April 10, 2025
In this paper, we introduce a class of functions that behave like classical Eisenstein series in many ways, but with a key distinction: only their non-holomorphic completions transform like (quasi)modular forms. We show how the partition rank generating function can be expressed in terms of partition traces of these functions. A key feature of our construction is that the completions satisfy a holomorphic anomaly equation–a phenomenon typically seen in the context of quantum field theory and string theory. We also show that the Fourier coefficients of these Eisenstein-type series are integral.
In this paper, we introduce a new class of Eisenstein-type series.These series share several structural features with classical Eisenstein series, but differ in a crucial respect: only their non-holomorphic completions transform as (quasi)modular forms. To motivate their definition, we first explore their connection to the rank generating function, before examining their further properties in detail.
A partition of \(n\in\mathbb{N}_0\) is a weakly decreasing sequence of positive integers that sum to \(n\). We denote by \(p(n)\) the number of partitions of \(n\). Recall the famous Ramanujan congruences \[p(5n+4) \equiv 0\ \, ( \mathrm{mod} \, 5 ),\quad p(7n+5) \equiv 0\ \, ( \mathrm{mod} \, 7 ),\quad p(11n+6) \equiv 0\ \, ( \mathrm{mod} \, 11 ).\] To explain the first two, Dyson [1] introduced the rank of a partition \(\lambda\), which is defined as \[\mathrm{rank}(\lambda) := \text{largest part of }\lambda - \text{number of parts of }\lambda.\] Dyson conjectured that reducing the rank (mod \(5\)) (resp.\(7\)) divides the partitions of \(5n+4\) (resp.\(7n+5\)) into \(5\) (resp.\(7\)) sets of equal size. This conjecture was proven by Atkin and Swinnerton-Dyer [2]. In the same paper, Dyson also conjectured the existence of another statistic, which he called the “crank” and which should explain all three partition congruences. Garvan [3] found a crank for vector partitions and Andrews–Garvan [4] defined a crank for ordinary partitions. Letting \(o(\lambda)\) denote the number of ones in a partition \(\lambda\), and \(\mu(\lambda)\) the number of parts strictly larger than \(o(\lambda)\), the crank is defined as \[\mathrm{crank}(\lambda) := \begin{cases} \text{largest part of }\lambda & \text{if }o(\lambda)=0,\\ \mu(\lambda)-o(\lambda) & \text{if }o(\lambda)>0. \end{cases}\]
Let \(N(m,n)\) denote the number of partitions of \(n\) with rank \(m\). Its generating function is \[\label{eq:rank} R(\zeta;q):= \sum_{\substack{n\ge0\\m\in\mathbb{Z}}} N(m,n) \zeta^m q^n = \sum_{n\ge0} \frac{q^{n^2}}{(\zeta q)_n\left(\zeta^{-1}q\right)_n},\tag{1}\] (see [2]) where \((a)_n=(a;q)_n:=\prod_{j=0}^{n-1} (1-aq^j)\) for \(n\in\mathbb{N}_0\cup\{\infty\}\). Let \(M(m,n)\) denote the number of partitions of \(n\) with crank \(m\), except for1 \(n = 1\) where \(M(-1,1) = -M(0,1) = M(1,1) := 1\) as given by the following generating function [4] \[\label{eq:crank} C(\zeta;q):= \sum_{\substack{n\geq 0\\m\in \mathbb{Z}}} M(m,n) \zeta^m q^n = \frac{(q)_\infty}{(\zeta q)_\infty\left(\zeta^{-1}q\right)_\infty}.\tag{2}\] Modularity properties of the rank and of the crank generating function differ significantly: the crank generating function is basically a meromorphic Jacobi form, whereas the rank generating function is a “mock Jacobi form” [5]: it only transforms like a Jacobi form after adding a non-holomorphic term (see Subsection 2.3 for the precise transformations of the rank generating function). We next consider the crank moments [6] \[C_{k}(q) := \sum_{\substack{n\geq 0\\m\in \mathbb{Z}}} m^{k} M(m,n) q^n.\] These moments can be written in terms of quasimodular forms. In [7], the authors expressed these as a so-called partition Eisenstein trace. Here, for a sequence of functions \(h=\{h_k\}_{k\in\mathbb{N}}\), define, for \(n\in\mathbb{N}\), the \(n\)-th partition trace with respect to \(h\) and a function \(\phi\) on partitions as \[\mathrm{Tr}_n(\phi,h;\tau):=\sum_{\lambda\vdash n} \phi(\lambda)h_\lambda(\tau),\] where the sum ranges over all partitions of \(n\) and for \(\lambda=(1^{m_1}, 2^{m_2}, \ldots, n^{m_n})\vdash n\), we set \[\begin{align} h_\lambda(\tau) := \prod_{j=1}^n h_j^{m_j}(\tau). \end{align}\] Partition Eisenstein traces are the partition traces with respect to the sequence of Eisenstein series \(G=\{G_k\}_{k\in\mathbb{N}}\). Here, \(G_{k}\) is the Eisenstein series2 of weight \(k\in2\mathbb{N}\), given by \[G_{k}(\tau) := -\frac{B_{k}}{2k} + \sum_{n,m\geq 1} m^{k-1} q^{nm} \qquad (q:=e^{2\pi i \tau},\, \tau\in\mathbb{H}:=\{w\in\mathbb{C}:\operatorname{Im}(w)>0\}),\] with \(B_k\) the \(k\)-th Bernoulli number. By convention, \(G_{k}:=0\) for \(k\) odd. Moreover, define \[\label{eq:phi-lambda} \phi(\lambda):=\prod_{j=1}^k \frac{2^{m_j}}{m_j!j!^{m_j}}.\tag{3}\] The following result was obtained in [7].
Theorem 1. We have \[\begin{align} \sum_{k\geq 0} C_{k}(q) \frac{z^{k}}{k!} &= \frac{2\sinh\!\left(\frac{z}{2}\right)}{z(q)_\infty} \sum_{k\geq 0} \mathrm{Tr}_k(\phi,G;\tau) z^{k}. \end{align}\]
The Eisenstein series \(G_k\) play a key role in the theory of modular forms and satisfy many interesting properties. They have a beautiful connection with Bernoulli numbers in that, for \(k\ge2\), we have \[\begin{align} \lim_{\tau\to i\infty} G_k(\tau) = -\frac{B_k}{2k}. \end{align}\] For \(k>2\), \(G_k\) is a modular form of weight \(k\) on \(\mathrm{SL}_2(\mathbb{Z})\) and \(G_2\) is quasimodular (see Subsection 2.1 for the definition). More precisely, we have for \(\left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right)\in\mathrm{SL}_2(\mathbb{Z})\), \[\label{eq:Gk} G_k\left(\frac{a\tau+b}{c\tau+d}\right) = \begin{cases} (c\tau+d)^{k} G_{k}(\tau)&\text{if k\neq2},\\ (c\tau+d)^2G_2(\tau) + \frac{ic}{4\pi}(c\tau+d) & \text{if k=2}. \end{cases}\tag{4}\] Another key property of the Eisenstein series is that the corresponding algebra \(\mathbb{Q}[G_2,G_4,G_6,\ldots]\) of quasimodular forms is closed under the action of \(D:=q\frac{\partial}{\partial q}\).
A result like Theorem 1 involving the rank moments \[\label{eq:rankmoments} R_{k}(q) := \sum_{\substack{n\geq 0\\m\in \mathbb{Z}}} m^{k} N(m,n) q^n\tag{5}\] may give rise to an interesting generalization of the Eisenstein series. In this paper, we show that this is indeed the case. We obtain a surprisingly simple definition of Eisenstein-type series3 \(g_\ell\) in 6 which seem not to be studied in the literature before. As mentioned above, the rank moments are related to mock modular forms. Hence, they do not admit an expansion as a trace of modular Eisenstein series like crank moments. As our first result, we write the rank moments as traces of the Eisenstein–type series \(f_k\) defined in Subsection 3.1. The aim of this paper is to understand these functions \(f_k\). We call a real-analytic function \(f^*(\tau,\overline{\tau})\) a quasi-completion of \(f(\tau)\), if \(f^*(\tau,\overline{\tau})\) transforms like a quasimodular form (see 8 ) and if4 \(\lim_{\overline{\tau}\to-i\infty}\) \(f^*(\tau,\overline{\tau})=f(\tau)\). If a quasi-completion \(f^*(\tau,\overline{\tau})\) transforms as a modular form, then we call \(f^*(\tau,\overline{\tau})\) a completion. If it is clear from the context, then we also just write \(f^*(\tau)\) instead of \(f^*(\tau,\overline{\tau})\). Our first result is the following theorem.
Theorem 2. There exists a family of functions \(f=\{f_{k}\}_{k\in\mathbb{N}}\) such that \[\begin{align} \label{eq:R95k-f95k} \sum_{k\ge 0} R_{k}(q) \frac{z^{k}}{k!} &= \frac{2\sinh\!\left(\frac{z}{2}\right)}{z(q)_\infty}\sum_{k\ge0} \mathrm{Tr}_k(\phi,f;\tau) z^{k}, \end{align}\qquad{(1)}\] where \(\phi\) is defined in 3 and \(f_{k}\) has the following properties:
For \(k\ge2\), we have \[\begin{align} \lim_{\tau\to i\infty} f_{k}(\tau) = -\frac{B_{k}}{2k}. \end{align}\]
The function \(f_{k}\) has a quasi-completion5 \(f^*_{k}\) which satisfies, for \(\left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right)\in\mathrm{SL}_2(\mathbb{Z})\), \[f^*_{k}\left(\frac{a\tau+b}{c\tau+d}\right) = \begin{cases} (c\tau+d)^{k} f^*_{k}(\tau)&\text{if k\neq2},\\ (c\tau+d)^2f^*_2(\tau) + \frac{3ic}{4\pi}(c\tau+d)&\text{if k=2}. \end{cases}\]
The algebra \(\mathcal{F}:=\mathbb{Q}[f_2,f_4,\ldots, G_2, G_4,\ldots]\) is closed under the action of \(D\).
Remark 1.
As for the Eisenstein series, we have \(f_k=0\) if \(k\) is odd; see Remark 15.
The functions \(f_k\) and \(f_k^*\) are of intrinsic interest, as they are a step up in complexity compared to quasimodular forms.
The transformation of \(f_k^*\) agrees with that of \(G_k\) up to a factor of \(3\) in front of \(c\tau+d\). This is explained by the (mock) Jacobi forms underlying the rank and crank statistics: namely, for the rank the corresponding index is \(-\frac{3}{2}\) and for the crank it is \(-\frac{1}{2}\).
