THE M -RANK OF PARTITIONS WITHOUT REPEATED ODD PARTS AS A MOCK 2 MODULAR FORM CHRISJENNINGS-SHAFFER 6 Abstract. While it is known that the M2-rank of partitions without repeated odd parts is the so-called 1 holomorphicpart of acertain harmonic Maass form,much morecan been done with this fact. We greatly 0 improvethestandingofthisfunctionasaharmonicMaassform,inparticularweshowtherelatedharmonic 2 Maass form transforms like the generating function for partitions without repeated odd parts (which is a c modularform). We then usetheseimprovements todetermine formulasforthe rankdifferences modulo7. e Additionally we give identities and formulas that allow one to determine formulas for the rank differences D moduloc,foranyc>2. 3 1 ] 1. Introduction T N Tobeginwerecallthatapartitionofanintegeris anon-increasingsequence ofpositiveintegersthatsum ton. Forexamplethepartitionsof5are5,4+1,3+2,3+1+1,2+2+1,2+1+1+1,and1+1+1+1+1. . h It is standard to let p(n) denote the number of partitions of n. From our example we see that p(5) = 7. t a Partitions have a rich history in number theory and combinatorics, going as far back as Euler. Today much m emphasis is put on Ramanujan’s work with partitions and his related work with q-series. [ One point of interest are the congruences of Ramanujan, 3 p(5n+4) 0 (mod 5), v ≡ p(7n+5) 0 (mod 7), 4 ≡ 7 p(11n+6) 0 (mod 11). 6 ≡ 6 Dysonin[8]proposedthefollowingcombinatorialexplanationofthefirsttwocongruences. Givenapartition 0 of n, we define the rank of the partition to be the largest part minus the number of parts. If we group the . 1 partitionsof5n+4accordingtothemodulo5valueoftheirrank,thenitturnsoutwehavefivesetsofequal 0 size. If we groupthe partitions of7n+5 accordingto the modulo 7 value oftheir rank,then it turns outwe 6 have seven sets of equal size. These two statements were later proved by Atkin and Swinnerton-Dyer in [2]. 1 Another point of interest are mock theta functions, which Ramanujan introduced in his famous last : v letter to Hardy. Mock theta functions are the topic of Watson’s final address as president of the London i X Mathematical Society [21]. We will save the more technical discussion for the next section, but we give two of Ramanujan’s examples: r a q q4 f(q)=1+ + +..., (1+q)2 (1+q)2(1+q2)2 q q4 φ(q)=1+ + +.... (1+q2) (1+q2)(1+q4) One of Ramanujan’s mock theta conjectures (which was provedby Watson in [21]) is that 2φ( q) f(q)= − − ϑ4(0,q) ∞n=1(1+qn)−1, where ϑ4(0,q) is a Jacobi theta function. These two areas are actually strongly connected. In particular, if we let N(m,n) denote the number of Q partitions of n with rank m, we have the generating function given by ∞ ∞ R(ζ;q)= N(m,n)ζmqn. n=0m= X X−∞ 2010 Mathematics Subject Classification. Primary11P81,11P82. Key words and phrases. Numbertheory,partitions,ranks,rankdifferences,maassforms,modularforms. 1 Of the seven functions commonly referred to as the third order mock theta functions, all seven can be expressed in terms of R(ζ;q) by replacing q by a power of q, letting ζ be a power of q times a root of unity, and possibly adjusting by a constant and a power of q. It is worth pointing out that it was Watson [21] who observedthe third orderfunctions could be written in terms of the rank,and he did so nearly a decade before the rank function was formally defined. The fifth and seventh order mock theta functions can also be obtained in such a manner, if one also allows adding on certain infinite products. Explicit versions of these statements can be found in section 5 of [12]. For this reason R(ζ;q) is called a universal mock theta function. With this we see a strong understanding of R(ζ;q) leads to a better understanding of the mock theta