1 Correction to: Ramanujan J (2015) 36:103–116 https://doi.org/10.1007/s11139-014-9621-4

This note corrects the proof of Theorem 1.1 of [1], and extends the statement of the result to odd m and also furnishes the missed statement with regard to the funding obtained from the European Research council and that provided to A. Folsom in the article note.

2 Introduction and statement of results

Let for \(m\in \mathbb {N}\)

$$\begin{aligned} \varphi _m(z)=\varphi _m(z;\tau ):=\left( \frac{\vartheta \left( z+\frac{1}{2}\right) }{\vartheta (z)}\right) ^m, \end{aligned}$$

where \((q:=e^{2\pi i\tau }, \zeta :=e^{2\pi iz}\) with \(\tau \in {\mathbb {H}}, z\in \mathbb {C}\))

$$\begin{aligned} \vartheta (z) = \vartheta (z;\tau ) := \sum _{\nu \in \frac{1}{2}+\mathbb {Z}} e^{\pi i \nu \tau + 2\pi i\nu \left( z+\frac{1}{2}\right) } \end{aligned}$$

is the Jacobi theta function. Note that in contrast to [1], we write \(\varphi _m\) in order to highlight the dependence on m. Denote the coefficients of the Fourier expansion (in z) by \(\chi _r,\) so that

$$\begin{aligned} \varphi _m(z;\tau ) =: \sum _{r\in \mathbb Z} \chi _r(\tau ) \zeta ^r. \end{aligned}$$
(1.1)

Define the Nebentypus character \(\psi _m\) for matrices \(\gamma = ({\begin{matrix}a&{}b\\ c&{}d\end{matrix}}) \in \Gamma _0(2)\) by

$$\begin{aligned} \psi _m(\gamma ) := e^{\frac{\pi i m}{2}\left( \frac{c}{2}d + d-1\right) }. \end{aligned}$$
(1.2)

Moreover, we require the well-known Eisenstein series \(E_{2j}(\tau )\). For \(j\ge 2\), they are holomorphic modular forms, while \(E_2(\tau )\) is a quasimodular form. The Bernoulli numbers \(B_{\ell }\) are defined for non-negative integers \(\ell \) by the generating function

$$\begin{aligned} \frac{t}{e^t-1}=\sum _{\ell \ge 0} B_\ell \frac{t^\ell }{\ell !}. \end{aligned}$$

Theorem 1.1

For \(r \in \mathbb {Z}\) and \(m \in \mathbb {N}\), we have

$$\begin{aligned} \begin{aligned} \chi _r(\tau )&= \frac{q^r}{1+(-1)^{m+1}q^{r}}\sum _{0\le \ell <\frac{m}{2}} {r^{m-2\ell -1}}\frac{D_{m-2\ell }(\tau )}{(m-2\ell -1)!}\\&\qquad \qquad (\text {assuming that } r\ne 0 \text { if } m \text { is even}), \\ \chi _0(\tau )&= D_0(\tau ) +\sum _{1\le j \le \frac{m}{2}} \frac{ B_{2j}}{(2j)!} D_{2j}(\tau )E_{2j}(\tau ) \quad \quad \ \text { for } m \text { even}, \end{aligned} \end{aligned}$$

where for each \(0\le j \le m\) such that \(j \equiv m \pmod {2}\), the function \(D_j\) is a modular form of weight \(-j\) on \(\Gamma _0(2)\) with Nebentypus character \(\psi _m\), as defined in (1.2).

Remark

Theorem 1.1 was given for even m in [1]; above, we have extended the statement to hold for odd m. Moreover, the proof in [1] had a mistake: the second displayed formula in the proof of Proposition 3.3 was incorrect. We thank Sander Zwegers for pointing out the mistake and for fruitful discussion.

3 Proof of Theorem 1.1

Using that, for \(\lambda , \mu \in \mathbb {Z}\), we have

$$\begin{aligned} \vartheta (z+\lambda \tau +\mu )&=(-1)^{\lambda +\mu } q^{-\frac{\lambda ^2}{2}} e^{-2\pi i\lambda z}\vartheta (z),\\ \vartheta \left( z+\frac{1}{2}+\lambda \tau +\mu \right)&=(-1)^\mu q^{-\frac{\lambda ^2}{2}}e^{-2\pi i\lambda z}\vartheta \left( z+\frac{1}{2}\right) , \end{aligned}$$

we obtain that

$$\begin{aligned} \varphi _m(z+\lambda \tau +\mu )=(-1)^{m\lambda }\varphi _m(z). \end{aligned}$$
(2.1)