Even though the \(f_k\) are uniquely determined in Theorem 2, properties (1), (2), and (3) itself do not determine them uniquely. For example, after adding a cusp form of weight \(k\) to each \(f_k\), the resulting functions still satisfy these properties. To describe the \(f_k\) uniquely without involving ?? , we give two recursive definitions for them involving divisor-like sums. For this, we define \(g_\ell\in \mathcal{F}\) by \(g_0:=1\), \(g_\ell:=0\) for \(\ell\in\mathbb{N}\) odd, and, for \(\ell\in\mathbb{N}\) even, \[\label{eq:gl} g_{\ell}(\tau) := \left(1-2^{\ell-1}\right)\frac{ B_{\ell}}{2\ell} + \sum_{\substack{2n-1\geq 3m\geq 3}} (2n-3m)^{\ell-1} q^{nm}-\sum_{\substack{n-1\geq 6m\geq 6}}(n-6m)^{\ell-1} q^{nm}.\tag{6}\] For \(r\in\mathbb{N}\), the \(r\)-th Fourier coefficient of \(g_\ell\) is a polynomial in some of the divisors of \(r\): namely those positive divisors \(n,m\) satisfying the inequality \(2n-1\geq3m\) or \(n-1\geq6m\). Hence, the \(g_k\) are in spirit of the “mock Eisenstein series” studied by Zagier [8] and Mertens–Ono–Rolen [9]. Moreover, compare \(g_\ell\) with \[F_\ell^{[2]}(\tau) := \sum_{\substack{n>m\ge1 \\n-m \text{ odd }}}(-1)^n m^{\ell-1} q^{\frac{nm}{2}}\] which can be expressed as the Rankin–Cohen bracket of a mock modular form and \(\eta^3\) as explained in [10]. This function goes back to [11], [12], where the authors studied characters of the \(\mathcal{N} = 4\) superconformal algebra in two dimensions and the elliptic genus of K3 surfaces.
Theorem 3. Let \(n\in \mathbb{N}\).
We have \[\begin{align} f_n(\tau) &= \frac{n}{2^{n-1}}g_n(\tau) - \sum_{\ell=2}^{n-2} \frac{\ell}{2^{\ell-2}}\binom{n-1}{\ell}f_{n-\ell}(\tau) g_{\ell}(\tau). \end{align}\]
We have \[\begin{align} \f_n(\tau) &= \sum_{\ell=2}^n \frac{(n-1)!\ell}{(\ell-1)!2^{\ell-1}} g_{\ell}(\tau) \mathrm{Tr}_{n-\ell}(\psi,f;\tau), \end{align}\] with \(f:=\{ f_{k}(\tau)\}_{k\in\mathbb{N}}\) and \(\psi(\lambda):= (-1)^{\sum_{j=1}^k m_j} \phi(\lambda)\).
As a direct corollary, we see that the algebra \(\mathcal{G}:=\mathbb{Q}[g_2,g_4,\ldots, G_2, G_4,\ldots]\) is closed under \(D\), and that the elements of \(\mathcal{G}\) admit a quasi-completion:
Corollary 4. We have \(\mathcal{G} = \mathcal{F}\), where the algebra \(\mathcal{F}\) is given by Theorem 2[it:iii].
The recursive formulas in Theorem \(\ref{T:Recursion}\) suggest that the Fourier coefficients of \(f_k\) may have large denominators. However, we show in the following theorem that, with the exception of the constant term, all Fourier coefficients of \(f_k\) are integers, similar as for \(G_k\).
Theorem 5. For \(k\geq 2\), the Fourier coefficients of \(f_k+\frac{B_k}{2k}\) are integers.
Finally, in part (1) of the next theorem, we give an explicit formula for \(D(f_k)\) and explain how the raising and lowering operators act on the completions \(\widehat{f}_k\) of the6 \(f_k\). More precisely, let \(\widehat{\mathcal{F}}:=\mathbb{C}[\widehat f_2,\widehat f_4, \widehat f_6,\ldots,\widehat{G}_2,G_4,G_6]\) and define the raising and the lowering operator by \[\mathcal{R}_k := 2i\frac{\partial{\partial\tau}}{+} \frac{k}{v}, \qquad L:=-2iv^2\frac{\partial}{\partial\overline{\tau}} \qquad\qquad (k\in \mathbb{Z}, \tau=u+iv).\] In parts (2) and (3), we obtain a recursive expression for the action of the raising and lowering operators on \(\widehat{f}_k\). In physics literature, the latter of these falls into the realm of “holomorphic anomaly equations”. In string theory, mirror symmetry, enumerative geometry, and topological field theory, a holomorphic anomaly refers to the phenomenon where a generating function (or its coefficients), instead of being purely holomorphic, acquires a controlled dependence on both \(\tau\) and \(\overline{\tau}\). A holomorphic anomaly equation describes this controlled dependency through a recursive relation that expresses the anti-holomorphic variation in terms of functions of lower genus or weight. While these may themselves carry non-holomorphic dependence, the anomaly equation organizes this structure in a precise and constrained way (see for example [13]–[16]). We let \(\delta_{\mathcal{S}}:=1\) if a statement \(\mathcal{S}\) holds and \(\delta_{\mathcal{S}}:=0\) otherwise.
Theorem 6. We have the following.
For \(k\ge2\), we have \[D(f_k(\tau)) = \frac{k!}{6} \mathrm{Tr}_{k+2}(\phi,3G-f;\tau)-\frac{k-1}{6(k+1)}f_{k+2}(\tau) -\frac{1}{3}\sum_{a=1}^{k-1} \binom{k}{a} f_{a+1}(\tau)f_{k-a+1}(\tau).\]
The algebra \(\widehat{\mathcal{F}}\) is closed under the raising operator. In particular, for \(k\ge2\) we have \[\begin{gather} -\frac{1}{4\pi}\mathcal{R}_k\!\left(\widehat{f_k}(\tau)\right) = \frac{k!}{6} \mathrm{Tr}_{k+2}\left(\phi,3\widehat G-\widehat f;\tau\right)-\frac{k-1}{6(k+1)} \widehat f_{k+2}(\tau)\\ -\frac{1}{3}\sum_{a=1}^{k-1} \binom{k}{a} \widehat f_{a+1}(\tau) \widehat f_{k-a+1}(\tau), \end{gather}\] where \(\widehat{G}=\{\widehat G_k\}_{k\geq 1}\) is defined by \(\widehat G_k=G_k\) for \(k\neq 2\) and \(\widehat{G}_2\) is the completion of \(G_2\), given in 7 .
We have \(L(\widehat{\mathcal{F}})\subseteq \widehat{\mathcal{F}} \oplus \sqrt{v} |\eta|^2 \widehat{\mathcal{F}}\). In particular, for \(k\ge2\) we have \[\begin{align} L\!\left(\widehat f_k(\tau)\right)&= -\frac{3}{8\pi}\delta_{k=2} + \frac{\sqrt{3}k!}{4\sqrt2\pi}\sqrt{v} |\eta(\tau)|^2 \mathrm{Tr}_{k-2}\left(\psi,\widehat{f};\tau\right), \end{align}\] where \(\psi\) is defined in Theorem 3 (2).
If \(\widehat f\in \widehat{\mathcal{F}}\) is of weight \(k\), then \((D +\frac{2k}{3} f_2^*)(\widehat f)\) transforms modular of weight \(k+2\).
Remark 2. We expect that the algebra \(\mathcal{F}\) is freely generated by the \(f_k\) for \(k\) even and \(G_2, G_4\), and \(G_6\) (see also question [q:2] in Section 6). In particular, this would imply that the subalgebra \(\mathbb{Q}[f_2,f_4,\ldots]\) of \(\mathcal{F}\) is not closed under applying \(D\), since Theorem 6 [it:Di] involves the modular Eisenstein series.
The paper is organized as follows. In Section 2, we provide certain preliminaries on quasimodular forms and crank moments, the completion of the rank generating function, the rank-crank PDE, Pólya cycle index polynomials, and finally, divisibility properties of multinomials. In Section 3 and Section 4, we prove our theorems. The last two sections are devoted to examples and open questions.
We thank Caner Nazaroglu, Kilian Rausch, and Don Zagier for helpful comments on a previous version of the manuscript. The first and the second authors were funded by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement No. 101001179) and the first and the third authors by the SFB/TRR 191 “Symplectic Structure in Geometry, Algebra and Dynamics”, funded by the DFG (Projektnummer 281071066 TRR 191).
The Eisenstein series \(G_k\) are modular forms of weight \(k\) for \(\mathrm{SL}_2(\mathbb{Z})\) for \(k\ge 4\) even. If \(k=2\), then we need to add a non-holomorphic part to make \(G_2\) modular. To be more precise, \[\label{eq:G242} \widehat{G}_2(\tau):=G_2(\tau)+\frac{1}{8\pi v}\tag{7}\] transforms like a modular form of weight \(2\). The holomorphic part \(G_2\) can be recovered by \[\lim_{\overline{\tau}\to-i\infty}\widehat{G}_2(\tau)=G_2(\tau).\] In general \(f:\mathbb{H}\to\mathbb{C}\) is an almost holomorphic modular form of weight \(k\in\mathbb{Z}\) and depth \(s\in \mathbb{N}_0\), if the following conditions hold:
We have, for \(\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\in\operatorname{SL}_2(\mathbb{Z})\), \[f\left(\frac{a\tau+b}{c\tau+d}\right) = (c\tau+d)^k f(\tau).\]
We have,7 for some holomorphic functions \(f_j:\mathbb{H}\to\mathbb{C}\) with \(f_s\neq0\), \[f(\tau) = \sum_{j=0}^s \frac{f_j(\tau)}{v^j}.\]
The function \(f\) grows at most polynomially in \(\frac{1}{v}\) as \(v\to0\).
By convention, the function zero is an almost holomorphic modular form of depth \(-\infty\).
The holomorphic part \(f_0\) is called a quasimodular form of weight \(k\) and depth \(s\). Note that \[f_0(\tau) = \lim_{\overline{\tau}\to-i\infty} f(\tau).\] More concretely, for a quasimodular form \(g\) of weight \(k\) and depth \(s\) there exist holomorphic functions \(g_j:\mathbb{H}\to\mathbb{C}\) for \(j\in\{0,\ldots,s\}\) with \(g_0=g\) such that, for all \(\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\in\operatorname{SL}_2(\mathbb{Z})\), \[\begin{align} (c\tau+d)^{-k}g\!\left(\frac{a\tau+b}{c\tau+d}\right) \: =\: \sum_{j=0}^s g_j(\tau)\Bigl(\frac{c}{c\tau+d}\Bigr)^j. \end{align}\] Similarly, we say an analytic function \(g^*(\tau,\overline{\tau})\) transforms like a quasimodular form if there exist real-analytic functions \(g_j^*(\tau,\overline{\tau})\) so that, for all \(\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\in\operatorname{SL}_2(\mathbb{Z})\), \[\begin{align} \label{eq:qmftrans2}(c\tau+d)^{-k}g^*\!\left(\frac{a\tau+b}{c\tau+d},\frac{a\overline{\tau}+b}{c\overline{\tau}+d}\right) \: =\: \sum_{j=0}^s g_j(\tau,\overline{\tau})\Bigl(\frac{c}{c\tau+d}\Bigr)^j. \end{align}\tag{8}\] The space of all quasimodular forms is a free algebra with generators \(G_2, G_4\), and \(G_6\). It follows from Ramanujan’s differential equations, [17] \[\begin{align} \label{eq:Ramanujandiff} D(G_2) &= - 2 G_2^2 + \frac{5}{6}G_4,\quad D(G_4) = -8G_2G_4 + \frac{7}{10}G_6,\quad D(G_6) = -12G_2G_6 + \frac{400}{7}G_4^2, \end{align}\tag{9}\] that this algebra is closed under differentiation. We also require the Serre derivative (see, e.g., [18]) \[\label{eq:Serrederivative} \vartheta_k := D + 2k G_2\tag{10}\] which acts on modular forms of weight \(k\), preserving the algebra of modular forms.
