Transformation properties for Dyson's rank function

0.4 MB·

by  Frank Garvan

TRANSFORMATION PROPERTIES FOR DYSON’S RANK FUNCTION
F.G.GARVAN
6 ABSTRACT. Atthe1987RamanujanCentenarymeetingDysonaskedforacoherentgroup-
1
theoreticalstructure for Ramanujan’smocktheta functionsanalogousto Hecke’s theoryof
0
modular forms. Many of Ramanujan’s mock theta functions can be written in terms of
2
R(ζ,q),whereR(z,q)isthetwo-variablegeneratingfunctionofDyson’srankfunctionand
p
ζ isarootofunity.BuildingonearlierworkofWatson,Zwegers,GordonandMcIntosh,and
e
S motivatedbyDyson’squestion,Bringmann,OnoandRhoadesstudiedtransformationprop-
ertiesofR(ζ,q). InthispaperwestrengthenandextendtheresultsofBringmann,Rhoades
8
andOno,andthelaterworkofAhlgrenandTreneer. Asanapplicationwegiveanewproof

] ofDyson’srankconjectureandshowthatRamanujan’sDysonrankidentitymodulo5from
T
theLostNotebookhasananalogueforallprimesgreaterthan3. Theproofofthisanalogue
N
wasinspiredbyrecentworkofJennings-Shafferonoverpartitionrankdifferencesmod7.
.
h
t
a
m
[