Let for \(z_0\in \mathbb {C},\ \tau \in \mathbb {H}\)

$$\begin{aligned} P_{z_0}:=\left\{ z_0+r\tau +s:0\le r,s\le 1 \right\} . \end{aligned}$$

Then, with \(z_0\) such that no pole of \(\varphi _m\) lies at the boundary of \(P_{z_0}\), we compute

$$\begin{aligned} \int _{\partial P_{z_0}} \varphi _m(w)e^{-2\pi irw}\mathrm{d}w= & {} \left( \int _{z_0}^{z_0+1} + \int _{z_0+1}^{z_0+1+\tau } +\int _{z_0+1+\tau }^{z_0+\tau } +\int _{z_0+\tau }^{z_0} \right) \varphi _m(w)e^{-2\pi irw} \mathrm{d}w\nonumber \\= & {} \int _0^1 \varphi _m(z_0+t)e^{-2\pi ir(z_0+t)} \mathrm{d}t\nonumber \\&+\,\tau \int _0^1 \varphi _m(z_0+1+t\tau )e^{-2\pi ir(z_0+t\tau )} \mathrm{d}t\nonumber \\&-\,\int _0^1 \varphi _m(z_0+\tau +t)e^{-2\pi ir(z_0+\tau +t)} \mathrm{d}t\nonumber \\&-\,\tau \int _0^1 \varphi _m(z_0+t\tau )e^{-2\pi ir(z_0+t\tau )} \mathrm{d}t. \end{aligned}$$
(2.2)

Using (2.1) gives

$$\begin{aligned} \varphi _m(z_0+1+t\tau )=\varphi _m(z_0+t\tau ),\quad \varphi _m(z_0+t+\tau )=(-1)^m\varphi _m(z_0+t). \end{aligned}$$

Thus (2.2) becomes

$$\begin{aligned} e^{-2\pi irz_0}\left( 1+(-1)^{m+1} e^{-2\pi ir\tau }\right) \int _0^1 \varphi _m(z_0+t) e^{-2\pi ir t}\mathrm{d}t. \end{aligned}$$

Inserting the Fourier expansion of \(\varphi _m\) yields

$$\begin{aligned} \int _0^1 \varphi _m\left( z_0+t\right) e^{-2\pi i rt}\mathrm{d}t=\sum _{\ell \in \mathbb {Z}}\chi _\ell (\tau )e^{2\pi i\ell z_0} \int _0^1 e^{2\pi i(\ell -r)t}\mathrm{d}t=\chi _r(\tau )e^{2\pi irz_0}. \end{aligned}$$

So (assuming \(r\ne 0\) if m is even)

$$\begin{aligned} \chi _r(\tau )=\frac{(-1)^{m+1}q^r}{1+(-1)^{m+1}q^r}\int _{\partial P_{z_0}}\varphi _m(w) e^{-2\pi irw}\mathrm{d}w. \end{aligned}$$
(2.3)

We now compute (2.2) in another way, picking \(z_0=-\frac{1}{2} -\frac{\tau }{2}\). Then the only pole of \(\varphi _m\) in \(P_{z_0}\) is at \(z=0\). So, using the Residue Theorem, (2.2) equals

$$\begin{aligned} 2\pi i{\text {Res}}_{z=0}\left( \varphi _m(z)e^{-2\pi irz}\right) . \end{aligned}$$
(2.4)

Write (noting that \(\varphi _m\) is even or odd, depending on the parity of m)

$$\begin{aligned} \varphi _m(z)=\sum _{m-2\ell >0}\frac{D_{m-2\ell }(\tau )}{(2\pi iz)^{m-2\ell }}+O(1). \end{aligned}$$
(2.5)

Inserting the series expansion of \(e^{-2\pi irz}\), (2.4) becomes

$$\begin{aligned} (-1)^{m+1}\sum _{0\le \ell <\frac{m}{2}} r^{m-2\ell -1}\frac{D_{m-2\ell }(\tau )}{(m-2\ell -1)!}. \end{aligned}$$

Thus, for \(r\in \mathbb {Z}\) (with the restriction that \(r\ne 0\) if m is even) we obtain by comparing with (2.3),

$$\begin{aligned} \chi _r(\tau )=\frac{q^r}{1+(-1)^{m+1} q^r}\sum _{0\le \ell < \frac{m}{2}}r^{m-2\ell -1}\frac{D_{m-2\ell }(\tau )}{(m-2\ell -1)!}. \end{aligned}$$

This gives the first equation in Theorem 1.1.