functions. Another universal mock theta function is R2(ζ;q), the generating function of the M -rank of partitions 2 without repeated odd parts, which appears among the tenth order mock theta functions. This also has an elegant combinatorial statement. The M -rank of a partition without repeated odd parts is given by 2 taking the ceiling of the largest part divided by 2, and then subtracting the number of parts. This rank was introduced by Berkovich and Garvan in [3] and further studied by Lovejoy and Osburn in [16]. We let N (m,n) denote the number of partitions of n without repeated odd parts and M -rank m. We denote the 2 2 generating function by ∞ ∞ R2(ζ;q)= N (m,n)ζmqn. 2 n=0m= X X−∞ InthisarticleweimprovetheunderstandingoftheM -rankastheholomorphicpartofaharmonicMaass 2 form. We determine a more precise formula for the transformation of the associated harmonic Maass form anddo soonalargergroupthanpreviouslyknown. Toaidindetermining dissectionformulasR2(e2πℓi;q)= ℓr−=10qrAr(qℓ),wefindadditionalharmonicMaassformswithholomorphicpartscorrespondingtothemock modularpartsoftheA (qℓ). Additionallywegivelowerboundsontheordersofthecuspsfortheholomorphic r P parts of the harmonic Maass forms. PartitionrankshavebeenstudiedintermsofharmonicMaassformsinanumberofrecentworks,perhaps the most influential being [5] and other important works are [1, 6, 7, 17] of which the second and third deal withtheM -rank. RecentlyGarvan[9]hasmaderemarkableimprovementstothetransformationsandlevels 2 from [5] for the harmonic Maass forms related to R(ζ;q). A key observation in Garvan’s work is that on a ratherlarge subgroup,the harmonic Maassformrelatedto R(ζ;q) has the same multiplier asη(τ) 1, which − isthegeneratingfunctionforpartitions. Thesamephenomenonwilloccurhere. Thatistosaytheharmonic Maass form related to R2(ζ;q) will transform like η(2τ) , which is the generating function for partitions η(4τ)η(τ) without repeated odd parts. This also occurs with the Dyson rank of overpartitions, R(ζ;q), which the author studied in [13]. Although not stated explicitly in that article, one can easily verify that on a certain subgroup the transformationformula for the Dyson rank of overpartitions agrees with η(2τ), the generating η(τ)2 function for overpartitions. With an improved understanding of the M -rank, we give new identities for 2 R2(ζ;q) when ζ is set equal to certain roots of unity. In the next section we formally state our definitions and results, and then give an outline of the rest of the article. For the reader thatwishes to comparethe variousappearancesof the the three rankfunctions mentioned here, we note the following. In the notation of Gordon and McIntosh [10], we have R(ζ;q) = h (ζ,q) = 3 (1 ζ)(1 + ζg (ζ,q)), R(ζ;q) = (1 ζ)(1 ζ 1)g (ζ,q), and R2(ζ;q) = h (ζ, q). In the notation of 3 − 2 2 − − − − HickersonandMortenson[12], we have R(ζ;q)=(1 ζ)(1+ζg(ζ,q)), R(ζ;q)=(1 ζ)(1 ζ 1)h(ζ,q), and − − − − R2(ζ;q)=(1 ζ)ζ12k(ζ21, q). − − 2. Statement of Results As in the introduction we let N (m,n) denote the number of partitions of n without repeated odd parts 2 with M -rank m. Furthermore we let N (k,c,n) denote the number of partitions of n without repeated 2 2 odd parts with M -rank congruent to k modulo c. To state out definitions and results we use the standard 2 product notation, n 1 − ∞ (ζ;q) = (1 ζqj), (ζ;q) = (1 ζqj), n − − ∞ j=0 j=0 Y Y 2 (ζ ,...,ζ ;q) =(ζ ;q) ...(ζ ;q) , (ζ ,...,ζ ;q) =(ζ ;q) ...(ζ ;q) . 