2 1. INTRODUCTION
v
1 Letp(n)denotethenumberofpartitionsofn. Therearemanyknowncongruencesforthe
8 partitionfunction. Thesimplestand mostfamouswere foundand provedbyRamanujan:
7
5
0 p(5n+4) 0 (mod 5),
. ≡
1 p(7n+5) 0 (mod 7),
0 ≡
6 p(11n+6) 0 (mod 11).
1 ≡
:
v In 1944, Dyson [14] conjectured striking combinatorial interpretations of the first two con-
i
X gruences. He defined the rank of a partition as the largest part minus the number of parts
r and conjectured that the rank mod 5 divided the partitions of 5n + 4 into 5 equal classes
a
and that the rank 7 divided the partitions of 7n+5 into 7 equal classes. He conjectured the
existence a partition statistic he called the crank which would likewise divide the partitions
of 11n + 6 into 11 equal classes. Dyson’s mod 5 and 7 rank conjectures were proved by
Atkinand Swinnerton-Dyer[7]. Themod11 crank conjecture was solvedbytheauthorand
Andrews[4].
Date:June17,2016.
2010MathematicsSubjectClassification. 05A19,11B65,11F11,11F37,11P82,11P83,33D15.
Key words and phrases. Dyson’s rank function, Maass forms, mock theta functions, partitions, Mordell
integral.
TheauthorwassupportedinpartbyagrantfromtheSimon’sFoundation(#318714).Apreliminaryversion
ofthispaperwasfirstgivenonMay10,2015attheInternationalConferenceonOrthogonalPolynomialsand
th
q-seriesatUniversityofCentralFlorida,Orlando,inhonourofMouradIsmail’s70 birthday.
1
2 F.G.GARVAN
Let N(m,n) denotethenumberofpartitionsofn withrank m. We let R(z,q) denotethe
two-variablegenerating functionfortheDysonrank functionso that
∞
R(z,q) = N(m,n)zmqn.
n=0 m
XX
Throughoutthispaper wewillusethestandardq-notation:
∞
(a;q) = (1 aqk),
∞ −
k=0
Y
(a;q)
(a;q)n = ∞ ,
(aqn;q)
∞
(a ,a ,...,a ;q) = (a ;q) (a ;q) ...(a ;q) ,
1 2 j 1 2 j
∞ ∞ ∞ ∞
(a ,a ,...,a ;q) = (a ;q) (a ;q) ...(a ;q) .
1 2 j n 1 n 2 n j n
Wehavethefollowingidentitiesfortherank generating functionR(z,q):
∞ qn2
(1.1) R(z,q) = 1+
(zq,z 1q;q)
− n
n=1
X
(1.2) = 1 1+ ∞ (−1)n(1+qn)(1−z)(1−z−1) q12n(3n+1) .
(q;q) (1 zqn)(1 z 1qn)
− !
∞ Xn=1 − −
See[16, Eqs(7.2), (7.6)].
Let N(r,t,n) denote the number of partitions of n with congruent to r mod t, and let
ζ = exp(2πi/p). Then
p
p 1
∞ −
(1.3) R(ζ ,q) = N(k,p,n)ζk qn.
p p
!
n=0 k=0
X X
Werestate
Dyson’sRank Conjecture 1.1(1944).
(1.4) N(0,5,5n+4) = N(1,5,5n+4) = = N(4,5,5n+4) = 1p(5n+4),
··· 5
(1.5) N(0,7,7n+5) = N(1,7,7n+5) = = N(6,7,7n+5) = 1p(7n+5).
··· 7
Dyson’s rank conjecture was first proved by Atkin and Swinnerton-Dyer [7]. As noted in
[16], [17], Dyson’srank conjecturefollowsfrom an identityin Ramanujan’s LostNotebook
[32, p.20],[3, Eq. (2.1.17)]. Welet ζ beaprimitive5throot ofunity. Then
5
(1.6) R(ζ ,q) = A(q5)+(ζ +ζ 1 2)φ(q5)+qB(q5)+(ζ +ζ 1)q2C(q5)
5 5 5− − 5 5−
ψ(q5)
(ζ +ζ 1)q3 D(q5) (ζ2 +ζ 2 2) ,
− 5 5− − 5 5− − q5
(cid:26) (cid:27)
where
(q2,q3,q5;q5) (q5;q5) (q5;q5) (q,q4,q5;q5)
A(q) = ∞, B(q) = ∞ , C(q) = ∞ , D(q) = ∞,
(q,q4;q5)2 (q,q4;q5) (q2,q3;q5) (q2,q3;q5)2
∞ ∞ ∞ ∞
DYSON’SRANKFUNCTION 3
and
∞ q5n2 ∞ q5n2
φ(q) = 1+ , ψ(q) = 1+ .
− (q;q5) (q4;q5) − (q2;q5) (q3;q5)
n+1 n n+1 n
n=0 n=0
X X
We recognize the functions A(q), B(q), C(q), D(q) as being modular forms except for a
powerofq. We rewriteRamanujan’sidentity(1.6)in termsofgeneralized eta-products:
η(z) := q214 ∞ (1 qn), q = exp(2πiz),
−
n=1
Y
and
t
η (z) := q2P(r/t) (1 qn),
t,r
−
n>0
Y
n r (modt)
≡±
whereP(t) := t 2 t +1/6and 0 < r < tareintegers. Wehave
{ } −{ }
(1.7) q−214 R(ζ5,q)−(ζ5 +ζ54−2)φ(q5)+(1+2ζ5 +2ζ54)q−2ψ(q5)
η(25z)η (5z) η(25z) η(25z) η(25z)η (5z)
= (cid:0) 5,2 + +(ζ +ζ4) (ζ +ζ4) (cid:1) 5,1 .
η (5z)2 η (5z) 5 5 η (5z) − 5 5 η (5z)2
5,1 5,1 5,2 5,2
Equation (1.6), or equivalently (1.7), give the 5-dissection of the q-series expansion of
R(ζ ,q). We observe that the function on the right side of (1.7) is a weakly holomorphic
5
modularform (withmultiplier)ofweight 1 on thegroupΓ (25) Γ (5).
2 0 ∩ 1
InTheorem1.2belowwegeneralizetheanalogueofRamanujan’sresult(1.7)toallprimes
p > 3. Forp > 3 primeand 1 a 1(p 1)define
≤ ≤ 2 −
∞ qpn2
, if0 < 6a < p,
(qa;qp) (qp a;qp)
(1.8) Φ (q) := n=0 n+1 − n
p,a X ∞ qpn2
 1+ , ifp < 6a < 3p,
− (qa;qp) (qp a;qp)
n+1 − n
n=0
 X

and 

(1.9)
1(p 1)
p(z) := q−214R(ζp,q) χ12(p) 2 − ( 1)a ζp3a+12(p+1) +ζp−3a−21(p+1)
R − −
Xa=1 (cid:16)
ζp3a+12(p−1) ζp−3a−12(p−1) qa2(p−3a)−p224 Φp,a(qp),
− −
(cid:17)
where
1 ifn 1 (mod 12),
12 ≡ ±
(1.10) χ (n) := = 1 ifn 5 (mod 12),
12 
n − ≡ ±
(cid:18) (cid:19) 0 otherwise,
andas usualq = exp(2πiz) with (z) >0. Oneofourmainresultsis

