Proof of a conjecture of Farkas and Kra

TL;DR

This paper proves a conjectured modular equation by Farkas and Kra involving modular forms for certain congruence subgroups, establishing its validity for all odd integers greater than or equal to 2, and introduces a new related modular equation.

Contribution

The paper confirms a longstanding conjecture of Farkas and Kra for all odd integers and introduces a novel modular equation of similar type.

Findings

Confirmed the conjectured modular equation for all odd k ≥ 2

Established a new modular equation of Farkas and Kra type

Extended understanding of modular forms for congruence subgroups

Abstract

In this paper we prove a conjectured modular equation of Farkas and Kra, which involving a half sum of certain modular form of weight $1$ for congruence subgroup $\Gamma_1(k)$ with any prime $k$. We prove that their conjectured identity holds for all odd integer $k\ge 2$. A new modular equation of Farkas and Kra type is also established.

Equations

$$\theta \left[\begin{array}{c}ϵ\\ {ϵ}^{\prime }\end{array}\right](z,\tau )=\sum _{n\in \mathbb{Z}}\exp \left\{2\pi \mathrm{i}\left(\frac{1}{2}{\left(n+\frac{ϵ}{2}\right)}^2\tau +\left(n+\frac{ϵ}{2}\right)\left(z+\frac{{ϵ}^{\prime }}{2}\right)\right)\right\},$$

$$\frac{d}{d\tau }\log \left(\frac{\eta (k\tau )}{\eta (\tau )}\right)+\frac{1}{2\pi \mathrm{i}(k-2)}\sum _{0\le \ell \le \frac{k-3}{2}}{\left(\frac{{\theta }^{\prime }\left[\begin{array}{c}1\\ \frac{1+2\ell }{k}\end{array}\right](0,\tau )}{\theta \left[\begin{array}{c}1\\ \frac{1+2\ell }{k}\end{array}\right](0,\tau )}\right)}^2$$

$$\theta^{\prime}\left[\begin{matrix}1\\ \frac{1+2\ell}{k}\end{matrix}\right](0,\tau)=\frac{\partial}{\partial z}\theta\left[\begin{matrix}1\\ \frac{1+2\ell}{k}\end{matrix}\right](z,\tau)\bigg{|}_{z=0}.$$

$${\left[\frac{\partial }{\partial z}\log \left(\theta \left[\begin{array}{c}1\\ \ell /k\end{array}\right](0,\tau )\right)\right]}^2$$

$$G(k)={\Gamma }_1(k):=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\mathrm{SL}}_2(\mathbb{Z}):\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\equiv \left(\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right)(\mathrm{mod}k)\right\}.$$

$${\theta }_2(z,q)=\sum _{n\in \mathbb{Z}}{q}^{(2n+1{)}^2/8}{e}^{\mathrm{i}(2n+1)z}.$$

$$\theta \left[\begin{array}{c}1\\ {ϵ}^{\prime }\end{array}\right](z,\tau )={\theta }_2\left(\pi z+\frac{{ϵ}^{\prime }\pi }{2},q\right)$$

$$4(k-2)q\frac{d}{dq}\log \left(\frac{\eta (k\tau )}{\eta (\tau )}\right)-\sum _{\begin{array}{c}0\le \ell <k\\ \ell \equiv 1(\mathrm{mod}2)\end{array}}{\left[\frac{\partial }{\partial z}\log {\theta }_2\left(\frac{\ell }{2k}\pi ,q\right)\right]}^2=0.$$

$$S_{\delta }(k):=\sum _{\begin{array}{c}0\le \ell <k\\ \ell -k\equiv \delta (\mathrm{mod}2)\end{array}}{\left[\frac{\partial }{\partial z}\log {\theta }_2\left(\frac{\ell }{2k}\pi ,q\right)\right]}^2$$

$$S_{\delta }(k)=4(k-2)q\frac{d}{dq}\log \left(\frac{\eta (k\tau )}{\eta (\tau )}\right),$$

$$S_{\delta }(k)=4q\frac{d}{dq}\log \left(\frac{\eta (2k\tau {)}^{2k-2}}{\eta (\tau {)}^k\eta (k\tau {)}^{k-2}}\right).$$