The Dedekind eta function \[\eta(\tau):=q^{\frac{1}{24}} \prod_{n\ge1} \left(1-q^n\right)\] is a modular form of weight \(\frac{1}{2}\). It is not hard to deduce the following lemma.
Lemma 7. For any holomorphic function \(f\) on \(\mathbb{H}\), we have \[\eta D\!\left(\frac{f}{\eta}\right)=G_2f+ D(f).\]
Set \(\zeta:=e^{2\pi i z}\). By 2 and [19], the crank generating function can be expressed in terms of the Eisenstein series. Recall that \(G_k=0\) if \(k\) is odd.
Lemma 8. We have \[C(\zeta;q) = \frac{\sin(\pi z)}{\pi z(q)_\infty} \exp\!\left(2\sum_{k\ge2}G_{k}(\tau)\frac{(2\pi iz)^{k}}{k!}\right).\]
Taking \(\tau \to i\infty\), we obtain the following lemma, which was also observed in [20].
Lemma 9. We have \[\frac{\zeta^\frac{1}{2}}{\zeta-1} = \frac{1}{2\pi iz} \exp\!\left(-\sum_{k\ge2}\frac{B_{k}}{k}\frac{(2\pi iz)^{k}}{k!}\right).\]
Recall that the Bernoulli numbers \(B_n\) are defined as the constant terms of the Bernoulli polynomials \(B_n(x)\) of degree \(n\), which satisfy \[\begin{align} \tag{11} \sum _{n\ge0} B_{n}(X){\frac{t^{n}}{n!}} &= {\frac{te^{Xt}}{e^{t}-1}}, \\ \tag{12} B_{n}(X+Y)&=\sum _{k=0}^{n}{n \choose k}B_{n-k}(X)Y^{k}, \\ \tag{13} B_n'(X) &= n B_{n-1}(X). \end{align}\]
The rank generating function can be written as a Lerch sum \[\label{eq:RankLerch} R(\zeta;q) = \frac{1-\zeta}{(q)_\infty} \sum_{n\in\mathbb{Z}} \frac{(-1)^nq^\frac{n(3n+1)}{2}}{1-\zeta q^n}.\tag{14}\] Note that this closely resembles the following representation of the crank generating function \[C(\zeta;q) = \frac{1-\zeta}{(q)_\infty} \sum_{n\in\mathbb{Z}} \frac{(-1)^n q^{\frac{n(n+1)}{2}}}{1-\zeta q^n}.\]
In Subsection 4.1 of [21], the first author defined (using different notation) \[\label{E:Rt} R^\#(z; \tau):= \left(\frac{R(\zeta; q)}{\zeta^{\frac{1}{2}}-\zeta^{-\frac{1}{2}}}q^{-\frac{1}{24}}+\frac{1}{2} q^{-\frac{1}{6}}\sum_{\pm} \pm \zeta^{\mp1} S(3z\pm\tau;3\tau)\right) e^{12\pi^2G_2(\tau)z^2},\tag{15}\] where (\(\tau = u+iv,z=x+iy, u,v,x,y\in \mathbb{R}\)) \[S(z; \tau):=\sum_{n\in\mathbb{Z}+\frac{1}{2}}\left(\mathop{\mathrm{sgn}}(n)-E\left(\left(n+\frac{y}{v}\right)\sqrt{2v}\right)\right) (-1)^{n-\frac{1}{2}} q^{-\frac{n^2}{2}} e^{-2\pi inz},\] with \(E(y):=2\int_0^y e^{-\pi t^2}dt\). She showed, building on work of Zwegers [22], the following transformation:
Lemma 10. For \(\begin{psmallmatrix}a&b\\c&d\end{psmallmatrix}\in\operatorname{SL}_2(\mathbb{Z})\), we have \[\label{E:Rt-mod-transform} \eta\!\left(\frac{a\tau+b}{c\tau+d}\right) R^\#\!\left(\frac{z}{c\tau+d}; \frac{a\tau+b}{c\tau+d}\right) = (c\tau+d)\eta(\tau) R^\#(z;\tau).\qquad{(2)}\]
The rank-crank PDE of Atkin and Garvan [6] relates the rank and crank generating functions by a differential equation. To state it, we define the heat operator 8 (of index \(-\frac{3}{2}\)) as \[H:=6q\frac{\partial{\partial q}}{+}\left(\zeta\frac{\partial{\partial \zeta}}{\right})^{\!2}.\] This heat operator maps Jacobi forms of weight \(\frac{1}{2}\) to weight \(\frac{5}{2}\) and does not change the index. We state a modified version of the rank-crank PDE which is more convenient for us.
Lemma 11. We have \[2 \left(\frac{\zeta^{\frac{1}{2}}(q)_\infty C(\zeta;q)}{1-\zeta}\right)^{\! 3} = (H+6G_2(\tau)) \left(\frac{\zeta^{\frac{1}{2}}(q)_\infty R(\zeta;q)}{1-\zeta}\right).\]
Proof. In [24], the rank-crank PDE of Atkin–Garvan was formulated as \[2\eta^2(\tau) \left(\frac{\zeta^{\frac{1}{2}}q^{-\frac{1}{24}}C(\zeta;q)}{1-\zeta}\right)^{\! 3} = H \left(\frac{\zeta^{\frac{1}{2}}q^{-\frac{1}{24}}R(\zeta;q)}{1-\zeta}\right).\] Multiplying both sides by \(\eta(\tau)=q^{\frac{1}{24}}(q)_\infty\) yields \[2 \left(\frac{\zeta^{\frac{1}{2}}(q)_\infty C(\zeta;q)}{1-\zeta}\right)^{\! 3} = \eta(\tau) H \left(\frac{1}{\eta(\tau)}\frac{\zeta^{\frac{1}{2}}(q)_\infty R(\zeta;q)}{1-\zeta}\right).\] We obtain the lemma, using the definition of \(H\) and Lemma 7 with \[f(z;\tau)=\frac{\zeta^{\frac{1}{2}}(q)_\infty R(\zeta;q)}{1-\zeta}.\qedhere\] ◻
We also require the following lemma, inspired by the observation in [25] that the non-holomorphic part of a certain non-holomorphic Jacobi form, closely related to \(R^\#\), is annihilated by the heat operator.
Lemma 12. We have \[\begin{align} H\!\left( R^{\#}(z;\tau)e^{-12\pi^2G_2(\tau)z^2}\right) = -H\!\left(\frac{\zeta^{\frac{1}{2}} R(\zeta;q)q^{-\frac{1}{24}}}{1-\zeta}\right). \end{align}\]
Proof. Using 15 , it is enough to show \[H\!\left(q^{-\frac{1}{6}} \zeta^{-1} S(3z+\tau;3\tau)\right) = 0.\] From (4.1) of [21], we have \[\label{E:S} q^{-\frac{1}{6}} \zeta^{-1} S(3z+\tau;3\tau) = \!\sum_{n\in\mathbb{Z}-\frac{1}{6}} \!\left(\mathop{\mathrm{sgn}}\!\left(n-\frac{1}{3}\right)-E\!\left(\left(n+\frac{y}{v}\right)\sqrt{6v}\right)\!\right) (-1)^{n-\frac{5}{6}} q^{-\frac{3n^2}{2}} \zeta^{-3n}.\tag{16}\] Thus the claim follows once we show that \[H\!\left(\!\left(\mathop{\mathrm{sgn}}\left(n-\frac{1}{3}\right)-E\left(\left(n+\frac{y}{v}\right)\sqrt{6v}\right)\right)q^{-\frac{3n^2}{2}}\zeta^{-3n}\right) = 0.\] Noting that \(H(q^{-\frac{3n^2}{2}}\zeta^{-3n})=0\) and \[H(fg) = H(f)g + fH(g) + 2\left(\zeta\frac{\partial{\partial\zeta}}{f}\right) \left(\zeta\frac{\partial{\partial\zeta}}{g}\right),\] it is enough to prove that \[0 = H\!\left(E\left(\left(n\!+\!\frac{y}{v}\right)\sqrt{6v}\right)\right) - \frac{3n}{\pi i} \frac{\partial{\partial z}}{E}\left(\left(n\!+\!\frac{y}{v}\right)\sqrt{6v}\right)= \frac{3}{8\pi^2v} [2\pi wE'(w)+E''(w)]_{w=\left(n+\frac{y}{v}\right)\sqrt{6v}}\raisebox{-5pt}{.}\] Thus we want to show that \[2\pi xE'(x) + E''(x) = 0,\] which holds since \(E'(x) = 2e^{-\pi x^2}\) and \(E''(x) = -4\pi xe^{-\pi x^2}\). ◻
We require a result about Pólya cycle index polynomials in the case of symmetric group \(S_n\), the set of permutations of the symbols \(x_1,x_2,\ldots,x_n\), see [26] and [7] for more details.
Lemma 13 (Example 5.2.10 of [26]). We have \[\sum_{{n\ge0}}\sum_{\lambda\vdash n}\prod_{k=1}^n\frac{x_k^{m_k}}{m_k!}w^n=\exp\left(\sum_{{k\ge1}}x_k w^k\right)\] as a formal power series in \(w\), where \(\lambda=(1^{m_1}, 2^{m_2}, \ldots, n^{m_n})\vdash n\).
Below, we require the following lemma regarding the divisibility of multinomial coefficients, which follows directly from Bézout’s Lemma.
Lemma 14. For \(a_1,\ldots,a_\ell\in\mathbb{N}\) with \(\sum_{j=1}^\ell a_j=n\), we have \[\frac{n}{\gcd(a_1,\ldots,a_\ell)}\,\Big|\,\binom{n}{a_1,a_2,\ldots,a_\ell}.\]
Motivated by Lemma 8, we define the Eisenstein–type series \(f_k\) in terms of the rank generating function as follows: \[\label{eq:f} R(\zeta;q) =: \frac{\sin\!\left(\pi z\right)}{\pi z(q)_\infty} \exp\!\left(2\sum_{k\geq 1} f_{k} (\tau)\frac{(2\pi iz)^{k}}{k!}\right).\tag{17}\]
Remark 15. Since \(R\) is invariant under \(\zeta\mapsto\zeta^{-1}\), i.e., \(z\mapsto-z\), we have \(f_{k}=0\) if \(k\) is odd.
