On the non-vanishing of theta lifting of Bianchi modular forms to Siegel modular forms

Abstract

In this paper we study the theta lifting of a weight 2 Bianchi modular form ${\mathcal {F}}$ of level $\Gamma _0({\mathfrak {n}})$ with ${\mathfrak {n}}$ square-free to a weight 2 holomorphic Siegel modular form. Motivated by Prasanna’s work for the Shintani lifting, we define the local Schwartz function at finite places using a quadratic Hecke character $\chi$ of square-free conductor ${\mathfrak {f}}$ coprime to level ${\mathfrak {n}}$. Then, at certain 2 by 2 g matrices $\beta$ related to ${\mathfrak {f}}$, we can express the Fourier coefficient of this theta lifting as a multiple of $L({\mathcal {F}},\chi ,1)$ by a non-zero constant. If the twisted L-value is known to be non-vanishing, we can deduce the non-vanishing of our theta lifting.

1 Introduction

Shimura initiated the systematic study of holomorphic modular forms of half-integral weight and provided a correspondence between certain modular forms of even weight and modular forms of half-integral weight. Later, in the other direction, Shintani [21] described a method in terms of weighted periods of holomorphic cusp forms to construct modular forms of half-integral weight. Waldspurger showed in [23] a proportional relation between special values of L-functions attached to an eigenform of even weight and the square of the square-free Fourier coefficients of the Shintani lifting. For the special case of modular forms on the full modular group, Kohnen-Zagier [14] proved a simple version of Waldspurger’s theorem with the constant of proportionality given explicitly. Inspired by their work we will analyse the theta lifting of Bianchi modular forms to Siegel modular forms and investigate the relationship between Fourier coefficients of this lifting and special L-values attached to the Bianchi modular forms. This can be used to describe the non-vanishing of the theta lifting, which is an open problem in general.

To construct the theta lifting of a weight 2 Bianchi modular form ${\mathcal {F}}$ for level $\Gamma _0({\mathfrak {n}})$ with ${\mathfrak {n}}$ a square-free ideal for an imaginary quadratic field F of class number one, following [3] and [13] we consider the 4-dimensional rational quadratic space V given by Hermitian matrices with entries in F. Its associated symmetric space D is isomorphic to the upper half space ${\mathbb {H}}_3$. In our theta integral we use the differential form $\eta _{\mathcal {F}}$ attached to ${\mathcal {F}}$ defined on the arithmetic quotient $\Gamma \backslash D$. We choose the Schwartz form $\varphi =\varphi _\infty ^{\text {KM}}\varphi _f$ defined on a pair of vectors in V so that the theta kernel is given by

$$\begin{aligned} \theta (g,h,z):=\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in V^2}\omega (g,h)\varphi ({\textbf{x}}_1,{\textbf{x}}_2;z)\quad \text { for }g\in {{\,\textrm{Sp}\,}}_4\subset {{\,\textrm{GL}\,}}_4, h\in {{\,\textrm{SO}\,}}(3,1). \end{aligned}$$

Then the theta lifting is constructed as

$$\begin{aligned} \Theta _\varphi (\eta _{\mathcal {F}})(g)=\int _{\Gamma \backslash D}\eta _{\mathcal {F}}(z)\wedge \theta (g,h,z) \end{aligned}$$

which by results of Kudla and Millson turns out to be a weight 2 holomorphic Siegel modular form. Its Fourier coefficient at a $2\times 2$ symmetric matrix $\beta >0$ is given here by

$$\begin{aligned} \sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma \backslash \Omega _\beta }\varphi _f({\textbf{x}}_1,{\textbf{x}}_2)\int _{C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\eta _{\mathcal {F}} \end{aligned}$$