$$S_{\delta }(k)=\sum _{\begin{array}{c}0\le \ell <k\\ \ell -k\equiv \delta (\mathrm{mod}2)\end{array}}{\left[\tan \left(\frac{\ell \pi }{2k}\right)-4\sum _{h=1}^{2k}(-1{)}^h\sin \left(\frac{\ell h\pi }{k}\right)\sum _{n\ge 1}\frac{q^{hn}}{1-q^{2kn}}\right]}^2.$$

$$\sum _{\begin{array}{c}0\le \ell <k\\ \ell -k\equiv \delta (\mathrm{mod}2)\end{array}}{\left[\tan \left(\frac{\ell \pi }{2k}\right)\right]}^2=\{\begin{array}{ll}\frac{(k-1)(k-2)}{6}& \text{i}f\delta =0,\\ \frac{k(k-1)}{6}& \text{i}f\delta =1.\end{array}$$

$${\left(1+2\sum _{n\ge 1}\frac{q^n+q^{2n}-q^{4n}-q^{5n}}{1-q^{6n}}\right)}^2=1+4\sum _{n\ge 1}\left(\frac{n{q}^n}{1-q^n}+\frac{n{q}^{3n}}{1-q^{3n}}-\frac{8n{q}^{6n}}{1-q^{6n}}\right).$$

$${\left(\frac{\partial }{\partial z}\log {\theta }_2(z,q)\right)}^2={\mathrm{T}}_{z,q}\left(\log {\theta }_2(z,q)\right),$$

$${\mathrm{T}}_{z,q}=-8q\frac{\partial }{\partial q}-\frac{{\partial }^2}{\partial z^2}.$$

$$\left(8q\frac{\partial }{\partial q}+\frac{{\partial }^2}{\partial z^2}\right){\theta }_2(z,q)=0,$$

$$\frac{1}{{\theta }_2(z,q)}\frac{{\partial }^2}{\partial z^2}{\theta }_2(z,q)=-8q\frac{\partial }{\partial q}\log {\theta }_2(z,q).$$

$$\frac{{\partial }^2}{\partial z^2}\log {\theta }_2(z,q)=\frac{1}{{\theta }_2(z,q)}\frac{{\partial }^2}{\partial z^2}{\theta }_2(z,q)-{\left(\frac{\partial }{\partial z}\log {\theta }_2(z,q)\right)}^2$$

$${\theta }_2(z,q)=q^{1/8}{e}^{-\mathrm{i}z}\prod _{n\ge 1}(1-q^n)\left(1+e^{-2\mathrm{i}z}{q}^n\right)\left(1+e^{2\mathrm{i}z}{q}^{n-1}\right).$$

$$-\frac{\partial }{\partial z}\log {\theta }_2\left(\frac{\ell }{2k}\pi ,q\right)=\tan \left(\frac{\ell \pi }{2k}\right)-4\sum _{h=1}^{2k}(-1{)}^h\sin \left(\frac{\ell h\pi }{k}\right)\sum _{n\ge 1}\frac{q^{hn}}{1-q^{2kn}}.$$

$$\frac{\partial }{\partial z}\log {\theta }_2(z,q)=-\tan (z)+4\sum _{n\ge 1}\frac{(-1{)}^n{q}^n}{1-q^n}\sin (2nz).$$

$$\sum _{n\ge 1}\frac{(-1{)}^n{q}^n}{1-q^n}\sin \left(2n\frac{\ell \pi }{2k}\right)$$

$$=\sum _{h=1}^{2k}(-1{)}^h\sin \left(\frac{\ell h\pi }{k}\right)\sum _{n\ge 0}\sum _{\ell \ge 1}{q}^{(2nk+h)\ell }$$

$$\frac{\partial }{\partial z}\log {\theta }_2\left(\frac{\ell }{2k}\pi ,q\right)=-\tan \left(\frac{\ell \pi }{2k}\right)+4\sum _{h=1}^{2k}(-1{)}^h\sin \left(\frac{\ell h\pi }{k}\right)\sum _{n\ge 1}\frac{q^{hn}}{1-q^{2kn}}.$$

$${\rm T}_{z,q}\left(\log\theta_{2}\left(kz,q^{k}\right)\right)\bigg{|}_{z=0}=8(k-1)q\frac{\,d}{\,dq}\log\left(\frac{\eta(2k\tau)^{2}}{\eta(k\tau)}\right)$$