1 k n 1 n k n 1 k 1 k ∞ ∞ ∞ Also we let q = exp(2πiτ) for τ , that is Im(τ) > 0. For c a positive integer and a an integer, we let ∈ H ζa =exp(2πia/c). c As noted in [16], one may use Theorem 1.2 of [15] to find that the generating function for N (m,n) is 2 given by ∞ ∞ ∞ qn2 q;q2 R2(ζ;q)= N (m,n)ζmqn = − n 2 (ζq2,ζ 1q2;q2) n=0m= n=0 (cid:0)− (cid:1) n X X−∞ X = (−q2q;;qq22)∞ 1+ ∞ (1−ζ)((11−ζζ−q12)n()−(11)nqζ2n12q+2nn()1+q2n) . (cid:0) (cid:1) n=1 − − − ! ∞ X Next for a and c integers, c>0, and c∤2a, we define R2(a,c;τ)=q−18R2(ζca;e2πiτ), i g (4w)dw i g (4w)dw R2(a,c;τ)=R2(a,c;τ)−i(1−ζca)ζ2−ca eπ4i Z−τ∞ 43,21−−i2c(aw+τ) +e−π4i Z−τ∞ 41,12−−i2c(aw+τ) !, wherefg (τ) is a theta function of Zwegers [23], the depfinition of which we give in tphe next section. The a,b factor of q−81 is necessary for certain transformations to work correctly. We notice that it is immediately clear that 2(a +c,c;τ) = 2(a,c;τ); it is also true that 2(a +c,c;τ) = 2(a,c;τ) and this follows R R R R immediately from properties of g (τ). a,b We will find R2, 2, and 2 are related to the following fufnctions. For k anfd c integers, 0 k < c, we R R ≤ define f ( 1)c+21q−(c−k)(2c−2k−1) ∞ ( 1)nq2c2n(n+1)−cn − − if c is odd, S(k,c;τ)= ((cid:0)−qc1,)q2c4qc2−−(cc−,qk4)(c22c;−q24kc2−(cid:1)1∞) n∞=X−∞ 1(−−1q)4ncq22nc+2n(2(kn−+c1))2c if c is even, 1, q4c2,q4c2;q4c2 1 q4c2n+(1+4k 2c)c − − n= − −  (cid:0) ((cid:1)∞1)cX+2−1i∞q−(1+4k8−4c)2 ∞ ( 1)nq2c2n(n+1)−cn − − if c is odd, S(k,c;τ)=ic2q−18S(k,c;τ)= (cid:0)qc(,−q41c)22c−qc−,q(14+c24k;8−q44cc)22(cid:1)∞ n=X∞−∞ 1−(−q14)cn2nq2+c(22nk(−nc+)12)c if c is even, 1, q4c2,q4c2;q4c2 1 q4c2n+(1+4k 2c)c (k,c;τ)= (k,c;τ) ( 1)ki1+ce(cid:0)−−π4ic− i∞ g1+4c4k,2c(4(cid:1)c∞2wn)=Xd−w∞. − − S S − − Z−τ −i(w+τ) e p We will see in Section 3 that we can define the functions (k,c;τ) and (k,c;τ) for all integer values of k, S S however some care must be taken in doing so as the above definitions would not be the proper choice and they do not satisfy the properties that (k+c,c;τ)= (k,c;τ) and (ke+c,c;τ)= (k,c;τ). S S S S Our main theorem is the following. This theorem relates the M -rank generating function to harmonic 2 Maass forms and modular forms. e e Theorem 2.1. Suppose a and c are integers, c>0, and c∤2a. Let t= c . gcd(4,c) (1) 2(a,c;8τ) is a harmonic Maass form of weight 1/2 and 2(a,c;8τ) is a mock modular of weight R R 1/2 on Γ (256t2) Γ (8t). 0 1 ∩ (2) Tfhe function c 1 − 2(a,c;τ) i c(1 ζa)ζ a ( 1)k(ζ 2ak ζ2ak+a) (k,c;τ) R − − − c c− − c− − c S k=0 X f 3 e is holomorphic in τ and has at worst poles at the cusps. Furthermore, c 1 − 2(a,c;τ) i c(1 ζa)ζ a ( 1)k(ζ 2ak ζ2ak+a) (k,c;τ) R − − − c c− − c− − c S k=0 X f c 1 e − = 2(a,c;τ) i c(1 ζa)ζ a ( 1)k(ζ 2ak ζ2ak+a) (k,c;τ). R − − − c c− − c− − c S k=0 X (3) The function c 1 η(4τ)η(τ) − η(2τ) R2(a,c;τ)−i−c(1−ζca)ζc−a (−1)k(ζc−2ak−ζc2ak+a)S(k,c;τ)! k=0 X is a weakly holomorphic modular form of weight 1 on Γ (4c2 gcd(c,2)) Γ (4c). 0 1 · ∩ (4) If c is odd then the function η(4c2τ)η(c2τ) c−1 2(a,c;τ) i c(1 ζa)ζ a ( 1)k(ζ 2ak ζ2ak+a) (k,c;τ) η(2c2τ) R − − − c c− − c− − c S ! k=0 X is a weakly holomorphic modular form of weight 1 on Γ (4c2) Γ (4c). 0 1 ∩ Part(1)ofTheorem2.1 is notentirelynew. In[6]Bringmann,Ono,andRhoadesstudied variousgeneral harmonic Maass forms, one of which has essentially R2(ζa,8t2z) as the holomorphic part. However in this c case the subgroup is Γ (210t4). Our construction of the harmonic Maass forms greatly differs from that of 1 [6]. Additionally, Hickerson and Mortenson in [11] considered the function D (a;M)= ∞ N (r,M,n) po(n) qn, 2 2 − t nX=0(cid:16) (cid:17) where p (n) is the number of partitions without repeated odd parts. There they showed that D (a;M) can o 2 be expressed as the sum of two Appell-Lerch sums of a form similar to S(k,c;τ) and a theta function. In Section 5 we give various formulas so that we may determine the orders of these modular forms at cusps. With this we canthen verifyvariousidentities for the M -rankfunction