Poincaré square series for the Weil representation

Abstract

We calculate the Jacobi Eisenstein series of weight $k \ge 3$ for a certain representation of the Jacobi group, and evaluate these at $z = 0$ to give coefficient formulas for a family of modular forms $Q_{k,m,\beta }$ of weight $k \ge 5/2$ for the (dual) Weil representation on an even lattice. The forms we construct have rational coefficients and contain all cusp forms within their span. We explain how to compute the representation numbers in the coefficient formulas for $Q_{k,m,\beta }$ and the Eisenstein series of Bruinier and Kuss p-adically to get an efficient algorithm. The main application is in constructing automorphic products.

Appendices

Appendix A: Averaging operators

For applications to automorphic products, we do not need the Eisenstein series $E_{k,\beta }$ for any nonzero $\beta \in {\varLambda }'/{\varLambda }$ with $q(\beta ) \in {\mathbb {Z}}$. This is essentially because the constant terms ${\mathfrak {e}}_{\beta }$, $\beta \ne 0$ are not counted towards the principal part of the input function F in Borcherds’ lift. However, the $E_{k,\beta }$ are still necessary in order to span the full space of modular forms.

It seems difficult to apply the formula for $E_{k,\beta }$ in [5] directly since the Kloosterman sums there do not reduce to Ramanujan sums. A brute-force way to find $E_{k,\beta }$ is to search for ${\varDelta } \cdot E_{k,\beta }$ as a linear combination of Poincaré square series, but this is usually messy. Instead, we mention here that averaging over Schrödinger representations allows one in many (but not all) cases to read off the coefficients of all $E_{k,\beta }$ from those of $E_{k,0}$.

Definition 33

Let $\beta \in {\varLambda }'/{\varLambda }$ have denominator $d_{\beta }$. The averaging operator attached to $\beta$ is

$$\begin{aligned} A_{\beta } : M_k(\rho ^*)&\rightarrow M_k(\rho ^*) \\ A_{\beta } F(\tau )&= \frac{1}{d_{\beta ^2}} \sum _{\lambda ,\mu \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} \sigma _{\beta }^*(\lambda ,\mu ,0) F(\tau ). \end{aligned}$$

This is well defined because $\sigma _{\beta }^*(\lambda ,\mu ,0)$ depends only on the remainder of $\lambda$ and $\mu$ mod $d_{\beta }^2$, and it defines a modular form because

$$\begin{aligned} \Big ( \sigma _{\beta }^*(\zeta ) F \Big ) |_{k,\rho ^*} M (\tau ) = \sigma _{\beta }^*(\zeta \cdot M) F(\tau ) \end{aligned}$$

for all $\zeta \in {\mathcal {H}}$ and $M \in \tilde{\varGamma }.$

Explicitly, if the components of F are written out as

$$\begin{aligned} F(\tau ) = \sum _{\gamma \in {\varLambda }'/{\varLambda }} f_{\gamma }(\tau ) {\mathfrak {e}}_{\gamma }, \end{aligned}$$

then

$$\begin{aligned} A_{\beta } F(\tau ) = \frac{1}{d_{\beta }^2} \sum _{\gamma \in {\varLambda }'/{\varLambda }} \sum _{\lambda \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} \sum _{\mu \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} {\mathbf {e}}\Big ( -\mu \langle \beta , \gamma \rangle + \lambda \mu \cdot q(\beta ) \Big ) f_{\gamma }(\tau ) {\mathfrak {e}}_{\gamma - \lambda \beta }. \end{aligned}$$

The sum over $\mu$ is nonzero exactly when $\langle \beta , \gamma \rangle - \lambda q(\beta ) \in {\mathbb {Z}}$, in which case it becomes $d_{\beta }^2$; therefore,

$$\begin{aligned} A_{\beta } F(\tau ) = \sum _{\lambda \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} \sum _{\begin{array}{c} \gamma \in {\varLambda }'/ {\varLambda } \\ \langle \beta , \gamma \rangle - \lambda q(\beta ) \in {\mathbb {Z}} \end{array}} f_{\gamma }(\tau ) {\mathfrak {e}}_{\gamma - \lambda \beta }. \end{aligned}$$

In the special case that $q(\beta ) \in {\mathbb {Z}}$, this is a constant multiple of the modified averaging operator

$$\begin{aligned}A'_{\beta } F(\tau ) = \sum _{\begin{array}{c} \gamma \in {\varLambda }'/{\varLambda } \\ \langle \beta , \gamma \rangle \in {\mathbb {Z}} \end{array}} \left( \sum _{\lambda \in {\mathbb {Z}}/d_{\beta } {\mathbb {Z}}} f_{\gamma + \lambda \beta }(\tau ) \right) {\mathfrak {e}}_{\gamma }, \end{aligned}$$

making this easier to compute.