ℑ
4 F.G.GARVAN
Theorem 1.2. Let p > 3 beprime. Then thefunction
η(p2z) (z)
p
R
isa weaklyholomorphicmodularformofweight 1 onthegroupΓ (p2) Γ (p).
0 1
∩
Remark. The form of this result is suggested by Ramanujan’s identity (1.7). The proof of
thisresultusesthetheoryofweakharmonicMaassformsandwasinspiredbyarecentresult
of Jennings-Shaffer [24] on overpartition rank differences mod 7. Jennings-Shaffer was the
firstto provearesultofthistypeusingthetheoryofweak harmonicMaass forms.
As aconsequencewehave
Corollary1.3. Let p > 3 beprimeands = 1 (p2 1). Then thefunction
p 24 −
p 1
∞ ∞ −
(1 qpn) N(k,p,pn s )ζk qn
− − p p
!
nY=1 n=⌈X1p(sp)⌉ Xk=0
isa weaklyholomorphicmodularformofweight 1 onthegroupΓ (p).
1
Remark. Corollary 1.3isthecasem = 0 ofProposition6.2(i)below.
In Ramanujan’s identity(1.6) we see that thecoefficient ofq5n+4 is zero, and this implies
Dyson’s rank 5 conjecture (1.4) in view of (1.3). The analog of (1.6) for the prime 7 does
not appear in Ramanujan’s lost notebook although Ramanujan wrote the left side [32, p.19]
andwrotesomeofthefunctionsinvolvedincodedformon[32,p.71]. Thecompleteidentity
is given by Andrews and Berndt [3, Eq. (2.1.42)]. As noted by the author [17, Theorem 4,
p.20] Ramanujan’s identity (1.6) is actually equivalent to Atkin and Swinnerton-Dyer’s [7,
Theorem 4, p.101]. In Section 6 we give a new proof of Dyson’s Conjecture 1.1 as well as
some of Atkin and Hussain’s [6] results on the rank mod 11 and O’Brien’s [28] results on
therank mod13.
We now explain the connection between Ramanujan’s mock theta functions and Dyson’s
rankfunctions. On[32,p.20]Ramanujangivesfouridentitiesforsomemockthetafunctions
oforder 5. Forexample,
χ (q) = 2+3φ(q) A(q),
0
−
where
∞ qn ∞ q2n+1
χ (q) = = 1+ .
0
(qn+1;q) (qn+1;q)
n n+1
n=0 n=0
X X
The other three identities correspond to [5, Eqs (3.2), (3.6), (3.7)]. These identities are the
Mock Theta Conjectures which were later proved by Hickerson [20]. Thus Ramanujan’s
mockthetafunctionsoforder5arerelatedtoR(z,q)whenz = ζ . AllofRamanujan’sthird
5
ordermockthetafunctionscan also bewritteninterms ofR(z,q). Forexample,
qn2
f(q) = = R( 1,q),
( q;q)2 −
n 0 − n
X≥
q2n(n+1)
ω(q) = = g(q;q2),
(q;q2)2
n 0 n+1
X≥
DYSON’SRANKFUNCTION 5
where
1
(1.11) g(x,q) = x 1 1+ R(x,q) .
−
− 1 x
(cid:18) − (cid:19)
AcatalogueoftheseandanalogousidentitiesforRamanujan’smockthetafunctionsisgiven
byHickersonand Mortenson[21,Section 5].
ThemainemphasisofthispaperistransformationsforDyson’srankfunctionR(ζ,q)and
thusforRamanujan’smockthetafunctions. WebeginwithaquotefromFreeman Dyson.
The mock theta-functions give us tantalizing hints of a grand synthesis still
to be discovered. Somehow it should be possibleto build them into a coher-
ent group-theoretical structure, analogous to the structure of modular forms
which Hecke built around the old theta-functions of Jacobi. This remains a
challengeforthefuture.
Freeman Dyson,1987
Ramanujan Centenary Conference
InthispaperwecontinuepreviousworkonDyson’schallenge. Firstwedescribethegenesis
of this work. Watson [35] found transformation formulas for the third order functions in
termsofMordellintegrals. Forexample,
2π 3α sinhαx
q 1/24f(q) = 2 q4/3ω(q2)+4 ∞e 3αx2/2 dx,
− α 1 1 2π − sinh3αx/2
r r Z0
where
q = exp( α), q = exp( β), αβ = π2.
1
− −
ThederivationoftransformationformulasfortheotherRamanujanmockthetafunctionswas
carried out in a series of papers by Gordon and McIntosh. A summary of these results can
befound in[18].
The big breakthrough came when Zwegers [37], [38] realised how Ramanujan’s mock
thetafunctions occurred as the holomorphicpart of certain real analytic modularforms. An
examplefor thethirdorderfunctionsf(q),ω(q),ω( q) isgivenin
−
Theorem 1.4(Zwegers [37]). DefineF(z) = (f ,f ,f )T by
0 1 2
f (z) = q 1/24f(q), f (z) = 2q1/3ω(q1/2), f (z) = 2q1/3ω( q1/2),
0 − 1 1
−
whereq = exp(2πiz), z h. Define
∈
i (g (τ),g (τ),g (τ))T
G(z) = 2i√3 ∞ 0 1 2 dτ,
i(τ +z)
Z−z −
where p
1
g (τ) = ( 1)n n+ q3(n+1/3)2/2,
0
− 3
n (cid:18) (cid:19)
X
1 1
g (τ) = n+ q3(n+6)2/2,
1
− 6
n (cid:18) (cid:19)
X
6 F.G.GARVAN
1
g (τ) = n+ q3(n+1/3)2/2.
2
3
n (cid:18) (cid:19)
X
Then
H(τ) = F(τ) G(τ)
−
isa (vector-valued)realanalyticmodularformofweight 1/2satisfying
ζ 1 0 0
2−4
H(τ +1) = 0 0 ζ H(τ),
3
 