To determine \(\chi _0\) (for m even), we plug in to (1.1), which implies

$$\begin{aligned} \varphi _m(z)=\sum _{1\le \ell \le \frac{m}{2}} \frac{D_{2\ell }(\tau )}{(2\ell -1)!}\sum _{r\in \mathbb {Z}\setminus \{0\}}\frac{r^{2\ell -1}q^r \zeta ^r}{1-q^r}+\chi _0(\tau ). \end{aligned}$$
(2.6)

We now insert the Laurent expansions around \(z=0\) on both sides. We write the sum on r as

$$\begin{aligned} \sum _{r\ge 1} \frac{r^{2\ell -1}q^r \zeta ^r}{1-q^r}+\sum _{r\ge 1}\frac{r^{2\ell -1}\zeta ^{-r}}{1-q^r}. \end{aligned}$$
(2.7)

It is not hard to see that both sums converge absolutely for \(-v<y<0\), where \(v := {\text {Im}}(\tau ), y := {\text {Im}}(z)\). We write the second summand in (2.7) as

$$\begin{aligned} \sum _{r\ge 1}\frac{r^{2\ell -1}\zeta ^{-r}}{1-q^r}=\sum _{r\ge 1} r^{2\ell -1}\zeta ^{-r}+\sum _{r\ge 1}\frac{r^{2\ell -1}\zeta ^{-r}q^r}{1-q^r}. \end{aligned}$$
(2.8)

The first summand equals

$$\begin{aligned} \left( -\frac{1}{2\pi i}\frac{\partial }{\partial z}\right) ^{2\ell -1}{\sum _{r\ge 1} \zeta ^{-r}}&= \left( -\frac{1}{2\pi i}\frac{\partial }{\partial z}\right) ^{2\ell -1}\frac{1}{\zeta -1}\\&=\left( -\frac{1}{2\pi i}\frac{\partial }{\partial z}\right) ^{2\ell -1}\left( \frac{B_0}{2\pi iz}+\frac{B_{2\ell }(2\pi iz)^{2\ell -1}}{(2\ell )!}\right) +O\left( z^2\right) \\&=\frac{(2\ell -1)!}{(2\pi iz)^{2\ell }}-\frac{B_{2\ell }}{2\ell } +O\left( z^2\right) . \end{aligned}$$

The second summand combines with the first summand in (2.7) as using that \(\varphi _m\) is an even function of z,

$$\begin{aligned} 2\sum _{r\ge 1}\frac{r^{2\ell -1}q^r}{1-q^r}+O\left( z^2\right) . \end{aligned}$$

Thus the right hand side in (2.6) becomes

$$\begin{aligned}&{\sum _{1\le \ell \le \frac{m}{2}} \frac{D_{2\ell }(\tau )}{(2\ell -1)!} \left( {2\sum _{r\ge 1}\frac{r^{2\ell -1}q^r}{1-q^r}-\frac{B_{2\ell }}{2\ell }}+\frac{(2\ell -1)!}{(2\pi iz)^{2\ell }}\right) }+\chi _0(\tau )+O\left( z^2\right) \\&\quad {=-\sum _{1\le \ell \le \frac{m}{2}}\frac{D_{2\ell }(\tau )}{(2\ell )!}B_{2\ell }E_{2\ell }(\tau )+\sum _{1\le \ell \le \frac{m}{2}}\frac{D_{2\ell }(\tau )}{(2\pi iz)^{2\ell }}}+ \chi _0(\tau ) +O\left( z^2\right) . \end{aligned}$$

Picking off the constant term on both sides of (2.5) then gives

$$\begin{aligned} \chi _0(\tau )= D_0(\tau )+\sum _{1\le \ell \le \frac{m}{2}} \frac{D_{2\ell }(\tau )}{(2\ell )!}B_{2\ell }E_{2\ell }(\tau ), \end{aligned}$$

as claimed.

The proof of the modularity follows from the fact that for \(\gamma =({\begin{matrix}a&{}b\\ c&{}d\end{matrix}}) \in \Gamma _0(2)\), we have that

$$\begin{aligned} \varphi _m\left( \frac{z}{c\tau +d};\gamma \tau \right)&= \psi _m(\gamma ) \varphi _m(z;\tau ). \end{aligned}$$