by passingto modular forms 2 and using the valence formula. In particular one can use this to give an exactdescription of the c-dissection of R2(ζa;q) in terms of generalized Lambert series and modular forms. We give these identities for c = 7. c Similar formulas were determined by Lovejoy and Osburn for c = 3 and c = 5 in [16] and for c = 6 and c=10 by Mao in [18]. Rather than using harmonic Maass forms, those formulas use the q-series techniques developedbyAtkinandSwinnerton-Dyer[2]todeterminerankdifferenceformulasfortherankofpartitions. Furthermore in [11] Hickerson and Mortenson demonstrated how their identities for D (a;M) can be used 2 to prove the formulas for c =3 and c = 5. With this in mind, we see that there are at least three different proof techniques for determining the c-dissection of R2(ζa;q). One might ask for the pros and cons of the c approachwe use in this article. Our approachwith harmonic Maass forms shares a difficulty with the other two methods, which is that we must correctlyguess the identity before we canprove it. This approachwith harmonic Maass forms helps us to guess the identity in that we know the weight and level of the modular forms. Theorem 2.2. Let ζ be a primitive seventh root of unity, J = qa,q28 a;q28 for 1 a 14, J = 7 a − ≤ ≤ 0 q28;q28 , and A(w,x,y)=wζ +xζ2+yζ3+yζ4+xζ5+wζ6. Then ∞ 7 7 7 7 7 7 (cid:0) (cid:1) ∞ (cid:0) R2(cid:1)(ζ7;q)=R20(q7)+qR21(q7)+q2R22(q7)+q3R23(q7)+q4R24(q7)+q5R25(q7)+q6R26(q7). Hereeach R2 (q) can be written as a sum of terms with S(k,7;τ) and quotients of J . Duetothe complexity i 7 a of the expressions, we only state the definition of R2 (q) here and give the definitions of the other R2 (q) in 0 i Section 6. We have R2 (q)=A( 3, 2, 2)S(0,7;τ/7)+A( 3, 2, 2)S(3,7;τ/7)+ A(11,15,8)q−3J0J7J82 0 − − − − − − J12J22J32J42J52J9J11J12J123J124 + J1A2J(2−2J1321J,−42J1532,J−109J)q12−1J31J20JJ1273JJ8214 + J12JA22(J132,2J,420J)q52−J392JJ01J1J721J228JJ12130J14 + J12AJ22(9J,321J14,28J)52qJ−93JJ1201JJ7212J28J123 + J12AJ(22−J1324J,42−J1526J,−92J1100)Jq1−213JJ102JJ721J3J8214 + J12JA22(J−32J1,421J,−52J37)qJ−103JJ101JJ61J282JJ1239J124 + J12AJ22(−J326J,−42J4,52−J47)Jq12−0J3J1210JJ162JJ8212J392J14 + AJ(−12J1222J,−32J9,42−J652)Jq12−13JJ120JJ1623JJ91J2410 4 + AJ(1−2J722,−J329J,−42J4)52qJ−93JJ1202JJ6123JJ81J2410 + AJ(12−J2226J,3−2J1429J,−52J1161)Jq−1223JJ1203JJ61J48 + J12AJ22(8J,326J,542)Jq52−J31J00JJ1261JJ812J29J123 + J12J22AJ(323J542,3J452,J292J)q1−0J31J10JJ162JJ8213J124 + JA12(J−222J132,J−421J952,−J12101J)12q1−J31J20JJ136JJ1824 + JA12(J−223J,32−J342,−J521J)q8−J1231JJ012J26JJ1273JJ12104 + J1A2J(2326J,322J442,2J152)Jq9−J312J10JJ162JJ71J31J0124 + JA12(J1262,J1327J,1420J)52qJ−130JJ012J16JJ1273JJ11224 + AJ(12−J522,J−322J,4−2J352)Jq−92J3J100JJ1263JJ712J48 + JA12J(−22J33,2−J442,J−523J)92qJ−131JJ012J26JJ137JJ184 + JA12J(−22J83,2−J642,J−525J)9qJ−1230JJ01J1J6J1237JJ184 + J12AJ(222J732,2J442,J1952)Jq9−J31J00JJ1216JJ1722JJ813 + A(J−128J,22−J432,J−425J)5q2−J1221JJ012J26JJ1273JJ812J410 + J12JA22J(332,J14,22J)52qJ−923JJ1200JJ612J17JJ1822J14 + AJ(12−J1222J,32−J842,J−526J)q8−J923JJ101JJ612J372JJ12140 + J12AJ(22−J932,J−426J,−52J68)qJ−92J3J1210JJ1262JJ721J312J014 + J12AJ(22−J132,−J423J,−52J38)qJ−9J31J210JJ1263JJ7214 + J12JA22(J532,2J,423J)q52−J392JJ01J06JJ12172JJ1132J124 + J1A2J(2121J,326J,642)Jq52−J292JJ012J16JJ12722JJ81J31J0124 + J12AJ22(−J326J,−42J45,2−J392)Jq12−02JJ110JJ162JJ7212J382J14 + J12AJ(226J,432,J442)qJ−523JJ720JJ12622JJ1823JJ9124 + J12JA22(J3324J,2429J,522J27)2qJ−103JJ101JJ6212J28JJ12923J14 + A(−6,−4,−4)q−3J0J62J82J9 + A(−6,−4,−4)q−3J0J62J9J120 + A(3,2,2)q−3J0J62J92J10 J12J22J32J42J52J72J120J11J12J13J124 J12J22J32J42J52J7J8J11J122J123J124 J12J22J32J42J52J7J8J121J122J123J14 + JA12(J−222J832,−J422J55,2−J17J8)12q1−J132JJ01J362JJ1294 + AJ(−12J2227J,−32J2642,J−521J57)Jq12−03JJ1203JJ62124J8 + J12JA22J(532,J54,21J)52qJ−73JJ1200JJ6212J18JJ1292J13 + J12JA22(J132,−J421J,052)Jq8−J312J10JJ12622JJ1130J14 + J12J22AJ(322J,442,J15)2qJ−823JJ90JJ1612JJ172JJ11023J14 + J12JA22(J632,4J,424J)q52−J382JJ012J1J621J27J123 + J12JA22J(332,J242,2J)52qJ−1210JJ012J16J21J27JJ12823J124 + J1A2J(2−2J832,−J428J,−5J56)qJ−1213JJ102JJ8212J31J0124 + A(6,4,4)q−3J0J82J92 + A(2,1,1)J0J62J7J82 + A(3,1,2)qJ0J62J7J82 . J12J22J32J42J5J72J121J12J123J124 J12J22J32J42J5J9J120J121J12J123J124 J12J22J32J42J92J120J121J12J123J124 It is trivial to verify that the qαS(k,7;τ/7) appearing in the definitions of the R2 (q) are all series with i integralpowersof q. One advantageto writing the identity in this formis thatwe canalsoreadoffformulas involving the rank differences, ∞ R2 (d;q)= (N (r,c,cn+d) N (s,c,cn+d))qn. r,s,c 2 2 − n=0 X To explain this, we let ζ =ζa and use that 1+ζ +ζ2+ +ζc 1 =0 to find that c c c c ··· c− c 1 c 1 c 1 ∞ − − − R2(ζ ;q)= N (r,c,n)ζrqn = ζr qdR2 (d,qc). c 2 c c r,0,c n=0r=0 r=1 d=0 XX X X Since N (m,n)=N ( m,n), we know R2 (d,qc)=R2 (d,qc). When c is odd we have 2 2 r,0,c c r,0,c − − c−21 c−1 R2(ζc;q)= (ζcr+ζcc−r) qdR2r,0,c(d;qc), r=1 d=0 X X whereas for even c we have R2(ζc;q)=ζc2c c−1qdR2c,0,c(d;qc)+ c−22(ζcr +ζcc−r)c−1qdR2r,0,c(d;qc). 2 d=0 r=1 d=0 X X X In the case of c being an odd prime, we have that ζ , ζ2, ..., ζc 1 are linearly independent over Q, so if c c c− c−21 c−1 R2(ζ ;q)= (ζr +ζc r) qdS (d;qc), c c c− r r=1 d=0 X X andeachS (d;q) is aseriesinq withrationalcoefficients,then S (d;q)=R2 (d;q). As anexample,from r r r,0,c the formula for R2 (q), we have that 0 R21,0,7(0;q)=−3S(0,7;τ/7)−3S(3,7;τ/7)+ J12J22J321J142qJ−523JJ90JJ171JJ8122J123J124 − J12J22J32J1421Jq−52J31J00JJ1271JJ8212J123J14 + q−3J0J72J8J10 + 9q−3J0J72J8 14q−3J0J72J82 J12J22J32J42J52J92J11J122J123J14 J12J22J32J42J52J9J121J122J123 − J12J22J32J42J52J92J10J121J12J13J14 + . ··· WenoteitisinfactmoreconvenienttoverifytheidentitiesfortheR2 (d;q),asthenweareworkingwith r,0,7 integer coefficients rather than coefficients from Z[ζ ]. When c is composite, we can also deduce an identity 7 based on the minimal polynomial for ζ . c The rest of the article is organizedas follows. In Section 3 we recall the basics of modular and harmonic Maass forms and introduce the functions studied by Zwegers in [23]; these functions allow us to relate our 5 functions to harmonic Maass forms. In Section 4 we work out the transformationformulas for our functions andproveTheorem2.1. InSection5wegiveformulasfortheordersatcusps. InSection6weusetheresults of Sections 4 and 5 to prove Theorem 2.2. We end by giving a few remarks in Section 7. 3. Modular Forms, Harmonic Maass Forms, and Zwegers’ Functions We first recall some basic terminology and results for modular and Maass forms. For further details, one may consult [5, 19, 20, 23]. We have that SL (Z) is the multiplicative group of 2 2 integer matrices with 2 × determinant 1. The principal congruence subgroup of level N is α β Γ(N)= SL (Z):α δ 1 (mod N),β γ 0 (mod N) . γ δ ∈ 2 ≡ ≡ ≡ ≡ (cid:26)(cid:18) (cid:19) (cid:27) A subgroup Γ of SL (Z) is called a congruence subgroup if Γ Γ(N) for some N. Three congruence 2 ⊇ subgroups we will use are α β Γ (N)= SL (Z):γ 0 (mod N) , 0 γ δ ∈ 2 ≡ (cid:26)(cid:18) (cid:19) (cid:27) α β Γ (N)= SL (Z):α δ 1 (mod N),γ 0 (mod N) , 1 γ δ ∈ 2 ≡ ≡ ≡ (cid:26)(cid:18) (cid:19) (cid:27) α β Γ0(N)= SL (Z):β 0 (mod N) . γ δ ∈ 2 ≡ (cid:26)(cid:18) (cid:19) (cid:27) WerecallSL (Z)actson viaMobiustransformations,thatisifA= αβ thenAτ = ατ+β. Additionally 2 H γ δ γτ+δ we let j(A,τ)=γτ +δ. One can verify that j(AB,τ)=j(A,Bτ) j(B(cid:16),τ).