where $\Omega _\beta :=\{({\textbf{x}}_1,{\textbf{x}}_2)\in V^2:(({\textbf{x}}_i,{\textbf{x}}_j))=\beta \}$ and $U({\textbf{x}}_1,{\textbf{x}}_2):=\text {Span}\{{\textbf{x}}_1,{\textbf{x}}_2\}\subset V$. For an auxiliary quadratic Hecke character $\chi$ with its conductor coprime to ${\mathfrak {n}}$ we define the Schwartz form as $\varphi =\varphi _\infty ^{\text {KM}}\varphi _f^\chi$. The choice of the Schwartz function $\varphi _f^\chi$ in Sect. 4 is crucial for us to get the period integral related to some twisted L-values. With this choice we take certain $\beta >0$ (again depending on the conductor of $\chi$) at which the coefficient of $\Theta _\varphi (\eta _{\mathcal {F}})$ is expressed as the above weighted sum of period integrals over infinite geodesics joining two cusps. By [24, Theorem 1.8], the period integral over infinite geodesics ending in $\infty$ can be related to $L({\mathcal {F}},\chi ,1)$. We apply Atkin-Lehner operators to transform other infinite geodesics and reduce to this case. In Sect. 5 I compute the coefficient at such a $\beta$ as a multiple of $L({\mathcal {F}},\chi ,1)$ by a non-vanishing number. By Friedberg-Hoffstein’s theorem [8, Theorem B], we can deduce that there always exists a character $\chi$ such that the twisted L-value is non-vanishing which implies the non-vanishing of the corresponding theta lifting.

Theorem 1.1

(Theorem 5.17) Let $F={\mathbb {Q}}(\sqrt{d})$ (square-free $d<0$) be an imaginary quadratic field of class number one and denote by ${\mathcal {O}}$ its ring of integers. Consider a weight 2 Bianchi cusp form ${\mathcal {F}}$ of level $\Gamma _0({\mathfrak {n}})$ with ${\mathfrak {n}}$ a square-free ideal away from ramified primes in $F/{\mathbb {Q}}$. Given a square-free product m of split or inert primes in $F/{\mathbb {Q}}$ such that $(m,{\mathfrak {n}})=1$, for ${\mathfrak {m}}=m{\mathcal {O}}$ we choose a quadratic Hecke character $\chi _{\mathfrak {m}}$ of conductor ${\mathfrak {m}}\sqrt{d}$. Then, at certain $\beta >0$ related to m, the Fourier coefficient of the theta lifting $\Theta _\varphi (\eta _{\mathcal {F}})$ can be computed as

$$\begin{aligned} (*)\cdot L({\mathcal {F}},\chi _{\mathfrak {m}},1) \end{aligned}$$

where the non-zero constant $(*)$ is given explicitly in (38).

Constructions of congruences between automorphic forms are one of important tools, particularly those between cusp forms and theta lifts on ${{\,\textrm{GSp}\,}}_4({\mathbb {A}})$. [17] gave an explicit construction of the theta lifts of automorphic forms for imaginary quadratic fields to further study the congruence on ${{\,\textrm{GSp}\,}}_4({\mathbb {A}})$. Different to our approach to construct the Schwartz form (defining the finite part), its choice is to define a distinguished infinite part while at each finite place the local Schwartz function is the characteristic function in Sect. 4.3.

It is known that Böcherer formulates an equality between sums of Fourier coefficients of Siegel modular forms and certain L-values. [5] proved a precise formula relating the Bessel period of lifted automorphic forms on ${{\,\textrm{GSp}\,}}_4({\mathbb {A}})$ to central L-values, where the Bessel period is the Fourier coefficient considered by Böcherer. We have not calculated the Bessel period of our theta lifts but in the future we expect to relate some combinations of the Fourier coefficients of our lifts to special L-values. This would lead us to investigate the connections between our calculations and those on the Bessel period, and how our further result would be related to Böcherer’s conjecture.

2 Binary Hermitian form

In this section we recall some basics from linear algebra about Hermitian matrices and Hermitian binary forms from [7, Chapter 9].

For an complex matrix A, the matrix ${\bar{A}}$ is obtained from A by applying complex conjugation to all entries and the matrix $A^t$ is the transpose of A. An $n\times n$ matrix A with complex entries is called Hermitian if ${\bar{A}}^t=A$. By the definition we see that an Hermitian matrix is unchanged by taking its conjugate transpose. Note that any Hermitian matrix must have real diagonal entries.