When $F = E_{k,0}$ is the Eisenstein series

$$\begin{aligned} E_{k,0}(\tau ) = \frac{1}{2} \sum _{c,d} (c \tau + d)^{-k} \rho ^*(M)^{-1} {\mathfrak {e}}_0, \end{aligned}$$

then we get

$$\begin{aligned} A_{\beta } E_{k,0}(\tau )&= \frac{1}{2 d_{\beta }^2} \sum _{c,d} (c \tau + d)^{-k} \sum _{\lambda ,\mu \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} \sigma _{\beta }^* (\lambda ,\mu ,0) \rho ^*(M)^{-1} {\mathfrak {e}}_0 \\&= \frac{1}{2 d_{\beta }^2} \sum _{c,d} (c \tau + d)^{-k} \sum _{\lambda , \mu \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}}} {\mathbf {e}}\Big ( \lambda \mu q(\beta ) \Big ) \rho ^*(M)^{-1} {\mathfrak {e}}_{-\lambda \beta } \\&= \sum _{\begin{array}{c} \lambda \in {\mathbb {Z}}/d_{\beta }^2 {\mathbb {Z}} \\ \lambda q(\beta ) \in {\mathbb {Z}} \end{array}} E_{k,\lambda \beta }. \end{aligned}$$

In many cases this makes it possible to find all the Eisenstein series $E_{k,\beta }.$

Example 34

Let S be the Gram matrix $S = \begin{pmatrix} -8 \end{pmatrix}.$ The Eisenstein series of weight 5 / 2 is

$$\begin{aligned} E_{5/2,0}(\tau )&= \Big ( 1 - 24q - 72q^2 - 96q^3 - \ldots \Big ) {\mathfrak {e}}_{0} \\&\quad + \Big ( -\frac{1}{2}q^{1/16} - 24q^{17/16} - 72 q^{33/16} - \ldots \Big ) ({\mathfrak {e}}_{1/8} + {\mathfrak {e}}_{7/8}) \\&\quad + \Big ( -5 q^{1/4} - 24q^{5/4} - 125q^{9/4} - \ldots \Big ) ({\mathfrak {e}}_{1/4} + {\mathfrak {e}}_{3/4} ) \\&\quad + \Big ( -\frac{25}{2} q^{9/16} - \frac{121}{2} q^{25/16} - 96 q^{41/16} - \ldots \Big ) ({\mathfrak {e}}_{3/8} + {\mathfrak {e}}_{5/8}) \\&\quad + \Big ( -46 q - 48q^2 - 144q^3 - \ldots \Big ) {\mathfrak {e}}_{1/2}. \end{aligned}$$

Averaging over the Schrödinger representation attached to $\beta = (1/2)$ gives

$$\begin{aligned} E_{5/2,0}(\tau ) + E_{5/2,1/2}(\tau )&= \Big ( 1 - 70q - 120q^2 - 240q^3 - \ldots \Big ) ({\mathfrak {e}}_0 + {\mathfrak {e}}_{1/2}) \\&\quad + \Big ( -10q^{1/4} - 48q^{5/4} - 250q^{9/4} - \ldots \Big ) ({\mathfrak {e}}_{1/4} + {\mathfrak {e}}_{3/4}), \end{aligned}$$

from which we can read off the Fourier coefficients of $E_{5/2,1/2}(\tau ).$

Appendix B: Calculating the Euler factors at $p = 2$

We will summarize the calculations of Appendix B in [7] as they apply to our situation.

