3 π 3 5 =−8 n 4 L − π (| |) k 4 β (| | ) 3 m m,n 2 cn k m Lm,k k Y > ≡ ( ) 4 0 k 1 24 3 + k 4 5 = p (n) − 2π cn (|k|) L β (|k|Y) . m m m,k 4 0>k≡1 (24) Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 15 of 16 By (5.2) and (5.3) we see that the sum goes to zero as Y →∞. So by Lemma 12, (1.4), and Theorem 1 we obtain 3 6 2 + g , g = p (n). πm n m reg m 6 Proof of Theorem 5 ≥ ! (ε) ! (ε) Let 5beprime.OnthespacesM− 1 and H 5 we have the Hecke operators 2 2 2 2 T− 1 ( ) and T 5 ( ).For 2 2 = ( ) n/24 ∈ ! (ε) f a n q M− 1 2 n≡23 (24) we have 2 2 −2 −3n −3 2 n/24 f T− 1 ( ) = a( n) + a(n) + a(n/ ) q . 2 n≡23 (24) ! (ε) There are also Hecke operators on the space H 5 (see, e.g. [13] or ([12], Section 7)). 2 2 For primes ≥ 5, our Hecke operator T 5 ( ) is the same as the usual Hecke operator of 2 2 ! ( ) 12 24 0 index on the space H 5 0 576 , • ,conjugatedby .Inparticular,for 2 01 = ( ) n/24 ∈ ! (ε) g b n q M 5 2 n≡1 (24) we have 2 2 3n 3 2 n/24 g T 5 ( ) = b( n) + b(n) + b(n/ ) q , 2 n≡1 (24) ( 2) ∈ ! (ε) and the action of T 5 on the holomorphic part of a form h H 5 is given by the 2 2 ∈ ! (ε) same formula. For h H 5 we have the relation 2 2 3 2 ξ 5 h T 5 ( ) = ξ 5 h T− 1 ( ) (6.1) 2 2 2 2 (see [12], Proposition 7). > = −m/24 + ( 1/24) ! (ε) For m 0, we have gm q O q .Sincethegm form a basis for M− 1 , 2 comparing principal parts gives the formula ( 2) = −3 + 3m −2 gm T− 1 g 2m gm. (6.2) 2 For m < 0wefindinasimilarwaythat ( 2) = 3 + 3m hm T 5 h 2m hm, 2 which gives the theorem in this case. 2 Suppose that m > 0. By Theorem 1 we have hm T 5 ( ) = cnhn forsomecoefficients 2 cn,fromwhichweobtain 3 2 2 6 ξ 5 hm T 5 ( ) = cn gn. 2 2 πn Ahlgren and Andersen Research in Number Theory (2015) 1:10 Page 16 of 16 On the other hand, the relation (6.1) together with (6.2) gives 3 2 ξ ( 2) = 6 + 3m 5 hm T 5 g 2m gm . 2 2 πm { } Since the gm are linearly independent, we see that the only nonzero coefficients in this = 3 = 3m linear combination are c 2m and cm . The theorem follows. Competing interests The authors declare that they have no competing interests. Acknowledgements We thank Kathrin Bringmann, Jan Bruinier, and Karl Mahlburg for their comments, and we thank the referee for suggestions which improved our exposition. The first author was supported by a grant from the Simons Foundation (#208525 to Scott Ahlgren). Received: 12 March 2015 Accepted: 3 June 2015 References 1. Ahlgren, S: Hecke relations for traces of singular moduli. Bull. Lond. Math. Soc. 44(1), 99–105 (2012). doi:10.1112/blms/bdr072 2. 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