0 ζ 0
3
 
0 1 0
1 1
H − = 1 0 0 H(τ).
√ iτ τ  
− (cid:18) (cid:19) 0 0 1
−
 
Bringmann and Ono [10], [9], [29] extended this theorem to R(ζ,q) when ζ is a more
general rootofunity.
Theorem 1.5(BringmannandOno,Theorem3.4[10]). Letcbeanoddpositiveintegerand
let 0 < a < c. Then q 1/24R(ζa,q) is the holomorphic part of a component of a vector
− c
valuedweak Maassformofweight 1/2forthefullmodulargroupSL (Z).
2
Remark. By component of a vector valued weak Maass form we mean an element of the
set V defined on [10, p.441]. We give this result explicitly below in Corollary 3.2. The
c
definitionofavectorvaluedMaassform ofweightk forSL (Z) isgivenon [10, p.440].
2
Bringmann and Ono used this theorem to obtain a subgroup of the full modulargroup on
which q 1/24R(ζ,q) is the holomorphic part a weak Maass form of weight 1/2. We state
−
theirtheoremin thecase whereζ is apthrootofunity.
Theorem 1.6(Bringmann and Ono[10]). Let p > 3 beprime,and0 < a < p. Define
θ(α,β;z) := nqn2 q = exp(2πiz),
n α (modβ)
≡ X
a aπ(6m+1) z
Θ ;z := ( 1)msin θ 6m+1,12p, ,
p − p 24
(cid:18) (cid:19) m (Xmod2p) (cid:18) (cid:19) (cid:16) (cid:17)
S a;z := isin πa 2√2p i∞ Θ ap;24p2τ dτ,
1 p − p (cid:16) i(τ +(cid:17)z)
(cid:18) (cid:19) (cid:18) (cid:19) Z−z −
Then
p
a a
D ;z := q p2R ζa;q24p2 S ;z
p − p − 1 p
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:16) (cid:17)
isa weakMaassformof weight1/2 onΓ (576 p4).
1
·
Bringmann,Ono and Rhoades [11]applied thistheorem toprove
Theorem 1.7 (Bringmann, Ono and Rhoades; Theorem 1.1 [11]). Suppose t 5 is prime,
≥
0 r ,r < tand0 d < t. Then thefollowingaretrue:
1 2
≤ ≤
DYSON’SRANKFUNCTION 7
(i) If 1 24d = 1, then
−
t −
(cid:0) (cid:1) ∞
(N(r ,t,tn+d) N(r ,t,tn+d))q24(tn+d) 1
1 2 −
−
n=0
X
isa weight1/2 weaklyholomorphicmodularformonΓ (576 t6).
1
·
(ii) Supposethat 1 24d = 1. Ifr ,r 1( 1 α) (mod t),whereαisanyintegerfor
−t 1 2 6≡ 2 ± ±
which0 α < 2tand 1 24d α2 (mod 2t), then
≤ (cid:0) (cid:1) − ≡
∞
(N(r ,t,tn+d) N(r ,t,tn+d))q24(tn+d) 1
1 2 −
−
n=0
X
isa weight1/2 weaklyholomorphicmodularformonΓ (576 t6).
1
·
Ahlgren andTreneer[1] strengthenedthistheorem toincludethecase24d 1 (mod t).
≡
Theorem 1.8(Ahlgren and Treneer[1]). Supposep 5 isprimeandthat0 r < t. Then
≥ ≤
∞ pn+1 1 pn+1
N r,p, p qn
24 − t 24
n=1(cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
X
isa weight1/2weakly holomorphicmodularformonΓ (576 p4).
1
·
Remark. AhlgrenandTrenneer[1]alsoderivedmanyanalogousresultswhentisnotprime.
InthispaperwestrengthenmanyoftheresultsofAhlgren,Bringmann,Ono,Rhoadesand
Trenneer[1], [10], [11]. In particular
(i) We make Theorem 1.5 (Bringmann and Ono [10, Theorem 3.4] more explicit. See
Theorem3.1below. InthistheoremwehavecorrectedBringmannandOno’sdefini-