Proposition 35

Let $f(X) = \bigoplus _{i \in {\mathbb {N}}_0} 2^i Q_i(X) \oplus L + c$ be a ${\mathbb {Z}}_2$-integral quadratic polynomial in normal form, and assume that all $Q_i$ are given by $Q_i(v) = v^T S_i v$ for a symmetric (not necessarily even) ${\mathbb {Z}}_2$-integral matrix $S_i$. For any $j \in {\mathbb {N}}_0$, define

$$\begin{aligned} \mathbf {Q}_{(j)} := \bigoplus _{\begin{array}{c} 0 \le i \le j \\ i \equiv j \, (2) \end{array}} Q_i, \; \; {\mathbf {r}}_{(j)} = {\mathrm {rank}}(\mathbf {Q}_{(j)}), \; \; {\mathbf {p}}_{(j)} = 2^{\sum _{0 \le i < j} {\mathbf {r}}_{(i)}}. \end{aligned}$$

Let $\omega \in {\mathbb {N}}_0$ be such that $Q_i = 0$ for all $i > \omega .$ Then:

• (i) If $L = 0$ and $c = 0$, let $r = \sum _i {\mathrm {rank}}(Q_i)$; then the Igusa zeta function for f at 2 is

  $$\begin{aligned} \zeta _{Ig}(f;2;s)&= \sum _{0 \le \nu < \omega - 1} \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I_0(\mathbf {Q}_{(\nu )}, \mathbf {Q}_{(\nu +1)}, Q_{\nu +2}) \; \\&\quad \quad \quad + \left[ \frac{2^{-s(\omega - 1)}}{{\mathbf {p}}_{(\omega - 1)}} I_0(\mathbf {Q}_{(\omega -1)}, \mathbf {Q}_{(\omega )}, 0) + \frac{2^{-\omega s}}{{\mathbf {p}}_{(\omega )}} I_0(\mathbf {Q}_{(\omega )}, \mathbf {Q}_{(\omega - 1)}, 0) \right] \cdot \\&\qquad \qquad \times (1 - 2^{-2s - r})^{-1}. \end{aligned}$$

• (ii) If $L(x) = bx$ for some $b \ne 0$ with $v_2(b) = \lambda$ and if $v_2(b) \le v_2(c)$, then

  $$\begin{aligned} \zeta _{Ig}(f;2;s)&= \sum _{0 \le \nu< \lambda - 2} \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I_0({\mathbf {Q}}_{(\nu )}, {\mathbf {Q}}_{(\nu +1)}, Q_{\nu +2}) \\&\quad \quad \quad + \sum _{\max \{0,\lambda - 2\} \le \nu < \lambda } \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I_0^{\lambda - \nu }({\mathbf {Q}}_{(\nu )}, {\mathbf {Q}}_{(\nu +1)}, Q_{i+2})\\&\quad \quad \quad + \frac{2^{-\lambda s}}{{\mathbf {p}}_{(\lambda )}} \cdot \frac{1}{2 - 2^{-s}}. \end{aligned}$$

• (iii) If $L(x) = bx$ with $b \ne 0$ and $v_2(c) < v_2(b) \le v_2(c) + 2$, let $\kappa = v_2(c)$; then

  $$\begin{aligned} \zeta _{Ig}(f;2;s)&= \sum _{0 \le \nu < \lambda - 2} \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I_{c / 2^{\nu }}({\mathbf {Q}}_{(\nu )}, {\mathbf {Q}}_{(\nu + 1)}, Q_{\nu + 2}) \\&\quad \quad \quad + \sum _{\max \{0,\lambda - 2\} \le \nu \le \kappa } \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I^{\lambda - \nu }_{c / 2^{\nu }}( {\mathbf {Q}}_{(\nu )}, {\mathbf {Q}}_{(\nu + 1)}, Q_{\nu + 2})\\&\quad \quad \quad + \frac{1}{{\mathbf {p}}_{(\kappa + 1)}} 2^{-\kappa s}. \end{aligned}$$

• (iv) If $L = 0$ or $L(x) = bx$ with $v_2(b) > v_2(c) + 2$, let $\kappa = v_2(c)$; then

  $$\begin{aligned} \zeta _{Ig}(f;2;s)&= \sum _{0 \le \nu \le \kappa } \frac{2^{-\nu s}}{{\mathbf {p}}_{(\nu )}} I_{c / 2^{\nu }}({\mathbf {Q}}_{(\nu )}, {\mathbf {Q}}_{(\nu + 1)}, Q_{\nu + 2}) + \frac{1}{{\mathbf {p}}_{(\kappa + 1)}} 2^{-\kappa s}. \end{aligned}$$

