Projections of modular forms on Eisenstein series and its application to Siegel’s formula

Abstract

Let $k \ge 2$ and N be positive integers and let $\chi$ be a Dirichlet character modulo N. Let f(z) be a modular form in $M_k(\Gamma _0(N),\chi )$. Then we have a unique decomposition $f(z)=E_f(z)+S_f(z)$, where $E_f(z) \in E_k(\Gamma _0(N),\chi )$ and $S_f(z) \in S_k(\Gamma _0(N),\chi )$. In this paper, we give an explicit formula for $E_f(z)$ in terms of Eisenstein series whose coefficients are sum of divisors function. Then we apply our result to certain families of eta quotients and to representations of positive integers by 2k–ary positive definite quadratic forms in order to give an alternative version of Siegel’s formula for the weighted average number of representations of an integer by quadratic forms in the same genus. Our formula for the latter is in terms of sum of divisors function and does not involve computation of local densities.

Appendix A The SAGE functions for computing the constant terms of eta quotients at a given cusp

Let $r_d \in {\mathbb {Z}}$, not all zeros, $N \in {\mathbb {N}}$ and define

$$\begin{aligned} f(z)=\prod _{d \mid N} \eta ^{r_d}(dz). \end{aligned}$$

Assuming f(z) to be a modular form the following SAGE functions (written using version 9.1 of the software [20]) help computing $[0]_{a/c} f$, the constant term of f(z) at the cusp a/c.

By Lemma 6 it will be sufficient to compute the constant terms of the eta quotient $f_k(z)$ defined by (17) at a set of inequivalent cusps of $\Gamma _0(24)$, which is done below with the help of this code. The set

$$\begin{aligned} \{ 1/1, 1/2, 1/3, 1/4, 1/6, 1/8, 1/12, 1/24 \} \end{aligned}$$

gives a complete set of inequivalent cusps of $\Gamma _0(24)$, see [5, Corollary 6.3.23]. Note that if k is fixed then the code can handle the vanishing order analysis. For instance the output for the code

will be 0. However, here we are working with a general k, and therefore, the order analysis has to be done manually. When $k \ge 1$, the vanishing order of $f_k(z)$ is greater than 0 at cusps $\{1/2,1/3,1/4,1/6,1/8,1/12\}$. Thus, we have

$$\begin{aligned}{}[0]_{1/2}f_k=0,~[0]_{1/3}f_k=0,~[0]_{1/4}f_k=0 ,~[0]_{1/6}f_k= 0,~[0]_{1/8}f_k= 0 ,~[0]_{1/12}f_k=0. \end{aligned}$$

To compute $[0]_{1/1} f_k$ and $[0]_{1/24} f_k$, we run the following code:

The output will be

Simplifying these we obtain

$$\begin{aligned}&[0]_{1/1} f_k=- \frac{i^{2k+1} \sqrt{6}}{3^{k+1} 2^{3k+2}},~ [0]_{1/24} f_k=1. \end{aligned}$$

Putting everything together, for all $k \ge 1$ we have

$$\begin{aligned}&[0]_{1/1} f_k =- \frac{i^{2k+1}\sqrt{6}}{3^{k+1} 2^{3k+2}},~ [0]_{1/2} f_k =0,~ [0]_{1/3}f_k =0,~ [0]_{1/4} f_k=0,\\&[0]_{1/6}f_k =0,~ [0]_{1/8}f_k =0,~ [0]_{1/12}f_k =0, [0]_{1/24}f_k =1. \end{aligned}$$