In the remainder of this section, we show that the \(f_k\) satisfies all of the properties in Theorem 2. To define their quasi-completions, we let \[\begin{align} R^\circ(z,\overline{z};\tau,\overline{\tau}) :=\;& 2i\sin(\pi z)q^{\frac{1}{24}} R^{\#}(z;\tau)e^{-12\pi^2G_2(\tau)z^2} \nonumber \\ =\;&R(\zeta;q) + \frac{1}{2} q^{-\frac{1}{8}} \left(\zeta^\frac{1}{2}-\zeta^{-\frac{1}{2}}\right) \sum_\pm \pm\zeta^{\mp1} S(3z\pm\tau;3\tau). \label{eq:widehatR} \end{align}\tag{18}\]
Lemma 16. We have \[\lim_{\overline{\tau}\to-i\infty} R^\circ(z,\overline{z};\tau,\overline{\tau}) = R(\zeta;q).\]
Proof. By 18 , we have \[\lim_{\overline{\tau}\to-i\infty} R^\circ(z,\overline{z};\tau,\overline{\tau}) = R(\zeta;q) + \frac{1}{2} q^{-\frac{1}{8}} \left(\zeta^\frac{1}{2}-\zeta^{-\frac{1}{2}}\right) \sum_\pm \pm\zeta^{\mp1} \lim_{\overline{\tau}\to-i\infty} S(3z\pm\tau;3\tau).\] From equation (4.1) of [21], we directly obtain \[\begin{gather} \lim_{\overline{\tau}\to-i\infty} q^{-\frac{1}{6}} \zeta^{-1} S(3z+\tau;3\tau)\\ = \sum_{n\in\mathbb{Z}-\frac{1}{6}} \left(\mathop{\mathrm{sgn}}\left(n-\frac{1}{3}\right)-\lim_{\overline{\tau}\to-i\infty}E\left(\left(n+\frac{y}{v}\right)\sqrt{6v}\right)\right) (-1)^{n-\frac{5}{6}} q^{-\frac{3n^2}{2}} \zeta^{-3n}. \end{gather}\] Now \[\lim_{\overline{\tau}\to-i\infty}E\left(\left(n+\frac{y}{v}\right)\sqrt{6v}\right) = \lim_{v\to\infty} E\left(n\sqrt{6v}+\frac{\sqrt6y}{\sqrt v}\right) = \mathop{\mathrm{sgn}}(n).\] As \(\mathop{\mathrm{sgn}}(n-\frac{1}{3}) = \mathop{\mathrm{sgn}}(n)\) for \(n\in \mathbb{Z}-\frac{1}{6}\) and \(S(3z-\tau;3\tau) = S(-3z+\tau;3\tau)\), the claimed statement follows. ◻
We next introduce quasi-completions \(f_k^*\) of \(f_{k}\). First, we define \[\label{eq:defRhat} \frac{(q)_\infty R^\circ(z,\overline{z}; \tau, \overline{\tau})}{2i\sin(\pi z)} =: \frac{1}{2\pi iz}\exp\!\left(2\sum_{k,\ell\ge0}f^*_{k,\ell}(\tau, \overline{\tau}) \frac{(2\pi iz)^k}{k!}\frac{(2\pi i\overline{z})^\ell}{\ell!}\right).\tag{19}\] We let9 \(R^*(z;\tau)\) be the constant term of \(R^\circ(z,\overline{z}; \tau, \overline{\tau})\) in the Taylor expansion in \(\overline{z}\) and set \[f^*_k(\tau) := f^*_{k,0}(\tau,\overline{\tau}).\] In terms of these, we define10 \[\label{Rs} \mathbb{F}(z;\tau) := \frac{(q)_\infty R^*(z;\tau)}{2i\sin(\pi z)} = \frac{1}{2\pi iz}\exp\!\left(2\sum_{k\ge0}f^*_k(\tau) \frac{(2\pi iz)^k}{k!}\right).\tag{20}\] Note that \(f^*_k=0\) for \(k\) odd. We are now ready to prove Theorem 2 (2).
Lemma 17. The function \(f_{k}^*\) is a quasi-completion of \(f_k\). In particular, for \(\left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right)\in\mathrm{SL}_2(\mathbb{Z})\), \[f^*_{k}\left(\frac{a\tau+b}{c\tau+d}\right) = \begin{cases} (c\tau+d)^{k} f^*_{k}(\tau)&\text{if k\neq2},\\ (c\tau+d)^2f^*_2(\tau) + \frac{3ic}{4\pi}(c\tau+d)&\text{if k=2}. \end{cases}\]
Proof. We start by showing the transformation law of \(f^*_k\). Combining 19 and 18 with 4 and Lemma 10, we obtain \[\begin{align} &\frac{c\tau+d}{2\pi iz} \exp\!\left(2\sum_{k,\ell\ge0} f^*_{k,\ell}\left(\frac{a\tau+b}{c\tau+d}, \frac{a\overline{\tau}+b}{c\overline{\tau}+d}\right) \frac{\left(\frac{2\pi iz}{c\tau+d}\right)^k}{k!} \frac{\left(\frac{2\pi i\overline{z}}{c\overline{\tau}+d}\right)^\ell}{\ell!}\right)\\ &=\eta\left(\frac{a\tau+b}{c\tau+d}\right)R^\#\left(\frac{z}{c\tau+d}; \frac{a\tau+b}{c\tau+d}\right) e^{-12\pi^2G_2\left(\frac{a\tau+b}{c\tau+d}\right)\left(\frac{z}{c\tau+d}\right)^2}\\ &= (c\tau+d)\eta(\tau) R^\#(z;\tau) e^{-12\pi^2\left(G_2(\tau)+\frac{ic}{4\pi(c\tau+d)}\right)z^2}\\ &= \frac{c\tau+d}{2\pi iz}\exp\!\left(2\left(\frac{3 ic}{4\pi(c\tau+d)}\frac{(2\pi iz)^2}{2!} + \sum_{k,\ell\ge0} f^*_{k,\ell}(\tau,\overline{\tau})\frac{(2\pi iz)^k}{k!}\frac{(2\pi i\overline{z})^\ell}{\ell!}\right)\right). \end{align}\] Hence, we get \[f^*_{k,\ell} \left(\frac{a\tau+b}{c\tau+d}, \frac{a\overline{\tau}+b}{c\overline{\tau}+d}\right) = \begin{cases} (c\tau+d)^k (c\overline{\tau}+d)^\ell f^*_{k,\ell}(\tau,\overline{\tau})&\text{if } (k,\ell)\neq(2,0),\\ (c\tau+d)^2f^*_{2,0}(\tau,\overline{\tau}) + \frac{3ic}{4\pi}(c\tau+d)&\text{if } (k,\ell)=(2,0). \end{cases}\] So the transformation formula for \(f^*_k\) follows.
We next show that \[\label{E:Limit} \lim_{\overline{\tau}\to-i\infty} f^*_k(\tau) = f_k(\tau).\tag{21}\] By 19 , Lemma 16, and 17 we have \[\begin{align} \frac{1}{2\pi iz} \lim_{\overline{\tau}\to-i\infty} \exp\!\left(2\sum_{k,\ell\ge0}f^*_{k,\ell}(\tau,\overline{\tau})\frac{(2\pi iz)^k}{k!}\frac{(2\pi i\overline{z})^\ell}{\ell!}\right) = \frac{(q)_\infty}{2i\sin(\pi z)} \lim_{\overline{\tau}\to-i\infty} R^\circ(z,\overline{z};\tau,\overline{\tau})\\ = \frac{(q)_\infty}{2i\sin(\pi z)} R(\zeta;q) = \frac{1}{2\pi i z}\exp\!\left(2\sum_{k\ge 0}f_k(\tau)\frac{(2\pi iz)^k}{k!}\right). \end{align}\] Comparing coefficients gives 21 . ◻
We also define \[\begin{align} \label{eq:F-hat} \widehat{\mathbb{F}}(z;\tau) := \frac{(q)_\infty R^*(z;\tau)e^{-\frac{3\pi z^2}{2v}}}{2i\sin(\pi z)} =: \frac{1}{2\pi iz} \exp\!\left(2\sum_{k\ge1}\widehat f_k(\tau)\frac{(2\pi iz)^k}{k!}\right). \end{align}\tag{22}\] Then, by the same argument used to prove Lemma 17, we obtain the following result.
Lemma 18. For \(k\in\mathbb{N}\), we have \[\widehat f_k(\tau) = \begin{cases} f_k^*(\tau) & \text{if k\neq 2},\\ f_2^*(\tau) + \frac{3}{8\pi v} & \text{if k=2}. \end{cases}\] In particular \[\widehat f_k\left(\frac{a\tau+b}{c\tau+d}\right) = (c\tau+d)^k\widehat f_k(\tau) .\]
Next, we determine the behavior of \(f_k(\tau)\) as \(\tau\to i\infty\).
Lemma 19. For \(k\ge2\), we have \[\lim_{\tau\to i\infty} f_k(\tau) = \lim_{\tau\to i\infty} f^*_k(\tau) = \lim_{\tau\to i\infty} \widehat f_k(\tau) = -\frac{B_k}{2k}.\]
Proof. Using 14 , we first compute \[\frac{(q)_\infty R(\zeta;q)}{2i\sin(\pi z)} = -\frac{\zeta^\frac{1}{2}(q)_\infty}{1-\zeta} R(\zeta;q) = -\zeta^\frac{1}{2} \sum_{n\in\mathbb{Z}} \frac{(-1)^nq^\frac{n(3n+1)}{2}}{1-\zeta q^n} \to -\frac{\zeta^\frac{1}{2}}{1-\zeta}\] as \(\tau\to i\infty\). Thus, by 17 , \[\frac{\zeta^\frac{1}{2}}{\zeta-1} = \frac{1}{2\pi iz} \exp\!\left(2\sum_{k\ge1}\lim_{\tau\to i\infty}f_k(\tau)\frac{(2\pi iz)^k}{k!}\right).\] By Lemma 9, we obtain the claim for \(f_k\).