$${\rm T}_{z,q}\left(\log\left(\frac{\theta_{2}\left(kz-\frac{\pi}{2},q^{k}\right)}{\theta_{2}\left(z-\frac{\pi}{2},q\right)}\right)\right)\bigg{|}_{z=0}=8q\frac{\,d}{\,dq}\log\left(\eta(k\tau)^{k-3}\eta(\tau)^{2}\right).$$

$$\frac{{\partial }^2}{\partial z^2}\log {\theta }_2(z,q)=-{\tan }^2(z)-1+8\sum _{n\ge 1}\frac{(-1{)}^{n}n{q}^n}{1-q^n}\cos (2nz)$$

$$\frac{{\partial }^2}{\partial z^2}\log {\theta }_2(z-\pi /2,q)=-{\cot }^2(z)-1+8\sum _{n\ge 1}\frac{n{q}^n}{1-q^n}\cos (2nz).$$

$$\displaystyle\frac{\partial^{2}}{\partial z^{2}}\log\theta_{2}(z,q)\bigg{|}_{z=0}=$$

$$=$$

Full text

This research was supported by the National Science Foundation of China (Grant No. 11571114).

1. Introduction and statement of results

In this paper, we let $z\in{\mathbb{C}}$, $\tau\in{\mathbb{C}}$ with $\Im(\tau)>0$ and $q=e^{2\pi\mathrm{i}\tau}$. The theta function with characteristic $\left[\begin{matrix}\epsilon\\ \epsilon^{\prime}\end{matrix}\right]\in\mathbb{R}^{2}$ is defined by

[TABLE]

which is a generalization of the Jacobi theta functions. The theory of above theta function was systematically studied by Farkas and Kra [1], which play an important role in combinatorial number theory, algebraic geometry and physics.

In [1, Chapter 4], Farkas and Kra treated the theta function (1.1) with $\epsilon,\epsilon^{\prime}\in\mathbb{Q}$ and $z=0$, that is, the theta constants with rational characteristics. Their derived many interesting results, one of them is the following (see [1, Theorem 9.8, p.318] and [2]):

Theorem 1.1**.**

For each odd prime $k$ and all $\tau\in{\mathbb{C}}$ with $\Im(\tau)>0$,

[TABLE]

is a cusp $1$-form (cusp form of weight $1$) for the Hecke congruence subgroup $\Gamma_{o}(k)$. This form is identically zero provided $k\leq 19$. Here $\eta(\tau)=q^{1/24}\prod_{n\geq 1}(1-q^{n})$ is the Dedekind eta function and

[TABLE]

They then in [1, Conjecture 9.10, p.320] (see also [2]) conjectured that (1.2) is identically zero for each odd prime $k$ and all $\tau\in{\mathbb{C}}$ with $\Im(\tau)>0$.

Remark 1.1*.*

We remark that for odd integers $k,\ell$ with $k\geq 3$,

[TABLE]

is a modular $1$-form (modular form of wight $1$) for the group:

[TABLE]

This fact and more related results can be found in [1, 2].

The aim of this paper is give a proof of the conjecture of Farkas and Kra of above. For the simplicity of the proof, we shall introduce the Jacobi theta function $\theta_{2}(z,q)$, which is defined by (see for example [3]):

[TABLE]

Hence it is clear that

[TABLE]

and the conjecture of we concerned is equivalent to the following.

Conjecture 1**.**

For each odd prime $k$ and all $\tau\in{\mathbb{C}}$ with $\Im(\tau)>0$,

[TABLE]

We shall prove a more general result than Conjecture 1. To statement our main result, we shall consider the following half sum:

[TABLE]

for each integer $k\geq 2$ and each $\delta\in\{0,1\}$. Our main result is the following two modular equations.

Theorem 1.2**.**

For all $\tau\in{\mathbb{C}}$ with $\Im(\tau)>0$, we have if $\delta=0$ then

[TABLE]

and if $\delta=1$ then

[TABLE]

We immediately obtain the proof of Conjecture 1 by setting $k\in 2\mathbb{Z}_{+}+1$ and $\delta=0$ in Theorem 1.2.

Corollary 1.3**.**

Conjecture 1 holds for all odd integer $k\geq 2$. In particular, Conjecture 1 is true.