(cid:17) · We recallaweaklyholomorphicmodularformofintegralweightk ona congruencesubgroupΓ ofSL (Z) 2 is a holomorphic function f on such that H (1) if A= αβ Γ, then f(Aτ)=(γτ +δ)kf(τ), γ δ ∈ (2) if B ∈(cid:16)SL2(Z(cid:17)) then (B : τ)−kf(Bτ) has an expansion of the form ∞n=n0anexp(2πinz/N), where n Z and N is a positive integer. 0 ∈ P When k is a half integer, we require Γ Γ (4) and replace (1) with f(Aτ) = γ 2kǫ(δ) 2k(γτ +δ)kf(τ). ⊂ 0 δ − Here m is the Jacobi symbol extended to all integers n by n (cid:0) (cid:1) (cid:0) (cid:1) 0 =1, 1 (cid:18)± (cid:19) m m if m>0 or n>0, n = | | n  (cid:16) m(cid:17) if m<0 and n<0, (cid:16) (cid:17)  − n | | (cid:16) (cid:17) and ǫ(δ) is 1 when δ 1 (mod 4) and is i otherwise.  ≡ A harmonic weak Maass form satisfies the transformationlaw in (1), but the condition of holomorphic is replaced with being smooth and annihilated by the weight k hyperbolic Laplacian operator, ∂2 ∂2 ∂ ∂ ∆ = y2 + +iky +i , k − ∂x2 ∂y2 ∂x ∂y (cid:18) (cid:19) (cid:18) (cid:19) whereτ =x+iy,andcondition(2)isreplacedwith(B :τ) kf(Bτ)havingatmostlinearexponentialgrowth − as τ i . We note that with ∂ = 1( ∂ i ∂ ) and ∂ = 1( ∂ +i ∂ ), we have ∆ = 4y2 k ∂ yk ∂ . → ∞ ∂τ 2 ∂x − ∂y ∂τ 2 ∂x ∂y k − − ∂τ ∂τ If f is a harmonic weak Maass form of weight 2 k, with k > 1, on Γ (N) and f satisfies additional 1 − growth conditions, then f can be written as f =f++f , where f+ and f have expansions of the form − − ∞ ∞ f+(τ)= a(n)qn, f−(τ)= b(n)Γ(k 1,4πny)q−n. − nX=n0 nX=1 Here Γ is the incomplete Gamma function given by Γ(y,x) = x∞e−tty−1dt. We call f+ the holomorphic part and f the non-holomorphic part. The non-holomorphic part is often written instead as an integralof − R 6 the form f−(τ)= ∞g(w)( i(w+τ))k−2dw, − Z−τ whereg =ξ (f)andξ =2iyk ∂ . Whenf isaharmonicMaassformonsomeothercongruencesubgroup, 2 k k ∂τ − similar expansions exist, but with q replaced by a fractional power of q. WenowrecallthevariousfunctionsneededfortheMaassforms. Foru,v,z C, τ ,andu,v Z+τZ ∈ ∈H 6∈ we let 1 ϑ(z;τ)= exp πin2τ +2πin z+ = iq81e−πiz e2πiz,e−2πizq,q;q , 2 − n∈XZ+21 (cid:18) (cid:18) (cid:19)(cid:19) (cid:0) (cid:1)∞ exp(πiu) ∞ ( 1)nexp(πin(n+1)τ +2πinv) µ(u,v;τ)= − . ϑ(v;τ) 1 exp(2πinτ +2πiu) n= − X−∞ Next for z C, y =Im(τ), and a=Im(z)/Im(τ) we define ∈ z E(z)=2 exp( πw2)dw, − Z0 R(z;τ)= sgn(n) E((n+a) 2y) ( 1)n−12 exp( πin2τ 2πinz). − − − − n∈XZ+12(cid:16) p (cid:17) We remark that R(z;τ) is not related to the rank generating function R(ζ;q) discussed in the introduction. For a,b R we set ∈ g (τ)= nexp(πin2τ +2πinb). a,b n Z+a ∈X Finally for u,v / Z+τZ we set ∈ i µ˜(u,v;τ)=µ(u,v;τ)+ R(u v;τ). 2 − InhisrevolutionaryPhDthesis[23],Zwegersstudiedthese functionsandgavetheirtransformationformulas which are essential to our results. Proposition 3.1. Let ζ =exp(πiu), then R2(ζ;τ)=i(1 ζ) ζ−1µ(u, τ;4τ) µ(u,τ;4τ) . − − − Proof. We have (cid:0) (cid:1) R2(ζ;τ)= (−q2q;;qq22)∞ 1+ ∞ (1−ζ)((11−ζζ−q12)n()−(11)nqζ2n12q+2nn()1+q2n) (cid:0) (cid:1) n=1 − − − ! ∞ X = (cid:0)(−q2q;;qq22)(cid:1)∞∞ 1+nX∞=1(−1)nq2n2+n(cid:18)(1(1−−ζqζ2)n) + (1(1−−ζ−ζ1−q12)n)(cid:19)! q;q2 ∞ ( 1)nq2n2+n(1 ζ)(1+ζq2n) = (−q2;q2)∞ − (1 ζ−2q4n) (cid:0) (cid:1) n= − ∞ X−∞ q;q2 ∞ ( 1)nq2n2+n q;q2 ∞ ( 1)nq2n2+3n =(1−ζ) (−q2;q2)∞ (−1 ζ2q4n) +(1−ζ)ζ (−q2;q2)∞ −(1 ζ2q4n) . (cid:0) (cid:1) n= − (cid:0) (cid:1) n= − ∞ X−∞ ∞ X−∞ We note that q;q2 q2;q4 1 − = = , (q2;q2)∞ (q,q2;q2∞) (q,q3,q4;q4) (cid:0) (cid:1) (cid:0) (cid:1) ∞ ∞ ∞ and ϑ(τ;4τ)= i q,q3,q4;q4 . − ∞ (cid:0)7 (cid:1) Thus q;q2 ∞ ( 1)nq2n2+3n µ(u,τ;4τ)=iζ (−q2;q2)∞ −1 ζ2q4n , (cid:0) (cid:1) n= − ∞ X−∞ and q;q2 ∞ ( 1)nq2n2+n µ(u,−τ;4τ)=−iζ (−q2;q2)∞ −1 ζ2q4n . (cid:0) (cid:1) n= − ∞ X−∞ With this we see R2(ζ;τ)=iζ 1(1 ζ)µ(u, τ;4τ) i(1 ζ)µ(u,τ;4τ). − − − − − (cid:3) We