To prove the claim for \(f^*_k\), by 17 and 20 , we have to show that \[\label{want} \lim_{\tau\to i\infty} \frac{(q)_\infty\left(R^*(z;\tau)-R(\zeta;q)\right)}{2i\sin(\pi z)} = 0.\tag{23}\] By 18 , we have \[R^*(z;\tau) - R(\zeta;q) = \frac{1}{2} q^{-\frac{1}{8}} \left(\zeta^\frac{1}{2}-\zeta^{-\frac{1}{2}}\right) \sum_\pm \pm\zeta^{\mp1} S(3z\pm\tau;3\tau).\] Next, note that \[S(3z\pm\tau;3\tau) = \sum_{n\in \mathbb{Z}+\frac{1}{2}} \left(\mathop{\mathrm{sgn}}(n)-E\left(\left(n\pm\frac{1}{3}+\frac{y}{v}\right)\sqrt{6v}\right)\right) (-1)^{n-\frac{1}{2}} q^{-\frac{3n^2}{2}} e^{-2\pi in(3z\pm\tau)}.\] Using Lemma 1.7 of [22], as \(\tau\to i\infty\) (so \(v\to\infty\)), \[\mathop{\mathrm{sgn}}(n)-E\left(\left(n\pm\frac{1}{3}+\frac{y}{v}\right)\sqrt{6v}\right) \ll e^{-\pi\left(n\pm\frac{1}{3}+\frac{y}{v}\right)^26v}.\] Thus \(q^{-\frac{1}{8}}S(3z\pm\tau;3\tau)\to 0\) as \(\tau\to i\infty\), from which we obtain 23 . The claim for \(\widehat f_k\) follows from the claim for \(f_k^*\) and Lemma 18. ◻
Next, we look at the action of \(D\) on the \(f_k\) as well as the action of the raising and the lowering operator on its completions.
Proof of Theorem 6. (1) It is well-known, and follows by \(\eqref{eq:Ramanujandiff}\), that \(D(G_\ell)\in\mathcal{F}\). Thus it remains to show the formula for \(D(f_k)\) for \(k\in2\mathbb{N}\) which directly implies that \(D(f_k)\in\mathcal{F}\). Using Theorem 11, Lemma 8, and 17 , we obtain \[\begin{align} &\frac{2}{(2\pi iz)^3} \exp\!\left(6\sum_{k\ge1}G_k(\tau)\frac{(2\pi iz)^k}{k!}\right) \nonumber\\ &= \left(6q\frac{\partial{\partial q}}{+}\left(\zeta\frac{\partial{\partial \zeta}}{\right})^2+6G_2(\tau)\right) \frac{1}{2\pi iz}\exp\!\left(2\sum_{k\ge1}f_k(\tau)\frac{(2\pi iz)^k}{k!}\right)\label{eq:C-R-diffeq}\\ &= \left(\!\left(12\sum_{k\ge1}D(f_k(\tau))\frac{(2\pi iz)^k}{k!} + 6G_2(\tau)\right) \frac{1}{2\pi iz} + \frac{2}{(2\pi i z)^3} + 2\sum_{k\ge1}f_k(\tau)\frac{(k-2)(2\pi iz)^{k-3}}{(k-1)!} \right. \nonumber\\ & \left.+2\left(-\frac{1}{(2\pi i z)^2} + 2\sum_{k\ge1}f_k(\tau)\frac{(2\pi iz)^{k-2}}{(k-1)!}\right)\sum_{\ell\ge1}f_\ell(\tau)\frac{(2\pi iz)^{\ell-1}}{(\ell-1)!}\right)\exp\!\left(2\sum_{k\ge1}f_k(\tau)\frac{(2\pi iz)^k}{k!}\right).\nonumber \end{align}\tag{24}\] Multiplying by \(\frac{(2\pi iz)^3}{2} \exp(-2\sum_{k\ge1}f_k(\tau)\frac{(2\pi iz)^k}{k!})\) and collecting all terms with \(D(f_k)\) on one side and the rest on the other, we have \[\begin{gather} 6\sum_{k\ge1}D(f_k(\tau))\frac{(2\pi iz)^{k+2}}{k!} =- 3G_2(\tau)(2\pi i z)^2 +\exp\!\left(2\sum_{k\ge1}(3G_k(\tau)-f_k(\tau))\frac{(2\pi iz)^k}{k!}\right)-1\\ -\sum_{k\ge1}f_k(\tau)(k-3)\frac{(2\pi iz)^k}{(k-1)!}-2\left(\sum_{k\ge1}f_k(\tau)\frac{(2\pi iz)^{k}}{(k-1)!}\right)^{2}.\label{D} \end{gather}\tag{25}\] Using Lemma 13 with \(w=2\pi i z\) and \({x_k}=\frac{2}{k!}(3G_k-f_k)\), we obtain \[\exp\!\left(2\sum_{k\ge1}(3G_k(\tau)-f_k(\tau))\frac{(2\pi iz)^k}{k!}\right) = \sum_{k\ge 0} \mathrm{Tr}_k(\phi,3G-f;\tau) (2\pi i z)^k.\] Plugging this in 25 and extracting the coefficient of \({(2\pi i z)^{k+2}}\) on both sides, we get, for \(k\geq 2\), \[\begin{align} \frac{6}{k!}D(f_k(\tau)) &= \mathrm{Tr}_{k+2}(\phi,3G-f;\tau) -\frac{k-1}{(k+1)!} f_{k+2}(\tau) - 2\sum_{\substack{a,b\ge0\\a+b=k}}\frac{1}{a!b!}f_{a+1}(\tau)f_{b+1}(\tau). \end{align}\] Multiplying with \(\frac{k!}{6}\), yields the claim.
(2) By 18 and Lemma 12, we have \[H\!\left(\frac{1}{\eta(\tau)}\frac{\zeta^{\frac{1}{2}}(q)_\infty R^\circ(z,\overline{z}; \tau, \overline{\tau})}{1-\zeta}\right) = H\!\left(\frac{1}{\eta(\tau)}\frac{\zeta^{\frac{1}{2}}(q)_\infty R(\zeta;q)}{1-\zeta}\right).\] Using Lemma 7 and then Theorem 11, we obtain \[(H+6G_2(\tau))\left(\frac{\zeta^{\frac{1}{2}}(q)_\infty R^\circ(z,\overline{z}; \tau, \overline{\tau})}{1-\zeta}\right) = 2 \left(\frac{\zeta^{\frac{1}{2}}(q)_\infty C(\zeta;q)}{1-\zeta}\right)^{\! 3}.\] Employing 19 for the left-hand side, we obtain \[2 \left(\frac{(q)_\infty C(\zeta;q)}{2i\sin(\pi z)}\right)^{\! 3} = \left(H+6G_2(\tau)\right)\left(\frac{1}{2\pi iz}\exp\!\left(2\sum_{k,\ell\ge0}f^*_{k,\ell}(\tau, \overline{\tau}) \frac{(2\pi iz)^k}{k!}\frac{(2\pi i\overline{z})^\ell}{\ell!}\right)\right).\] Hence, using Lemma 8 and the definition of \(H\), we obtain \[\begin{gather} \frac{2}{(2\pi iz)^3} \exp\!\left(6\sum_{k\ge1}G_k(\tau)\frac{(2\pi iz)^k}{k!}\right) \\ =\left(6q\frac{\partial{\partial q}}{+}\left(\zeta\frac{\partial{\partial \zeta}}{\right})^2+6G_2(\tau)\right)\left(\frac{1}{2\pi iz}\exp\!\left(2\sum_{k,\ell\ge0}f^*_{k,\ell}(\tau, \overline{\tau}) \frac{(2\pi iz)^k}{k!}\frac{(2\pi i\overline{z})^\ell}{\ell!}\right)\right). \end{gather}\] Taking the constant terms from both sides with respect to \(\overline{z}\) gives \[\begin{gather} \frac{2}{(2\pi iz)^3} \exp\!\left(\!6\sum_{k\ge1}G_k(\tau)\frac{(2\pi iz)^k}{k!}\!\right) \\ =\left(\!6q\frac{\partial{\partial q}}{+}\left(\zeta\frac{\partial{\partial \zeta}}{\right})^2+6G_2(\tau)\!\right)\!\left(\frac{1}{2\pi iz}\exp\!\left(\!2\sum_{k\ge0}f^*_{k}(\tau) \frac{(2\pi iz)^k}{k!}\!\right)\right)\!. \end{gather}\] Now the shape of the above equation is exactly the same as in 24 except that \(f_k\) is replaced by \(f^*_k\) on the right-hand side. Hence, the same calculation as in part (1) gives that \[\frac{6}{k!}D\left(f^*_k(\tau)\right) = \mathrm{Tr}_{k+2}\left(\phi,3G-f^*;\tau\right)-\frac{k-1}{(k+1)!}f^*_{k+2}(\tau) -2\sum_{\substack{a,b\ge0\\a+b=k}} \frac{1}{a!b!} f^*_{a+1}(\tau) f^*_{b+1}(\tau).\] Noting that \(3G-f^*=3\widehat G-\widehat f\), by Lemma 18, we have \[\frac{6}{k!}D\left(f^*_k(\tau)\right) = \mathrm{Tr}_{k+2}\left(\phi,3\widehat G-\widehat f;\tau\right)-\frac{k-1}{(k+1)!}f^*_{k+2}(\tau) -2\sum_{\substack{a,b\ge0\\a+b=k}} \frac{1}{a!b!} f^*_{a+1}(\tau) f^*_{b+1}(\tau).\] From this, it is not hard to conclude the claim.