tionofthefunctions a;z and (a,b,c;z).
G2 c G2
(ii) For the case p > 3 prime we strengthen Theorem 1.6 (Bringmann and Ono [10,
(cid:0) (cid:1)
Theorem1.2]). WeshowthatthegroupΓ (576 p4)canbeenlargedtoΓ (p2) Γ (p)
1 0 1
· ∩
but with a simplemultiplier. See Corollary 4.2. We provemore. In Theorem 4.1 we
basically determine the image of Bringmann and Ono’s function D a; z under
p 24p2
thegroup Γ (p). (cid:16) (cid:17)
0
(iii) WestrengthenTheorem1.7(i)(Bringmann,OnoandRhoades[11,Theorem1.1(i)]),
and Theorem 1.8 (Ahlgren and Treneer [1, Theorem 1.6, p.271]). In both cases we
enlarge the group Γ (576 p4) to the group Γ (p) but with a simple multiplier. See
1 1
·
Corollary6.5 below.
(iv) We take a different approach to Bringmann, Ono and Rhoades Theorem 1.7(ii). We
handletheresidueclasses 1 24d = 1inadifferentway. Fortheseresidueclasseswe
−
t
showthedefinitionofthefunctionsinvolvedmaybeadjustedtomakethemholomor-
(cid:0) (cid:1)
phic modular forms. In particular see equation (6.5), Theorem 6.3 and its Corollary
6.5.
Thepaperisorganizedasfollows. InSection2wegooverBringmannandOno’stransfor-
mationresultsforvariousLambertseriesandMordellintegrals. Wemakeallresultsexplicit
andcorrectsomeerrors. InSection3wedescribeBringmannandOno’svector-valuedMaass
formofweight 1,correctingsomedefinitionsandmakingallresultsexplicit. InSection4we
2
8 F.G.GARVAN
derive a Maass form multiplier for Bringmann and Ono’s function ℓ;z for the group
G1 p
Γ (p)whenp > 3isprime. InSection5weproveTheorem1.2whiche(cid:16)xtend(cid:17)sRamanujan’s
0
Dyson rank identity (1.6) or (1.7) to all primes p > 3. In Section 6 we give a new proof of
theDysonRankConjecture(1.4)–(1.5). WeextendAhlgren,Bringmann,Ono,Rhoadesand
Treneer’sresults,mentionedin(ii)–(iv)above,forallprimesp > 3. Wealsouseourmethod
to derive Atkin and Hussain’s [6] results for the rank mod 11 and O’Brien’s [28] results for
rank mod 13. In Section 7 we mention the recent results of Hickerson and Mortenson [23],
whohaveadifferentapproach and methodofproofofBringmannand Ono’swork.
2. PRELIMINARIES
Following [10] we define a number of functions. Suppose 0 < a < c are integers, and
assumethroughoutthatq := exp(2πiz). Wedefine
M ac;z := (q;1q) ∞ (−1 1)nqqnn++acac q23n(n+1)
(cid:16) (cid:17) ∞ nX=−∞ −
N a;z := 1 1+ ∞ (−1)n(1+qn) 2−2cos 2πca q21n(3n+1) .
c (q;q) 1 2cos 2πa qn +q2n
(cid:16) (cid:17) ∞ Xn=1 − (cid:0)c (cid:0) (cid:1)(cid:1) !
Forintegers0 a < c, 0 < b < c define (cid:0) (cid:1)
≤
M(a,b,c;z) := (q;1q)∞ nX=∞−∞ 1(−−1)ζncbqqnn++aacc q23n(n+1),
whereζ := exp(2πi/c). In additionfor b 0, 1, 1, 5 define
c c 6∈ { 6 2 6}
0 if0 < b < 1,
c 6
1 if 1 < b < 1,
k(b,c) :=  6 c 2
2 if 1 < b < 5,
 2 c 6
3 if 5 < b < 1,
6 c 6