note that we could also use functions from [22], rather than µ(u,v;τ). In particular, the modular transformation and elliptic properties of functions of the form eπiu ∞ (−1)mnqmn(2n+1)e2πinv, 1 e2πiuqn n= − X−∞ for m Z are well understood. ∈ Proposition 3.2. For a and c integers, c>0, and c∤2a, we have that R2(a,c;τ)=i(1−ζca) ζc−aq−81µ 2ca,−τ;4τ −q−81µ 2ca,τ;4τ , R2(a,c;τ)=i(1−ζca)(cid:16)ζc−aq−81µ˜(cid:0)2ca,−τ;4τ(cid:1)−q−81µ˜(cid:0)2ca,τ;4τ(cid:1)(cid:17). Proof. Weseethatthefirstidentityisatriv(cid:16)ialcorollar(cid:0)yofthepre(cid:1)viouspro(cid:0)position.(cid:1)F(cid:17)orthesecondidentity, f given the definitions of 2 and µ˜, we see we must show that R i(1−ζca)fζc−aq−182iR 2ca +τ;4τ −q−812iR 2ca −τ;4τ (cid:16) (cid:0) i g (cid:1) (4w)dw(cid:0) i(cid:1)(cid:17)g (4w)dw =−i(1−ζca)ζ2−ca eπ4i Z−τ∞ 43,21−−i2c(aw+τ) +e−π4i Z−τ∞ 41,12−−i2c(aw+τ) !. However, by Theorem 1.16 of [23] we have thapt p i g (z)dz i g (4w)dw R(2ca −τ;4τ)=−exp(cid:0)4π16iτ + 24πi(−c2a + 21)(cid:1)Z−4∞τ 41,−21i−(z2ca+4τ) =−2q81eπ4iζ2−caZ−τ∞ 41,12−−i2c(aw+τ) . Similarly we have p p i g (z)dz i g (4w)dw R(2ca +τ;4τ)=−exp(cid:0)4π16iτ − 24πi(−c2a + 12)(cid:1)Z−4∞τ 43,−12i−(z2ca+4τ) =−2q81e−4πiζ2acZ−τ∞ 43,21−−i2c(aw+τ) . Thus the proposition follows. p p (cid:3) Proposition 3.3. Suppose k and c are integers and 0 k <c. Then ≤ (k;c;τ)= q−(1+4k8−2c)2µ((2k−c)2cτ,c−21 −cτ;4c2τ) if c is odd, S ( q−(1+4k8−2c)2µ((1+4k−2c)cτ,c−21;4c2τ) if c is even, (k;c;τ)= q−(1+4k8−2c)2µ˜((2k−c)2cτ,c−21 −cτ;4c2τ) if c is odd, S ( q−(1+4k8−2c)2µ˜((1+4k−2c)cτ,c−21;4c2τ) if c is even. e Proof. When c is odd, we have that q−(1+4k8−2c)2µ (2k−c)2cτ,c−21 −cτ;4c2τ = q−ϑ((1c+41k8−2c)c2τ+;(42ck2−τc))c ∞ (1−1)nq4qc22cn2+n((n2k+1)c−)2ccn −2 − n= − − (cid:0) (cid:1) X−∞ 8 ( 1)c+21iq−(1+4k8−4c)2 ∞ ( 1)nq2c2n(n+1)−cn = − − qc,q4c2 c,q4c2;q4c2 1 q4c2n+(2k c)2c − n= − − ∞ X−∞ = (cid:0)(k,c;τ). (cid:1) S When c is even, we instead have that q−(1+4k8−2c)2µ (1+4k−2c)cτ,c−21,4c2τ = q−(1+ϑ4k(8−c2c1)2;+4c(12+τ4)k2−2c)c ∞ 1 q4qc22cn2n+((n1++14)k 2c)c −2 n= − − (cid:0) (cid:1) X−∞ ( 1)2cq−(1+4k8−4c)2 ∞ q2c2n(n+1) = − 1, q4c2,q4c2;q4c2 1 q4c2n+(1+4k 2c)c − − ∞ n=X−∞ − − = (cid:0)(k,c;τ). (cid:1) S This establishes the identities for (k,c;τ). To verify the identities for (k,c;τ), we need only verify that S S for 0 k <c we have ≤ 2iq−(1+4k8−2c)2R(cid:0)(1+4k−2c)cτ − c−21;4c2τ(cid:1)=(−1)k+1i1+ce−π4eicZ−∞τ g1+4−c4ki,(2cw(4+c2τw))dw. For this we use Theorem 1.6 of [23] to determine that p 2iq−(1+4k8−2c)2R (1+4k−2c)cτ − c−21;4c2τ =−2iq−(1+4k8−2(cid:0)c)2 exp(cid:16)πiτ(1+4k4−2c)2 −πi(1(cid:1)+44k−2c)(cid:17)Z−i4∞c2τ −g1i+4(c4zk+,2c(4zc)2τ)dz i g (4c2w) =( 1)k+1i1+ce−π4ic ∞ 1+4c4k,2c dw. p − Z−τ −i(4c2w+4c2τ) (cid:3) p Itisthroughtheseidentities thatwetakethe definitions of (k,c;τ) and (k,c;τ)whenk is anarbitrary S S integer; that is to say they are defined in terms of µ and µ˜ rather than series and integrals. Additionally we seethe definitionsdepends onthe parityofc becauseµ(u,v;τ)andµ˜(u,v;τ)erequireu,v Z+τZ. Onecan 6∈ easily check that (k+c,c;τ)=( 1)c (k,c;τ), however the relation for (k+c,c;τ) is not as elegant. In S − S S particular one can verify that (k+c,c;τ) = i−cq(1+4k2+c)c +( 1)c (k,c;τ). Lastly we note when c is odd S − S e e we can rewrite (k,c;τ) as S e (k,c;τ)=( 1)c−21q−(1+4k8−2c)2µ˜((2k c)2cτ, cτ;4c2τ). S − − − Workingwithµ˜(u,v;τ)isadvantageousinthatthetransformationundertheactionofthemodulargroup e SL (Z)isknownandquiteelegant. Withthisweareabletodetermineasimplemultiplierfor 2(a,c;τ)and 2 R (k,c;τ). Forthisreasonwedonotneedtoreplaceτ by8τ tohaveastrongunderstandingofthesefunctions. S Additionally the transformations of µ˜ allow us to determine the behavior of 2(a,c;τ) and f(k,c;τ) at the R S