Here, $I_a^b(Q_0,Q_1,Q_2)(s)$ are helper functions that we describe below, and we set $I_a(Q_0,Q_1,Q_2) = I^{\infty }_a(Q_0,Q_1,Q_2).$ Note that not every unimodular quadratic form $Q_i$ over ${\mathbb {Z}}_2$ can be written in the form $Q_i(v) = v^T S_i v$; but $2 \cdot Q_i$ can always be written in this form, and replacing f by $2 \cdot f$ only multiplies $\zeta _{Ig}(f;2;s)$ by $2^{-s}$, so this does not lose generality.

Every unimodular quadratic form over ${\mathbb {Z}}_2$ that has the form $Q_i(v) = v^T S_i v$ is equivalent to a direct sum of at most two one-dimensional forms $a \cdot \mathrm {Sq}(x) = ax^2$; at most one elliptic plane $\mathrm {Ell}(x,y) = 2x^2 + 2xy + 2y^2$; and any number of hyperbolic planes $\mathrm {Hyp}(x,y) = 2xy.$ This decomposition is not necessarily unique. It will be enough to fix one such decomposition.

The following proposition explains how to compute $I_a^b(Q_0,Q_1,Q_2)(s).$

Proposition 36

Define the function

$$\begin{aligned} \mathrm {Ig}(a,b,\nu ) = {\left\{ \begin{array}{ll} \frac{2^{-\nu s}}{2 - 2^{-s}}: &{} v_2(a) \ge \mathrm {min}(b,\nu ); \\ 2^{-v_2(a) s}: &{} v_2(a) < \mathrm {min}(b,\nu ). \end{array}\right. } \end{aligned}$$

(Here, $v_2(0) = \infty$.) For a unimodular quadratic form Q of rank r, fix a decomposition into hyperbolic planes, at most one elliptic plane and at most two square forms as above. Let $\varepsilon =1$ if Q contains no elliptic plane and $\varepsilon = -1$ otherwise. Define functions $H_1(a,b,Q)$, $H_2(a,b,Q)$, and $H_3(a,b,Q)$ as follows:

• (i) If Q contains no square forms, then

  $$\begin{aligned} H_1(a,b,Q)&= (1 - 2^{-r}) \mathrm {Ig}(a,b,1); \\ H_2(a,b,Q)&= \Big ( 1 - 2^{-r/2} \varepsilon \Big ) \cdot \Big ( \mathrm {Ig}(a,b,1) + 2^{-r/2}\varepsilon \mathrm {Ig}(a,b,2) \Big ); \\ H_3(a,b,Q)&= 0. \end{aligned}$$

• (ii) If Q contains one square form $cx^2$, then

  $$\begin{aligned} H_1(a,b,Q)&= \mathrm {Ig}(a,b,0) - 2^{-r} \mathrm {Ig}(a,b,1); \\ H_2(a,b,Q)&= (1 - 2^{-(r-1)/2} \varepsilon ) \mathrm {Ig}(a,b,0) - 2^{-r} \mathrm {Ig}(a,b,2) \\&\quad \quad \quad +2^{-(r+1)/2}\varepsilon (\mathrm {Ig}(a,b,2) + \mathrm {Ig}(a+c,b,2)); \\ H_3(a,b,Q)&= 2^{-r} (\mathrm {Ig}(a+c,b,3) - \mathrm {Ig}(a+c,b,2)). \end{aligned}$$

• (iii) If Q contains two square forms $cx^2$, $dx^2$, and $c + d \equiv 0 \, (4)$, then

  $$\begin{aligned} H_1(a,b,Q)&= \mathrm {Ig}(a,b,0) - 2^{-r} \mathrm {Ig}(a,b,1); \\ H_2(a,b,Q)&= \mathrm {Ig}(a,b,0) - 2^{-r/2}\varepsilon \mathrm {Ig}(a,b,1) + (2^{-r/2}\varepsilon - 2^{-r}) \mathrm {Ig}(a,b,2); \\ H_3(a,b,Q)&= (-1)^{(c+d)/4} 2^{-r} (\mathrm {Ig}(a,b,3) - \mathrm {Ig}(a,b,2)). \end{aligned}$$