Let R be a subring of ${\mathbb {C}}$ with $R={\bar{R}}$. We write ${\mathcal {H}}(R)$ for the set of Hermitian $2\times 2$ matrices with entries in R, i.e.

$$\begin{aligned} {\mathcal {H}}(R)=\{A\in M_2(R)\mid {\bar{A}}^{t}=A\}. \end{aligned}$$

Every $f\in {\mathcal {H}}(R)$ defines a binary Hermitian form with coefficients in R. If $f=\left( \begin{array}{ll}a&{}b\\ {\bar{b}}&{}d\end{array}\right)$ then the associated binary Hermitian form is the semi quadratic map $f:{\mathbb {C}}\times {\mathbb {C}}\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} f(u,v)=(u,v)\left( \begin{array}{ll}a&{}b\\ {\bar{b}}&{}d\end{array}\right) ({\bar{u}},{\bar{v}})^{t}=au{\bar{u}}+bu{\bar{v}}+{\bar{b}}{\bar{u}}v+dv{\bar{v}}. \end{aligned}$$

We shall often call an element $f\in {\mathcal {H}}(R)$ a binary hermitian form with coefficients in R. The discriminant $\Delta (f)$ of $f\in {\mathcal {H}}(R)$ is defined as $\Delta (f)=\det (f)$. Set $|a|=(a{\bar{a}})^{1/2}$ for $a\in {\mathbb {C}}$ where $^{-}$ denotes the complex conjugation. We define the ${{\,\textrm{GL}\,}}_2(R)$-action on ${\mathcal {H}}(R)$ given by the formula

$$\begin{aligned} \sigma \cdot f=(|\det (\sigma )|^{-1/2}\sigma ) f(|\det ({\bar{\sigma }}^t)|^{-1/2}{\bar{\sigma }}^t)=|\det (\sigma )|^{-1}\sigma f{\bar{\sigma }}^{t} \end{aligned}$$

(1)

for $\sigma \in {{\,\textrm{GL}\,}}_2(R)$ and $f\in {\mathcal {H}}(R)$. If $\sigma =\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \in {{\,\textrm{GL}\,}}_2(R)$ we have

$$\begin{aligned} \sigma \cdot f=|\det (\sigma )|^{-1}\left( \begin{matrix}(\alpha ,\beta )f({\bar{\alpha }},{\bar{\beta }})^t&{}(\alpha ,\beta )f(\gamma ,\delta )^t\\ (\gamma ,\delta )f({\bar{\alpha }},{\bar{\beta }})^t&{}(\gamma ,\delta )f({\bar{\gamma }},{\bar{\delta }})^t\end{matrix}\right) . \end{aligned}$$

Note that $\Delta (\sigma \cdot f)=\Delta (f)$ for every $\sigma \in {{\,\textrm{GL}\,}}_2(R)$ and $f\in {\mathcal {H}}(R)$. Two elements $f,g\in {\mathcal {H}}(R)$ are called ${{\,\textrm{GL}\,}}_2(R)$-equivalent if $g=\sigma \cdot f$ for some $\sigma \in {{\,\textrm{GL}\,}}_2(R)$; ${{\,\textrm{SL}\,}}_2(R)$-equivalence is defined analogously.

A binary Hermitian form $f\in {\mathcal {H}}(R)$ is positive definite if $f(u,v)>0$ for all $(u,v)\in {\mathbb {C}}\times {\mathbb {C}}\setminus \{(0,0)\}$. If $-f$ is positive definite f is called negative definite. If $\Delta (f)<0$ then f is called indefinite.

We define

$$\begin{aligned} \begin{aligned} {\mathcal {H}}^+(R)=\{f\in {\mathcal {H}}(R)\mid f \text { is positive definite}\}\\ {\mathcal {H}}^-(R)=\{f\in {\mathcal {H}}(R)\mid f \text { is indefinite}\}. \end{aligned} \end{aligned}$$

Clearly the group ${{\,\textrm{GL}\,}}_2(R)$ leaves the ${\mathcal {H}}^{\pm }$ invariant. It is easy to see that $f\in {\mathcal {H}}^+(R)$ if and only if $a>0$ and $\Delta (f)>0$. The group ${\mathbb {R}}_{>0}$ acts on ${\mathcal {H}}^+({\mathbb {C}})$ by scalar multiplication. Similarly ${\mathbb {R}}^\times$ acts on ${\mathcal {H}}^-({\mathbb {C}})$. We define

$$\begin{aligned} \tilde{{\mathcal {H}}}^+({\mathbb {C}}):={\mathcal {H}}^+({\mathbb {C}})/{\mathbb {R}}_{>0}, \quad \tilde{{\mathcal {H}}}^-({\mathbb {C}}):={\mathcal {H}}^-({\mathbb {C}})/{\mathbb {R}}^\times . \end{aligned}$$