We shall give some consequence of Theorem 1.2. For this purpose we first use Lemma 2.2 of below deduce the proposition as follows.

Proposition 1.4**.**

We have

[TABLE]

By setting $q=0$ in Theorem 1.2, application Proposition 1.4 and (2.3) of below we obtain the following trigonometric identity, which has been appeared in [1, 2].

Corollary 1.5**.**

For each integer $k\geq 2$,

[TABLE]

From Theorem 1.2, Proposition 1.4 and (2.3), by choose different pair $(k,\delta)$ one can obtain many Lambert series identities. For example, if we pick $(k,\delta)=(3,1)$, then it is easy to see that:

Corollary 1.6**.**

We have

[TABLE]

2. Primaries

We shall need the following primary results, which will be used to prove main results of this paper.

Proposition 2.1**.**

We have:

[TABLE]

where and throughout, ${\rm T}_{z,q}$ is a linear operator be defined as

[TABLE]

Proof.

By (1.3) it is clear that

[TABLE]

which means that

[TABLE]

Then from the basic fact that

[TABLE]

we complete the proof of the proposition. ∎

We need the Jacobi triple product identity for $\theta_{2}(z,q)$ (see for example [4, 3]),

[TABLE]

Lemma 2.2**.**

For each $\ell,k\in\mathbb{Z}$ with $\ell\neq k$ and $k>0$,

[TABLE]

Proof.

Taking the logarithmic derivative of $\theta_{2}(z,q)$ respect to $z$ by (2.1), we have the well known Fourier expansion:

[TABLE]

Notice that

[TABLE]

and (2.2) we immediately obtain that

[TABLE]

This completes the proof of the lemma. ∎

The following lemma will be used to proof Theorem 1.2 in next section.

Lemma 2.3**.**

We have:

[TABLE]

and

[TABLE]

Proof.

By (2.2) we have:

[TABLE]

and

[TABLE]

Hence we obtain that

[TABLE]

and

[TABLE]

By using of the fact that

[TABLE]

and the above we obtain

[TABLE]

and

[TABLE]

Moreover, by (2.1) and the definition of $\eta(\tau)$, it is easy to see that

[TABLE]

and

[TABLE]

Thus for integer $k\geq 1$, application of (2.4) and (2.6) implies that

[TABLE]

and application of (2.5) and (2.7) implies that

[TABLE]

which completes the proof of the lemma. ∎

We need the following half product formula for Jacobi theta function $\theta_{2}$, which will be used to proof Theorem 1.2 in next section.

Lemma 2.4**.**

For integer $k\geq 1$ and $\delta\in\{0,1\}$,

[TABLE]

where $C_{k,\delta}=e^{\frac{\mathrm{i}\pi}{2}(\delta-k+{\bf 1}_{k\not\equiv\delta~{}(\bmod 2)})}$. Here and throughout, ${\bf 1}_{condition}=1$ if the ’condition’ is true, and equals to [math] if the ’condition’ is false.

Proof.

From (2.1) we have

[TABLE]

It is easy to check that

[TABLE]

and

[TABLE]

Thus we obtain that

[TABLE]

with

[TABLE]

which completes the proof of the lemma. ∎

3. The proof of Theorem 1.2

First of all, we shall define

[TABLE]

then from (1.4) we have $S_{\delta}(k)=G_{\delta,k}(0,q)$. By Proposition 2.1 we get

[TABLE]

We shall define the auxiliary function as follows:

[TABLE]

We claim that

[TABLE]

In fact, by $\theta_{2}(z+\pi,q)=-\theta_{2}(z,q)$ and $\theta_{2}(z,q)=\theta_{2}(-z,q)$ we have

[TABLE]

By the definition of ${\rm T}_{z,q}$ and above we immediately obtain the proof of (3.1). From Lemma 2.4 we find that

[TABLE]

which implies that

[TABLE]

Further by Lemma 2.3 we obtain that

[TABLE]

and

[TABLE]

which completes the proof of Theorem 1.2 by note that $S_{\delta}(k)=G_{\delta,k}(0,q)=\frac{1}{2}A_{1,k}(0,q)$.

Acknowledgment

The author would like to thank his advisor Zhi-Guo Liu for consistent encouragement and useful suggestions.