ceusps. f e 4. Transformations Formulas For a matrix A= αβ SL (Z), we have, ν(A), the η-multiplier defined by γ δ ∈ 2 (cid:16) (cid:17) η(Aτ)=ν(A) γτ +δη(τ), where η(τ) is Dedekind’s eta-function, p η(τ)=q214 (q;q) . ∞ Aconvenientformforthe η-multiplierwhenγ =0,whichcanbefoundasTheorem2inChapter4of[14],is 6 δ exp πi (α+δ)γ βδ(γ2 1) 3γ if γ 1 (mod 2), ν(A)= γ 12 − − − ≡ (4.1) γ| |exp πi (α+δ)γ βδ(γ2 1)+3δ 3 3γδ if δ 1 (mod 2). (cid:26) (cid:0)δ (cid:1) (cid:0)12 (cid:0) − − −(cid:1)(cid:1) − ≡ (cid:0) (cid:1) (cid:0) (cid:0) 9 (cid:1)(cid:1) For an integer m we let α mβ A = . m γ/m δ (cid:18) (cid:19) The utility of A is in the fact that mAτ =A (mτ). m m Our transformation formulas for 2(a,c;τ) and (k,c;τ) are easily deduced by the transformations of R S µ˜(u,v;τ). The following essential properties are from Theorem 1.11 of [23]. If k,l,m,n are integers then f e µ˜(u+kτ +l,v+mτ +n;τ)=( 1)k+l+m+nexp πiτ(k m)2+2πi(k m)(u v) µ˜(u,v;τ). (4.2) − − − − If A= αβ SL (Z) then (cid:0) (cid:1) γ δ ∈ 2 (cid:16) (cid:17) u v πiγ(u v)2 µ˜ , ;Aτ =ν(A)−3exp − γτ +δµ˜(u,v;τ). (4.3) γτ +δ γτ +δ − γτ +δ (cid:18) (cid:19) (cid:18) (cid:19) p The propositions and formulas of this section are organizedas follows. We begin with 2(a,c;τ) by first R investigatingthemostgeneraltransformationthatfollowsfrom(4.3),wethenrefinethisresultbyrestricting to smaller congruence subgroups of SL (Z) until we have a subgroup that yields a simpfle multiplier. We 2 additionally need a formula for the transformation of 2(a,c;τ) under the full modular group to deduce R that 2(a,c;τ) behaviors correctly at the cusps and to determine the orders at the cusps. We then verify R 2(a,c;τ) is annihilated by ∆ . With this we can dedufce that replacing τ by 8τ yields a harmonic Maass 1 fRorm fof weight 1 on a certain c2ongruence subgroup, however we find it beneficial to work with the original 2 f 2(a,c;τ) and the subgrouponwhich it has a simple multiplier. In particularwe find this multiplier agrees R withthatofacertaineta-quotient,whichallowsustoobtainfunctionsthattransformlikeweight1formsbut tfhatarenotnecessarilyannihilatedby∆1 or∆1. Wethendetermineaformulaforthenon-holomorphicpart 2 of 2(a,c;τ) that allows us to determine the non-holomorpic parts of the c-dissections of 2(a,c;τ). We R R then introduce (k,c;τ) as the functions whose non-holomorphic parts agree with those of the c-dissections S of f2(a,c;τ). After this we give (k,c;τ) the same treatment as 2(a,c;τ). Lastly we quicfkly deduce the R S R necessary transfeormationformulas for various generalizedeta quotients using the work of Biagioli in [4]. To begfin we will find it convenient toedefine the functions f M±(a,c;τ)=q−312µ˜(2ca,±τ4;τ). Proposition 4.1. Suppose a and c arge integers, c>0, and c∤2a. Let A= αγ βδ ∈Γ0(4), then (cid:16) (cid:17) M (a,c;4Aτ) ± g=(−1)βζ2−ca22γδζ2±caγβexp −π4iαβ ν(A4)−3 γτ +δq−18(α∓2acγ)2µ˜ 2acγτ + 2acδ,±ατ;4τ . Proof. We note that A SL (Z) a(cid:16)nd so b(cid:17)y (4.3) wephave that (cid:0) (cid:1) 4 2 ∈ M (a,c;4Aτ) ± =exp πiAτ µ˜ 2a, Aτ;A (4τ) g −4 c ± 4 =exp(cid:0)−4πiAτ(cid:1)µ˜(cid:0)2caγγττ++δδ,±γαττ+±δβ;(cid:1)αγ((44ττ))++4δβ 4 =exp(cid:0)−4πiAτ(cid:1)ex(cid:16)p 4(−γπτ+iγδ) 2ca(γτ +δ)−((cid:17)±ατ ±β) 2 ν(A4)−3 γτ +δµ˜ 2acγτ + 2acδ,±ατ ±β;4τ =exp(cid:0)−4πiAτ(cid:1)exp(cid:16)−π4iγ(γ(ατ+τ+(cid:0)δ)β)2 exp −π4iγ 4ca22(γτ (cid:1)+(cid:17)δ)∓ 4ca(ατp+β) ν((cid:0)A4)−3 γτ +δ (cid:1) µ˜ (cid:0)2aγτ + 2(cid:1)aδ, (cid:16)ατ β;4τ .(cid:17) (cid:16) (cid:16) (cid:17)(cid:17) p · c c ± ± However(cid:0), since αδ βγ =1, we s(cid:1)ee that − exp −4πiAτ exp −π4iγ(γ(ατ+τ+δ)β)2 =exp −4π(iγ(ατ+τ+δ)β)(1+αγτ +βγ) (cid:0) (cid:1) (cid:16) (cid:17)=exp(cid:16)−πi(ατ+β)(αγτ +αδ) (cid:17) 4(γτ+δ) 10 (cid:16) (cid:17)