For $f\in {\mathcal {H}}^{\pm }({\mathbb {C}}),[f]$ stands for the class of f in $\tilde{{\mathcal {H}}}^{\pm }({\mathbb {C}})$. The action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ on ${\mathcal {H}}^{\pm }({\mathbb {C}})$ clearly induces an action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ on $\tilde{{\mathcal {H}}}^{\pm }({\mathbb {C}})$. The centre of ${{\,\textrm{SL}\,}}_2({\mathbb {C}})$ acts trivially on ${\mathcal {H}}({\mathbb {C}})$, so we get an induced action of ${{\,\textrm{PSL}\,}}_2({\mathbb {C}})$ on ${\mathcal {H}}({\mathbb {C}})$ and $\tilde{{\mathcal {H}}}^{\pm }({\mathbb {C}})$.

Recall the upper half space ${\mathbb {H}}_3={\mathbb {C}}\times {\mathbb {R}}_{>0}$, elements of which can be written as (z, r) with $z=x+iy$ for $x,y\in {\mathbb {R}},r\in {\mathbb {R}}_{>0}$.

Definition 2.1

The map $\Phi :{\mathcal {H}}^+({\mathbb {C}})\rightarrow {\mathbb {H}}_3$ is defined as

$$\begin{aligned} \phi :f=\left( \begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}\right) \rightarrow \frac{b}{d}+\frac{\sqrt{\Delta (f)}}{d}\cdot j \end{aligned}$$

In fact $\phi$ induces a map $\phi :\tilde{{\mathcal {H}}}^+({\mathbb {C}})\rightarrow {\mathbb {H}}_3$.

This map is a bijection since for a point $(z,r)\in {\mathbb {H}}_3$ there exists $f=\left( {\begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}}\right)$ such that $\phi (f)=z+rj\in {\mathbb {H}}_3$. Therefore, this map gives a one to one correspondence between equivalence classes of positive definite Hermitian forms and points in the upper half space. Note that $\Phi$ is the analogue of identification of the set of equivalence classes of binary positive definite quadratic forms with points of the upper half plane.

Proposition 2.2

The map $\phi :\tilde{{\mathcal {H}}}^+({\mathbb {C}})\rightarrow {\mathbb {H}}_3$ is a ${{\,\textrm{PSL}\,}}_2({\mathbb {C}})$-equivariant bijection; that is $\phi (\sigma \cdot f)=\sigma \cdot \phi (f)$ for every $\sigma \in {{\,\textrm{PSL}\,}}_2({\mathbb {C}})$ and $f\in \tilde{{\mathcal {H}}}^+$.

Proof

See [7, Proposition 9.1.2, Chapter 9]. $\square$

Definition 2.3

For a binary Hermitian form $f=\left( {\begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}}\right) \in {\mathcal {H}}^-({\mathbb {C}})$ we define

$$\begin{aligned} \psi (f)=\{z+rj\in {\mathbb {H}}_3\mid a-{\bar{b}}z-b{\bar{z}}+dz{\bar{z}}+r^2d=0\} \end{aligned}$$

and ${\textbf{G}}=\{\psi (f)\mid f\in {\mathcal {H}}^-({\mathbb {C}})\}$ which is a set of geodesic planes in ${\mathbb {H}}_3$.

Remark 2.4

This map $\psi$ is slightly different to the map in [7, Definition 1.3, Chapter 9] which is given by

$$\begin{aligned} f\mapsto \{z+rj\in {\mathbb {H}}_3: a+{\bar{b}}z+b{\bar{z}}+dz{\bar{z}}+r^2d=0\}. \end{aligned}$$

The above map $\psi$ is chosen for us to prove Proposition 3.4. In addition we will consider the cycle $D_U$ for positive definite U generated by f with $f\in {\mathcal {H}}^-({\mathbb {C}})$.