• (iv) If Q contains two square forms $cx^2$, $dx^2$, and $c + d \not \equiv 0 \, (4)$, then

  $$\begin{aligned} H_1(a,b,Q)&= \mathrm {Ig}(a,b,0) - 2^{-r} \mathrm {Ig}(a,b,1); \\ H_2(a,b,Q)&= (1 - 2^{-(r-2)/2} \varepsilon ) \mathrm {Ig}(a,b,0) + 2^{-r/2} \varepsilon (\mathrm {Ig}(a,b,1)\\&\quad \quad \quad + \mathrm {Ig}(a+c,b,2)) - 2^{-r} \mathrm {Ig}(a,b,2); \\ H_3(a,b,Q)&= -2^{1-r} \mathrm {Ig}(a,b,1) + 2^{-r} (\mathrm {Ig}(a,b,2) + \mathrm {Ig}(a+c+d,b,3) ). \end{aligned}$$

Let $\varepsilon _1 = 1$ if $Q_1$ contains no elliptic plane and $\varepsilon _1 = -1$ otherwise, and let $r_1$ denote the rank of $Q_1$. Then $I_a^b(Q_0,Q_1,Q_2)$ is given as follows:

• (1) If both $Q_1$ and $Q_2$ contain at least one square form, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2) = H_1(a,b,Q_0). \end{aligned}$$

• (2) If $Q_1$ contains no square forms but $Q_2$ contains at least one square form, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2) = H_2(a,b,Q_0). \end{aligned}$$

• (3) If both $Q_1$ and $Q_2$ contain no square forms, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2) = H_2(a,b,Q_0) + 2^{-r_1 / 2} \varepsilon _1 H_3(a,b,Q_0). \end{aligned}$$

• (4) If $Q_1$ contains one square form $cx^2$, and $Q_2$ contains no square forms, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2)= & {} H_1(a,b,Q_0) \\&\quad + 2^{-(r_1 + 1)/2} \varepsilon _1 (H_3(a,b,Q_0) + H_3(a+2c,b,Q_0)). \end{aligned}$$

• (5) If $Q_1$ contains two square forms $cx^2$ and $dx^2$ such that $c+d \equiv 0\, (4)$, and $Q_2$ contains no square forms, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2) = H_1(a,b,Q_0) + 2^{-r_1 / 2} \varepsilon _1 H_3(a,b,Q_0). \end{aligned}$$

• (6) If $Q_1$ contains two square forms $cx^2$ and $dx^2$ such that $c +d \not \equiv 0 \, (4)$, and $Q_2$ contains no square forms, then

  $$\begin{aligned} I_a^b(Q_0,Q_1,Q_2) = H_1(a,b,Q_0) + 2^{-r_1/2} \varepsilon _1 H_3(a+c,b,Q_0). \end{aligned}$$

Proof

In the notation of [7],

$$\begin{aligned} \mathrm {Ig}(a,b,\nu ) = \mathrm {Ig}(z^{a + 2^b {\mathbb {Z}}_2 + 2^{\nu } {\mathbb {Z}}_2}) \end{aligned}$$

and

$$\begin{aligned} H_1(a,b,Q) = \mathrm {Ig}\Big (z^{a + 2^b {\mathbb {Z}}_2} \hat{H}_Q(z)\Big ) \end{aligned}$$

and

$$\begin{aligned} H_2(a,b,Q) = \mathrm {Ig}\Big (z^{a + 2^b {\mathbb {Z}}_2} \tilde{H}_Q(z)\Big ) \end{aligned}$$

and

$$\begin{aligned} H_3(a,b,Q) = \mathrm {Ig}\Big (z^{a + 2^b {\mathbb {Z}}_2} (H_Q(z) - \tilde{H}_Q(z))\Big ). \end{aligned}$$

This calculation of $I_a^b(Q_0,Q_1,Q_2)$ is available in Appendix B of [7]. Finally, the calculation of $\zeta _{Ig}(f;2;s)$ is given in Theorem 4.5 loc. cit. $\square$