(3) Since \(\widehat{\mathcal{F}}\) is generated by \(\widehat{f}_k\) for \(k\in\mathbb{N}\), \(\widehat G_2, G_4\), and \(G_6\), it is enough to show that \(L(\widehat{f}_k), L(\widehat G_2)\), \(L(G_4),L(G_6)\in \widehat{\mathcal{F}}\oplus \sqrt{v} |\eta|^2\widehat{\mathcal{F}}\) and prove the formula for \(L(\widehat f_k)\). First, we have \(L(G_4)=L(G_6) = 0\in \widehat{\mathcal{F}}\oplus \sqrt{v} |\eta|^2\widehat{\mathcal{F}}\) since \(G_4\) and \(G_6\) are holomorphic. Next, by 7 , we have \(L(\widehat{G}_2)=-\frac{1}{8\pi}\in \widehat{\mathcal{F}}\oplus \sqrt{v} |\eta|^2\widehat{\mathcal{F}}\). Finally, we look at \(L(\widehat{f}_k)\). We compute, using 22 , \[\label{eq:L-on-F-hat} L\!\left(\widehat{\mathbb{F}}(z;\tau)\right)= 2\widehat{\mathbb{F}}(z;\tau) \sum_{k\ge0} L\!\left(\widehat f_k(\tau)\right) \frac{(2\pi iz)^k}{k!}.\tag{26}\] Using 22 again, the left-hand side can be written as \[L\!\left(\frac{(q)_\infty R^*(z;\tau) e^{-\frac{3\pi z^2}{2v}}}{2i\sin(\pi z)}\right) =\frac{(q)_\infty}{2i\sin(\pi z)} \left(L\!\left(R^*(z;\tau)\right)+R^*(z;\tau) \frac{3\pi z^2}{2}\right)e^{-\frac{3\pi z^2}{2v}}.\] Using the above for the left-hand side and 22 for the right-hand side of 26 gives \[\label{eq:L-on-R42} L\!\left(R^*(z;\tau)\right) = R^*(z;\tau)\sum_{k\ge0} \left(2L\!\left(\widehat f_k(\tau)\right) + \frac{3}{4\pi}\delta_{k=2} \right) \frac{(2\pi iz)^k}{k!}.\tag{27}\]
We next compute, using 18 , \[\label{eq:L-on-R429543-} L\!\left(R^*(z;\tau)\right) = \frac{1}{2} q^{-\frac{1}{8}} \left(\zeta^{\frac{1}{2}} - \zeta^{-\frac{1}{2}}\right) \sum_{\pm} \pm \zeta^{\mp 1} L\!\left(S(3z\pm\tau;3\tau)\right).\tag{28}\]
We first consider the plus sign. Applying the lowering operator to 16 and using that by [21], we have \[L\!\left(E\left(\left(n+\frac{y}{v}\right)\sqrt{6v}\right)\right) = \sqrt{6} v^{\frac{3}{2}} \left(n-\frac{y}{v}\right) e^{-6\pi v\left(n+\frac{y}{v}\right)^2}\] yields \[L\!\left(S(3z+\tau;3\tau)\right) = -\sqrt6 v^\frac{3}{2} q^{\frac{1}{6}}\zeta e^{- \frac{6\pi y^2}{v}} \sum_{n\in\mathbb{Z}-\frac{1}{6}} \left(n-\frac{y}{v}\right) (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}} \overline{\zeta}^{3n}.\]
Next, we turn to the minus sign. Using that \(S\) is an even function and changing \(z\mapsto -z\) in the above, yields \[\begin{align} L\!\left(S(3z-\tau;3\tau)\right) = L\!\left(S(-3z+\tau;3\tau)\right) &= -\sqrt6 v^\frac{3}{2} q^{\frac{1}{6}}\zeta^{-1} e^{- \frac{6\pi y^2}{v}}\!\!\!\sum_{n\in\mathbb{Z}-\frac{1}{6}}\!\! \left(n+\frac{y}{v}\right) (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}}\overline{\zeta}^{-3n}. \end{align}\] Plugging the above two equations back into 28 gives \[\begin{align} L\!\left(R^*(z;\tau)\right) &= -\sqrt{\frac{3}{2}} v^\frac{3}{2} e^{- \frac{6\pi y^2}{v}} q^{\frac{1}{24}} \left(\zeta^{\frac{1}{2}} - \zeta^{-\frac{1}{2}}\right) \sum_{\pm} \pm \!\sum_{n\in\mathbb{Z}-\frac{1}{6}} \left(n\mp\frac{y}{v}\right) (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}} \overline{\zeta}^{\pm 3n}\\ &=-i\sqrt{6} v^\frac{3}{2} e^{ \frac{3\pi (z-\overline{z})^2}{2v}} q^{\frac{1}{24}} \sin(\pi z) \sum_\pm \pm \sum_{n\in \mathbb{Z}-\frac{1}{6}} \left(n\mp\frac{z-\overline{z}}{2iv}\right) (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}} \overline{\zeta}^{\pm3n}, \end{align}\] using that \(y=\frac{z-\overline{z}}{2i}\). Using 27 and 22 , we obtain \[\begin{gather} \sum_{k\ge0} \left(2L\!\left(\widehat f_k(\tau)\right) + \frac{3}{4\pi}\delta_{k=2} \right)\! \frac{(2\pi iz)^k}{k!} = \frac{L\!\left(R^*(z;\tau)\right)}{ R^*(z;\tau)} \\ = - \frac{i\sqrt{6} v^\frac{3}{2} e^{ \frac{3\pi (z-\overline{z})^2}{2v}} q^{\frac{1}{24}} \sin(\pi z)}{R^*(z;\tau)} \sum_\pm \pm \sum_{n\in \mathbb{Z}-\frac{1}{6}} \left(n\mp\frac{z-\overline{z}}{2iv}\right) (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}} \overline{\zeta}^{\pm3n}. \end{gather}\] Collecting the constant terms in the Taylor expansions with respect to \(\overline{z}\) on both sides yields \[\begin{align} &\sum_{k\ge0} \!\left(\!2L\Bigl(\widehat f_k(\tau)\Bigr) \!+\! \frac{3}{4\pi}\delta_{k=2} \!\right)\! \frac{(2\pi iz)^k}{k!} \!=\! - \frac{i\sqrt{6} v^\frac{3}{2} e^{ \frac{3\pi z^2}{2v}} q^{\frac{1}{24}} \sin(\pi z)}{R^*(z;\tau)} \sum_\pm \pm \!\!\!\sum_{n\in \mathbb{Z}-\frac{1}{6}} \!\!\!\left(n \!\mp\! \frac{z}{2iv}\right)\! (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}}\\ &= \frac{\sqrt{6v} z e^{\frac{3\pi z^2}{2v}} q^\frac{1}{24}\sin(\pi z)}{R^*(z;\tau)} \!\sum_{n\in\mathbb{Z}-\frac{1}{6}}\! (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}} = -iz\sqrt\frac{3v}{2} \frac{\eta(\tau)}{\widehat{\mathbb{F}}(z;\tau)} \!\sum_{n\in \mathbb{Z}-\frac{1}{6}}\! (-1)^{n-\frac{5}{6}} \overline{q}^{\frac{3n^2}{2}}, \end{align}\] using 22 . The sum on the right-hand side equals \(-\eta(-\overline{\tau})\), so the above becomes \[iz \sqrt{\frac{3 v}{2}} \frac{|\eta(\tau)|^2}{\widehat{\mathbb{F}}(z;\tau)} =-\pi \sqrt{6v} |\eta(\tau)|^2 z^2 \exp\!\left(-2\sum_{k\ge1}\widehat f_k(\tau)\frac{(2\pi iz)^k}{k!}\right),\] by 22 again. Using Lemma 13 with \(w=2\pi i z\) and \(x_k=-\frac{2}{k!}\widehat{f}_k(\tau)\), we conclude \[\sum_{k\ge0} \!\left(\!2L\Bigl(\widehat f_k(\tau)\Bigr) + \frac{3}{4\pi}\delta_{k=2} \!\right)\! \frac{(2\pi iz)^k}{k!} = \frac{\sqrt{3v}}{2\sqrt2\pi} |\eta(\tau)|^2 (2\pi iz)^2 \sum_{n\ge 0} \mathrm{Tr}_n\bigl(\psi,\!\widehat{f};\!\tau\bigr)(2\pi i z)^n\!.\] Comparing the coefficient of \((2\pi iz)^k\) gives the claim for \(k\ge 2\).
(4) The Serre derivative \(\vartheta_k\) defined in 10 increases the weight of a modular object by \(2\) (note that the proof in [18] goes through if \(f\) is non-holomorphic). We have \[\begin{align} D+\frac{2k}{3} f_2^*-\vartheta_k = 2k\left(\frac{1}{3} f_2^*-G_2\right). \end{align}\] By Lemma 17 and 4 this is a non-holomorphic modular form of weight \(2\). Hence \(D+\frac{2k}{3} f_2^*\) maps a modular object of weight \(k\) to a modular object of weight \(k+2\). ◻
We are now ready to prove Theorem 2.
Proof of Theorem 2. Using 17 and Lemma 13 with \(w=2\pi i z\) and \(x_k = \frac{2}{k!}f_k\), we find \[\begin{align} \label{eq:r-trace} R(\zeta;q) = \frac{\sin\!\left(\pi z\right)}{\pi z(q)_\infty} \sum_{k\ge0} \mathrm{Tr}_k(\phi,f;\tau) (2\pi i z)^{k}. \end{align}\tag{29}\] From 1 and 5 we deduce that \[\label{rmom} R(\zeta;q) = \sum_{k\ge0 } R_{k}(q) \frac{(2\pi iz)^{k}}{k!}.\tag{30}\] Substituting \(z \mapsto \frac{z}{2\pi i}\) and applying 30 and \(\sin(-\frac{iz}{2})=-i\sinh(\frac{z}{2})\) in 29 , we obtain \[\begin{align} \sum_{k\geq 0} R_k(q) \frac{z^k}{k!} = \frac{2\sinh\!\left(\frac{z}{2}\right)}{z(q)_\infty} \sum_{k\ge0} \mathrm{Tr}_k(\phi,f;\tau) z^{k}. \end{align}\] By Lemma 19 the first property is satisfied, by Lemma 17 the second, and by Theorem 6 (1) the third. ◻
The following lemma rewrites the rank moments \(R_k\) in terms of the \(g_\ell\), defined by 6 .