and 

N(a,b,c;z) := 1 iζ2−caq2bc + ∞ K(a,b,c,m;z)qm(3m2+1) ,
(q;q)∞ 2 1−ζc−aqcb mX=1 
 (cid:16) (cid:17) 
where
K(a,b,c,m;z)
sin πa b +2k(b,c)m πz +sin πa b 2k(b,c)m πz qm
:= ( 1)m c − c c − c − .
− 1 2cos 2πa 2πbz qm +q2m
(cid:0) (cid:0) − (cid:1) (cid:1)c − c(cid:0) (cid:0) (cid:1) (cid:1)
Weneed thefollowingidentities.
(cid:0) (cid:1)
1 ∞ qn2 z ∞ ( 1)nq32n(n+1)
(2.1) 1+ = − ,
− 1 z (zq,z 1q;q) (q) 1 zqn
− n
− Xn=0 ∞ nX=−∞ −
DYSON’SRANKFUNCTION 9
and
(2.2) 1 ∞ (−1)nq32n(n+1) = z ∞ (−1)nqnq32n(n+1).
(q) 1 zqn (q) 1 zqn
∞ nX=−∞ − ∞ nX=−∞ −
Equation(2.1)is [16, Eq.(7.10), p.68]and (2.2)is an easy exercise. Replacing q by qc and z
byqa in (2.1), (2.2)wefind that
∞ qcn2 qa ∞ ( 1)nq32cn(n+1)
(2.3) = 1+ −
(qa;ac) (qc a;qc) (qc;qc) 1 qcn+a
n+1 − n
Xn=0 ∞ nX=−∞ −
a
= 1+qaM ;cz .
c
(cid:16) (cid:17)
By (1.2)we have
a
(2.4) R(ζa,q) = N ;z .
c c
(cid:16) (cid:17)
Extending earlier work of Watson [35], Gordon and McIntosh [19], Bringmann and Ono
[10] found transformation formula for all these functions in terms of Mordell integrals. We
definethefollowingMordellintegrals
a cosh 3a 2 αx +cosh 3a 1 αx
J ;α := ∞e−32αx2 c − c − dx,
c cosh(3αx/2)
(cid:16) (cid:17) Z0 (cid:0)(cid:0) (cid:1) (cid:1) (cid:0)(cid:0) (cid:1) (cid:1)
and
ζbe αx +ζ2be 2αx
J(a,b,c;α) := ∞ e−32αx2+3αxac c − c − dx.
cosh 3αx 3πib
Z−∞ (cid:0) 2 − c (cid:1)
Following[10]weadjustthedefinitionsoftheN-andM-functionssothatthetransformation
(cid:0) (cid:1)
formulasaretidy. We define
(2.5)
a aπ a
1
;z := csc q−24 N ;z ,
N c c c
(2.6) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
a a
;z := 2q32ac(1−ac)−214 M ;z ,
M c c
(2.7) (cid:16) (cid:17) (cid:16) (cid:17)
(a,b,c;z) := 2q32ac(1−ac)−214 M(a,b,c;z),
M
(2.8)
N(a,b,c;z) := 4exp −2πiack(b,c)+3πibc 2ca −1 ζc−bqbck(b,c)−23cb22−214 N(a,b,c;z).
(cid:18) (cid:18) (cid:19)(cid:19)
WenowrestateBringmannand Ono’s[10, Theorem 2.3,p.435]moreexplicitly.
Theorem2.1. Supposethatcisapositiveoddinteger,andthataandbareintegersforwhich
0 a < cand 0 < b < c.
≤
(1) Forz h wehave
∈
a a
(2.9) ;z +1 = ζ 1 ;z ,
N c 2−4 N c
(cid:16) (cid:17) (cid:16) (cid:17)
10 F.G.GARVAN
ζ3b2ζ 1 (a b,b,c;z) ifa b,
(2.10) (a,b,c;z +1) = 2c2 2−4 N − ≥
N ζ3b2 ζ 3bζ 1 (a b+c,b,c;z) otherwise,
(− 2c2 c− 2−4 N −
a
(2.11) ;z +1 = ζ5aζ 3a2ζ 1 (a,a,c;z),
M c 2c 2−c2 2−4 M
(cid:16) (cid:17)
(a,a+b,c;z) ifa+b < c,
M
(2.12) (a,b,c;z +1) = ζ5aζ 3a2ζ 1 a;z ifa+b = c,
M 2c 2−c2 2−4 M c
 (a,a+b c,c;z) otherwise,
M(cid:0) (cid:1) −
whereais assumedtobepositiveinthefirstandthirdformula.