Lemma 20. For \(k\geq 1\), we have \[R_k(q) = \frac{2^{2-k}}{(q)_\infty }\sum_{\substack{\ell=2 \\ \ell\equiv k\ \left( \mathrm{mod} \, 2 \right)}}^k \binom{k}{\ell-1}\left(g_{\ell}(\tau) + \left(2^{\ell-1}-1\right) \frac{B_{\ell}}{2\ell}\right).\]
Proof. Both sides of the lemma are zero for \(k\) odd. Namely, since \(N(m,n)=N(-m,n)\), we have that \(R_k\) is zero for \(k\) odd, and for \(\ell\) odd, we have \[\begin{align} \label{eq:B95l-g95l-odd} \left(2^{\ell-1}-1\right)\frac{ B_{\ell}}{2\ell} + g_{\ell}(\tau)=0. \end{align}\tag{31}\] So, we may assume that \(k\) is even. Recall that by [2] we have, for \(k\ge2\) even, \[\label{eq:formula95for95Rk} R_k(q) = \frac{2}{(q)_\infty} \sum_{n\ge1} (-1)^{n+1} q^\frac{n(3n-1)}{2}(1-q^n)\sum_{m\geq 0} m^k q^{nm}.\tag{32}\] Distinguishing between \(n\) even and \(n\) odd yields \[\begin{align} 2^{k-1}(q)_\infty R_k(q)&\!=\! \!\!\! \sum_{n\geq 1, m\geq 0} \!\!\! (2m)^k \!\left(\!- q^{n(6n+2m-1)} \!+\! q^{(2n-1)(3n+m-2)} \!+\! q^{n(6n+2m+1)} \!-\! q^{(2n-1)(3n+m-1)}\!\right)\\ &\!=\! \sum_{m\geq 3n\geq 3} (2m-6n)^k \!\left(\!- q^{n(2m-1)} + q^{(2n-1)(m-2)} + q^{n(2m+1)} - q^{(2n-1)(m-1)}\!\right)\!, \end{align}\] making the change of variables \(m\mapsto m-3n\). Interchanging the role of \(n\) and \(m\) in the second and fourth sum the above becomes \[\begin{gather} -\sum_{m\geq 3n\geq 3} (2m-6n)^k q^{n(2m-1)} + \sum_{n\geq 3m\geq 3} (2n-6m)^k q^{(n-2)(2m-1)} \\ + \sum_{m\geq 3n\geq 3} (2m-6n)^k q^{n(2m+1)} - \sum_{n\geq 3m\geq 3} (2n-6m)^k q^{(n-1)(2m-1)}. \end{gather}\] Making the change of variables \(n\mapsto n+2\) in the second sum, the change of variables \(m\mapsto m-1\) in the third sum, and the change of variables \(n\mapsto n+1\) in the fourth sum, the above becomes \[\begin{align} &- \sum_{2m-1\geq 6n-1\geq 5} (2m-1-6n+1)^k q^{n(2m-1)} + \sum_{2n+1\geq 3(2m-1)\geq 3} (2n+1-3(2m-1))^k q^{n(2m-1)} \nonumber\\ &+ \sum_{2m-1\geq 6n+1\geq 7} (2m-1-6n-1)^k q^{n(2m-1)} -\sum_{2n-1\geq 3(2m-1)\geq 3} (2n-1-3(2m-1))^k q^{n(2m-1)} \nonumber\\ &\qquad=\sum_{\substack{m\geq 6n+1\geq 7\\\text{m odd}}} \left((m-6n-1)^k - (m-6n+1)^k\right) q^{nm}\nonumber\\[-32pt]\nonumber\\ & + \sum_{\substack{2n-1\geq 3m\geq 3\\ \text{m odd}}} \left((2n-3m+1)^k - (2n-3m-1)^k\right) q^{nm}\nonumber\\ &\qquad= \sum_{\substack{m\geq 6n+1\geq 7}} \left((m-6n-1)^k - (m-6n+1)^k\right) q^{nm}\nonumber\\ &+ \sum_{\substack{2n-1\geq 3m\geq 3}} \left((2n-3m+1)^k - (2n-3m-1)^k\right) q^{nm}.\label{eq:all-sums} \end{align}\tag{33}\] We require the identity \[(x+1)^k-(x-1)^k = 2\sum_{\substack{1\leq \ell \leq k-1 \\ 2\nmid \ell}} \binom{k}{\ell} x^\ell = 2\sum_{\substack{2\leq \ell \leq k \\ 2\mid \ell}} \binom{k}{\ell-1} x^{\ell-1} .\] Using this and the fact that \(k\) is even, we obtain that 33 equals \[\begin{align} 2\sum_{\substack{\ell=2 \\ 2\mid \ell}}^k \binom{k}{\ell-1} \left(-\sum_{\substack{m\geq 6n+1\geq 7}}(m-6n)^{\ell-1} q^{nm} + \sum_{\substack{2n-1\geq 3m\geq 3}} (2n-3m)^{\ell-1} q^{nm}\right). \end{align}\] Now the lemma follows from the definition of \(g_\ell\). ◻
The generating function of the \(g_\ell\) is closely related to the rank generating function.
Lemma 21. We have \[\frac{ \pi z(q)_\infty}{\sin\left(\pi z\right)}R(\zeta;q) \\ = 1 + 4\sum_{k\ge1}g_{k}(\tau) \frac{(\pi iz)^{k}}{(k-1)!}.\]
Proof. By setting \(X=\frac{1}{2}\) and \(t=-2\pi i z\) in 11 we have \[\begin{align} \label{eq:sin-exp} \frac{\pi z}{\sin(\pi z)} &= \sum_{n\ge0} B_{n}\!\left(\frac{1}{2}\right) \frac{(2\pi i z)^{n}}{n!}. \end{align}\tag{34}\] Note that, by 5 , \(R_0(q)=\sum_{n\geq 0} \sum_{m\in \mathbb{Z}} N(m,n) q^{n} = \frac{1}{(q)_\infty}\) is the generating function of partitions. Hence, using 30 and Lemma 20, we obtain \[\begin{align} &\frac{\pi z(q)_\infty}{\sin\left(\pi z\right)} \left(R(\zeta;q)-\frac{1}{(q)_\infty}\right) \nonumber\\ &= \sum_{{n\ge0}} B_{n}\!\left(\frac{1}{2}\right) \frac{(2\pi i z)^{n}}{n!} \sum_{k\ge 1} 2^{2-k} \sum_{\substack{\ell= 1\\ \ell\not\equiv k\ \, ( \mathrm{mod} \, 2 )}}^k \binom{k}{\ell} \left( g_{\ell+1}(\tau) + \frac{\left(2^{\ell}-1\right) B_{\ell+1}}{2(\ell+1)} \right) \frac{(2\pi iz)^{k}}{k!} \nonumber\\ &= \sum_{\substack{n\ge0\\ \ell\ge1}} \sum_{\substack{k\geq \ell\\ k\not\equiv \ell\ \, ( \mathrm{mod} \, 2 )}} \binom{n}{k} \binom{k}{\ell} B_{n-k}\!\left(\frac{1}{2}\right) 2^{2-k} \left( g_{\ell+1}(\tau) + \frac{\left(2^{\ell}-1\right) B_{\ell+1}}{2(\ell+1)}\right) \frac{(2\pi iz)^{n}}{n!},\nonumber \end{align}\] where we make the change of variables \(n\mapsto n-k\). Using 12 , 13 , and the fact that \(B_m(\frac{1}{2})=0\) and \(B_m(0)=-\frac{\delta_{m=1}}{2}\) for \(m\) odd, for \(\ell\in\mathbb{N}\) odd and \(n\in \mathbb{N}\) we find \[\begin{align} \sum _{\substack{k=\ell\\ k\not\equiv\ell\ \, ( \mathrm{mod} \, 2 )}}^n {n \choose k}\binom{k}{\ell}B_{n-k}\!\left(\frac{1}{2}\right)2^{\ell-k} = -\binom{n}{\ell} B_{n-\ell}(0) = \begin{cases} \frac{n}{2} & \text{if } n=\ell+1, \\ 0 & \text{otherwise}.\end{cases} \end{align}\] Using the above and 31 , we find \[\begin{align} &\frac{\pi z(q)_\infty}{\sin\left(\pi z\right)}\left(R(\zeta;q)-\frac{1}{(q)_\infty}\right) = \sum_{n\ge1} \frac{n}{2^{n-2}} \left( \frac{\left(2^{n-1}-1\right) B_{n}}{2n} + g_{n}(\tau) \right) \frac{(2\pi iz)^{n}}{n!}. \end{align}\] The result follows by adding \(\frac{\pi z}{\sin(\pi z)}\) on both sides and using 34 and the fact that \(B_n(\frac{1}{2})=-(1-2^{1-n}) B_n\). ◻
Now, we are ready to prove Theorems 3 and 5.
Proof of Theorem 3. (1) Using Lemma 21 and then taking the derivative of 17 with respect to \(z\) yields \[4\pi i \sum_{k\geq 1} f_{k}(\tau)\frac{(2\pi iz)^{k-1}}{(k-1)!} \exp\!\left(\!2\sum_{k\geq 1} f_{k}(\tau)\frac{(2\pi iz)^{k}}{k!}\!\right) =2\pi i\sum_{k\ge0} \frac{kg_{k}(\tau)}{2^{k-2}} \frac{(2\pi iz)^{k-1}}{(k-1)!}. \label{eq:recf}\tag{35}\] Again using Lemma 21 we obtain \[2\sum_{k\geq 1} f_{k}(\tau) \frac{(2\pi iz)^{k-1}}{(k-1)!} + 2\sum_{n,\ell\geq 1} \binom{n+\ell-1}{n-1}f_n(\tau) \frac{\ell g_{\ell}(\tau)}{2^{\ell-2}} \frac{(2\pi iz)^{n+\ell-1}}{(n+\ell-1)!}=\sum_{k\ge0} \frac{kg_{k}(\tau)}{2^{k-2}} \frac{(2\pi iz)^{k-1}}{(k-1)!}.\] By extracting the coefficients of \((2\pi i z)^{k-1}\), we obtain (1).
(2) We send the exponential in 35 to the other side and apply Lemma 13 with \(w=2\pi i z\) and \(x_k=-\frac{2f_k}{k!}\) to obtain \[\begin{align} 2\sum_{k\geq 1} f_{k}(\tau)\frac{(2\pi iz)^{k-1}}{(k-1)!} =\sum_{n\ge0} \frac{ng_{n}(\tau)}{2^{n-2}} \frac{(2\pi iz)^{n-1}}{(n-1)!} \sum_{m\geq 0} \mathrm{Tr}_m(\psi,f;\tau) {(2\pi i z)^m}. \end{align}\] Comparing the coefficients of \((2\pi i z)^{k-1}\) gives (2). ◻
Proof of Theorem 5. By 17 , 30 , and Lemma 9, we find that \[\begin{align} \exp\!\left(2\sum_{k\geq 1} \left(f_{k} (\tau)+\frac{B_k}{2k}\right) \frac{z^{k}}{k!}\right) &=1+(q)_\infty\sum_{k\ge1}R_{k}(q) \frac{z^{k}}{k!}. \end{align}\] Taking the logarithm and expanding the right-hand side formally, we deduce that \[2\sum_{k\geq 1} \left(f_{k} (\tau)+\frac{B_k}{2k}\right)\frac{z^{k}}{k!} = \sum_{n \geq 1}\frac{(-1)^{n+1}}{n}\sum_{k_1,\ldots,k_n\geq 1}z^{k_1+\ldots+k_n}\prod_{j=1}^n (q)_\infty R_{k_j}(q) \frac{1}{k_j!} .\] Let \(\Omega\) be the linear map on \(\mathbb{C}[\![z]\!]\) given by \(\Omega(z^k) := k!z^k\) for \(k\geq 0\). Applying \(\Omega\), we obtain \[\begin{align} \!\!2\sum_{k\geq 1} \!\left(f_{k} (\tau)\!+\!\frac{B_k}{2k}\right)\! z^{k} &= \sum_{n \geq 1} \frac{(-1)^{n+1}}{n} \mathcal{R}_n(z;q), \label{eq:fklog} \end{align}\tag{36}\] where we set \[\mathcal{R}_n(z;q):=\sum_{k_1,\ldots,k_n\geq 1}\!\!\binom{k_1+\ldots+k_n}{k_1,\ldots,k_n} z^{k_1+\ldots+k_n}\prod_{j=1}^n (q)_\infty R_{k_j}(q).\]
To show that the Fourier coefficients of \(f_{k} (\tau)\!+\!\frac{B_k}{2k}\) are integral, it suffices to show that these on the right-hand side of 36 are even integers. By 32 the Fourier coefficients of \((q)_\infty R_k(q)\) are even. Hence, those of \(\frac{1}{n}\mathcal{R}_n(z;q)\) are even. Let \(k_1,\ldots,k_n\in\mathbb{N}\). There exists a unique partition \(\lambda\) of length \(n\) associated to these \(k_j\) (obtained by ordering the \(k_j\) in non-increasing order). Conversely, for a partition \(\lambda\) of length \(n\), there are \(\binom{n}{r_1(\lambda),r_2(\lambda),\ldots}\) many \(n\)-tuples \((k_1,\ldots,k_n)\) with associated partition \(\lambda\), where \(r_m(\lambda)\) denotes the number of parts of size \(m\) in \(\lambda\). Let \(\mathcal{P}\) be the set of partitions and \(|\lambda|=\sum_{j}\lambda_j\) the size of \(\lambda \in \mathcal{P}\). We replace the sum over \(k_1,\ldots,k_n\) by a sum over partitions. Then, we have \(\binom{k_1+\ldots+k_n}{k_1,\ldots,k_n} = \binom{|\lambda|}{\lambda_1,\ldots,\lambda_n}\) and \[\mathcal{R}_n(z;q)=\sum_{\substack{\lambda \in \mathcal{P}\\\ell(\lambda)=n}} {\binom{|\lambda|}{\lambda_1,\ldots,\lambda_n}} \binom{n}{r_1(\lambda),r_2(\lambda),\ldots}z^{|\lambda|}\prod_{j=1}^n (q)_\infty R_{\lambda_j}(q).\] The claim follows once we show that \[\label{show} n\,\Big|\, \binom{|\lambda|}{\lambda_1,\ldots,\lambda_n}\binom{n}{r_1(\lambda),r_2(\lambda),\ldots}.\tag{37}\]
Let \(d\) be the greatest common divisor of \(r_1(\lambda)\), \(r_2(\lambda), \ldots\). Since \(\sum_{m\geq 1} r_m(\lambda)=n\), we have that \(d\mid n\) and we write \(n=rd\). Moreover, note that one can write the partition \((\lambda_1,\ldots,\lambda_n)\) as \((\ell_1,\ldots,\ell_1,\ell_2,\ldots,\ell_2,\ldots,\ell_r,\ldots,\ell_r)\), where each \(\ell_j\) is repeated \(d\) times. First, we show that \[\label{eq:sdiv} d\,\Big|\,\binom{|\lambda|}{\lambda_1,\ldots,\lambda_n}.\tag{38}\] We factorize the multinomial coefficient as \[\label{split} \binom{|\lambda|}{\lambda_1,\ldots,\lambda_n} = \binom{|\lambda|}{\tfrac{|\lambda|}{d},\ldots, \tfrac{|\lambda|}{d}} \binom{\frac{|\lambda|}{d}}{\ell_1,\ldots,\ell_r}^{\!d},\tag{39}\] where \(\tfrac{|\lambda|}{d}\) occurs \(d\) times in the first multinomial coefficient on the right-hand side. All of the multinomial coefficients are integers and by Lemma 14 with \(n=|\lambda|\) and \(a_j=\frac{|\lambda|}{d}\) for \(j\in \{1,\ldots,d\}\), we find that \(d\) divides the first factor in 39 and hence, 38 holds.
Again using Lemma 14, in this case with \(a_m=r_m(\lambda)\) and \(d=\gcd(a_1,\ldots,a_\ell)\), we have \(r\,|\,\binom{n}{r_1(\lambda),r_2(\lambda),\ldots}.\) Combining this with 38 gives 37 . ◻
Finally, we refine Lemma 19 by computing the first Fourier coefficients of \(f_k\). In particular, the \(f_k\) are naturally normalized in the sense that the second Fourier coefficient is one.
Proposition 22. For \(k\geq 2\) even we have \[f_k(\tau) = - \frac{B_k}{2k}+q^2+\left(2^{k}-1\right)q^3+O\left(q^4\right).\]
Proof. Recall that \((q)_\infty R_k(q)=0\) for \(k\) odd. By 32 , we have for \(k\) even \[(q)_\infty R_k(q)=2q^2+2\left(2^{k}-1\right)q^3+O\left(q^4\right).\] Hence, we have \(\prod_{j=1}^n(q)_\infty R_{k_j}(q)=O(q^4)\) if \(n\geq 2\) and \(k_j\in \mathbb{N}\) even. Therefore, by 36 , \[\begin{align} 2\sum_{k\geq 1} \left(f_{k} (\tau)+\frac{B_k}{2k}\right) {z^{k}} &= 2\sum_{\substack{k\geq 2 \\ k \text{ even}}}\!\left(q^2+\left(2^k-1\right)q^3\right) z^k + O\!\left(q^4\right). \qedhere \end{align}\] ◻
Remark 3. Using a similar argument as in the proof of Propostion 22, one can show that the \(n\)-th Fourier coefficient of \(f_k\) is a linear combination of the \(k\)-th powers of \(\{1,\ldots,n-1\}\).
Here, we write down the first Fourier coefficients of the Eisenstein–type series \(f_k\). We have \[\begin{align} f_2(\tau) & = -\frac{1}{24}+q^2+3 q^3+5 q^4+7 q^5+9 q^6+10 q^7+13 q^8+ O\!\left(q^9\right)\!, \\ f_4(\tau) & = \frac{1}{240}+q^2+15 q^3+59 q^4+139 q^5+255 q^6+406 q^7+595 q^8+O\!\left(q^9\right)\!, \\ f_6(\tau) & = -\frac{1}{504}+q^2+63 q^3+635 q^4+2827 q^5+8199 q^6+18550 q^7+36043 q^8+O\!\left(q^9\right)\!, \\ f_8(\tau) & = \frac{1}{480}+q^2+255 q^3+6179 q^4+53179 q^5+253815 q^6+844966 q^7+2234875 q^8+O\!\left(q^9\right)\!. \end{align}\] Some examples for Theorem 6 (1) are given by \[\begin{align} D(f_2) &= -f_2 G_2-\frac{f_2^2}{2}-\frac{f_4}{12}+\frac{3 G_2^2}{2}+\frac{G_4}{12},\\ D(f_4) &= 6 f_2^2 G_2-18 f_2 G_2^2-f_2 G_4-f_4 G_2-\frac{2}{3} f_2^3-\frac{7 f_4 f_2}{3}-\frac{f_6}{9}+18 G_2^3+3 G_2 G_4+\frac{G_6}{30},\\ D(f_6) &= -60 f_2^3 G_2+270 f_2^2 G_2^2+15 f_2^2 G_4-540 f_2 G_2^3+30 f_4 f_2 G_2-90 f_2 G_2 G_4-f_2 G_6\\ &-45 f_4 G_2^2-f_6 G_2-\frac{5 f_4 G_4}{2}+5 f_2^4-5 f_4 f_2^2-\frac{11 f_6 f_2}{3}-\frac{25 f_4^2}{4}-\frac{f_8}{8}+405 G_2^4\\ &+\frac{21855 G_4^2}{3652}+135 G_2^2 G_4+3 G_2 G_6-\frac{39 G_8}{51128}, \\ D(f_8) &= 840 f_2^4 G_2-5040 f_2^3 G_2^2-280 f_2^3 G_4+15120 f_2^2 G_2^3-840 f_4 f_2^2 G_2+2520 f_2^2 G_2 G_4\\ &+28 f_2^2 G_6-22680 f_2 G_2^4+2520 f_4 f_2 G_2^2-\frac{305970}{913} f_2 G_4^2+56 f_6 f_2 G_2-7560 f_2 G_2^2 G_4\\ &+140 f_4 f_2 G_4-168 f_2 G_2 G_6+\frac{39 f_2 G_8}{913}-2520 f_4 G_2^3-84 f_6 G_2^2+70 f_4^2 G_2-f_8 G_2 \\ &-420 f_4 G_2 G_4-\frac{14 f_6 G_4}{3} -\frac{14 f_4 G_6}{3}-56 f_2^5+\frac{280}{3} f_4 f_2^3-\frac{28}{3} f_6 f_2^2-\frac{70}{3} f_4^2 f_2-5 f_8 f_2\\ &-\frac{322 f_4 f_6}{9}-\frac{2 f_{10}}{15}+13608 G_2^5+\frac{917910}{913} G_2 G_4^2+7560 G_2^3 G_4+252 G_2^2 G_6\\ &+\frac{19352886 G_4 G_6}{1983949}+\frac{36751 G_{10}}{1803590}-\frac{117 G_2 G_8}{913}. \end{align}\]
We end by raising some open questions.
The three properties of the \(f_k\) given in Theorem 2 do not determine them uniquely. Therefore, we provide two recursive definitions of \(f_k\) in Theorem 3. It would be interesting to find “nice” properties that define an Eisenstein–type series uniquely. For example, the fact that \(G_k\) is a normalized Hecke eigenform of weight \(k\) which does not vanish at \(i\infty\) determines it uniquely. As the coefficient of \(q\) in the Fourier expansion of the \(f_k\) vanishes by Proposition 22, one deduces that \(f_k\) cannot be a Hecke eigenform with respect to the usual action of the Hecke operators on \(q\)-series.
The functions \(f_2,f_4,f_6,\ldots\) together with \(G_2,G_4\), and \(G_6\), do not seem to satisfy any algebraic relations. This has been verified numerically up to weight 24 (and also up to mixed weight 12). Is it indeed the case that the algebra \(\mathcal{F}\) is free?
What variations of the \(g_\ell\), in particular in the range of summation, are also mock modular? More concretely, for which \(a, b\in\mathbb{N}\) is the function \[\sum_{\substack{an-1\geq bm\geq b}} (an-bm)^{\ell-1} q^{nm}-\sum_{\substack{n-1\geq abm\geq ab}}(n-abm)^{\ell-1} q^{nm}\] an Eisenstein–type series? Note that \(a=2\) and \(b=3\) yields the function \(g_\ell\) in this paper. Are other choices of \(a\) and \(b\) also of particular interest? Do these functions, together with those in [8] and [9], form the first examples of a theory of (higher level) Eisenstein–type series?
The correct combinatorial values for the anomalous case \(n=1\) are \(1\) if \(m=0\) and \(0\) for \(m\neq 0\).↩︎
In [7] the authors wrote \(G_{k}\) for what here is \(2G_{k}\) and \(\mathrm{Tr}_k(\phi;\tau)\) for what is \(\mathrm{Tr}_{2k}(\phi,G;\tau)\) here.↩︎
We emphasize that these functions are not constructed by averaging, as is done for classical Eisenstein series. They are called Eisenstein-type because they satisfy properties analogous to those of classical Eisenstein series.↩︎
Here and throughout, we consider \(\tau\) and \(\overline{\tau}\) as independent variables.↩︎
We consider quasi-completions, because they are more analogous to the functions in Theorem 1. In the sequel, we also study the completions \(\widehat{f_k}\) of \(f_k\); see 20 and 22 for their definitions.↩︎
See Subsection 2.1 for the definition; in particular, \(\widehat{f}_k\) and \(f_k^*\) agree if \(k\neq 2\).↩︎
The \(f_j\) here should not be confused with the Eisenstein–type series \(f_j\).↩︎
We point out that we re-normalize the standard heat operator by multiplying by a factor of \(\frac{1}{4\pi^2}\) (see [23] for more details).↩︎
Note that here we suppress the dependence on \(\overline{\tau}\).↩︎
Again \(\mathbb{F}(z;\tau)\) depends on \(\overline{\tau}\).↩︎