(2) Forz h wehave
∈
1 a 1 a a
(2.13) ; = ;z +2√3√ izJ ; 2πiz ,
√ iz N c −z M c − c −
− (cid:18) (cid:19) (cid:16) (cid:17) (cid:16) (cid:17)
1 1
(2.14) a,b,c; = (a,b,c;z)+ζ 5b√3√ izJ(a,b,c; 2πiz),
√ iz N −z M 2−c − −
− (cid:18) (cid:19)
1 a 1 a 2√3i a 2πi
(2.15) ; = ;z J ; ,
√ iz M c −z N c − z c z
− (cid:18) (cid:19) (cid:16) (cid:17) (cid:18) (cid:19)
1 1 √3i 2πi
(2.16) a,b,c; = (a,b,c;z) ζ 5b J a,b,c; ,
√ iz M −z N − 2−c z z
− (cid:18) (cid:19) (cid:18) (cid:19)
whereagainais assumedtobepositiveinthefirstandthirdformula.
WewillwriteeachoftheMordellintegralsasaperiodintegralofatheta-function. Before
wecan dothisweneed someresultsofShimura[34]. Forintegers0 k < N wedefine
≤
∞ πiz
∼θ(k,N;z) := (Nm+k)exp (Nm+k)2 .
N
m= (cid:18) (cid:19)
X−∞
Wenotethatthiscorrespondstoθ(z;k,N,N,P)inShimura’snotation[34,Eq.(2.0),p.454]
(withn = 1, ν = 1, and P(x) = x). Forintegers0 a,b < c wedefine
≤
6c 1
− π 2πima
Θ (a,b,c;z) := ζ3abζ a ( 1)msin (2m+1) exp − ∼θ(2mc 6b+c,12c2;z),
1 c2 2−c − 3 c −
mX=0 (cid:16) (cid:17) (cid:18) (cid:19)
and
2c 1
− πib
Θ (a,b,c;z) := ( 1)ℓ exp − (6ℓ+1) ∼θ(6cℓ+6a+c,12c2;z)
2
− c
ℓ=0 (cid:18) (cid:18) (cid:19)
X
πib
+( 1)ℓexp − (6ℓ 1) ∼θ(6cℓ+6a c,12c2;z) .
− c − −
(cid:18) (cid:19) (cid:19)
Aneasy calculationgives
(2.17)
∞ n 1 b π 2πina
Θ (a,b,c;z) = 6cζ3abζ a ( 1)n + sin (2n+1) exp −
1 c2 2−c − 3 6 − c 3 c
nX=−∞ (cid:18) (cid:19) (cid:16) (cid:17) (cid:18) (cid:19)