If $d\ne 0$ then $\psi (f)$ is the following geodesic hemisphere

$$\begin{aligned} \psi (f)=\{z+rj\in {\mathbb {H}}_3\mid |dz-b|^2+|d|^2r^2=-\Delta (f)\}. \end{aligned}$$

If $d=0$ then $\psi (f)$ is a vertical plane. The group ${{\,\textrm{PSL}\,}}_2({\mathbb {C}})$ acts on ${\textbf{G}}$ by its induced action on subsets of ${\mathbb {H}}_3$. Clearly $\psi$ induces a map $\psi :\tilde{{\mathcal {H}}}^-({\mathbb {C}})\rightarrow {\textbf{G}}$.

Proposition 2.5

The map $\psi :\tilde{{\mathcal {H}}}^-({\mathbb {C}})\rightarrow {\textbf{G}}$ is a ${{\,\textrm{PSL}\,}}_2({\mathbb {C}})$-equivariant bijection; that is $\psi (\sigma \cdot f))=\sigma \cdot \psi (f)$ for every $\sigma \in {{\,\textrm{PSL}\,}}_2({\mathbb {C}})$ and $f\in \tilde{{\mathcal {H}}}^-({\mathbb {C}})$.

Proof

We will prove the equivariance property only for the generators of ${{\,\textrm{PSL}\,}}_2({\mathbb {C}})$.

Let $\sigma =\left( {\begin{matrix}1&{}\beta \\ 0&{}1\end{matrix}}\right)$ where $\beta \in {\mathbb {C}}$. Then

$$\begin{aligned} \sigma \cdot f=\left( \begin{matrix}1&{}\beta \\ 0&{}1\end{matrix}\right) \left( \begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}\right) \left( \begin{matrix}1&{}0\\ {\bar{\beta }}&{}1\end{matrix}\right) =\left( \begin{matrix}a+\beta {\bar{b}}+{\bar{\beta }}b+\beta {\bar{\beta }}d&{}b+\beta d\\ {\bar{b}}+{\bar{\beta }}d&{}d\end{matrix}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} \psi (\sigma \cdot f)\!=\!\{z+rj\in {\mathbb {H}}_3\mid a\!+\!\beta {\bar{b}}\!+\!{\bar{\beta }}b+\beta {\bar{\beta }}d\!-\!({\bar{b}}+{\bar{\beta }}d)z-(b\!+\!\beta d){\bar{z}}+dz{\bar{z}}+r^2d=0\}. \end{aligned}$$

On the other hand, for $z+rj\in \psi (f)$, we have $\sigma \cdot (z+rj)=(z+\beta )+rj\in {\mathbb {H}}_3$. Setting $z^{'}=z+\beta$ and $r^{'}=r$, we observe that

$$\begin{aligned} a-{\bar{b}}(z^{'}-\beta )-b({\bar{z}}^{'}-{\bar{\beta }})+d(z^{'}-\beta )({\bar{z}}'-{\bar{\beta }})+r'^2d=0. \end{aligned}$$

Then it is not hard to see that $\psi (\sigma \cdot f))=\sigma \cdot \psi (f)$ for $\sigma =\left( {\begin{matrix}1&{}\beta \\ 0&{}1\end{matrix}}\right)$.

In the same way we prove this property for $\sigma =\left( {\begin{matrix}0&{}-1\\ 1&{}0\end{matrix}}\right)$. We have

$$\begin{aligned} \sigma \cdot f=\left( \begin{matrix}0&{}-1\\ 1&{}0\end{matrix}\right) \left( \begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}\right) \left( \begin{matrix}0&{}-1\\ 1&{}0\end{matrix}\right) =\left( \begin{matrix}d&{}-{\bar{b}}\\ -b&{}a\end{matrix}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} \psi (\sigma \cdot f)=\{z+rj\in {\mathbb {H}}_3\mid d+bz+{\bar{b}}{\bar{z}}+az{\bar{z}}+r^2a=0\}. \end{aligned}$$

For $z+rj\in \psi (f)$, we have $z^{'}+r^{'}j=\sigma \cdot (z+rj)=-\frac{{\bar{z}}}{|z|^2+r^2}+\frac{r}{|z|^2+r^2}j$. Then $|z'|^2+r'^2=\frac{1}{|z|^2+r^2}$. It follows that $z=-\frac{{\bar{z}}'}{|z'|^2+r'^2}$ and $r=\frac{r'}{|z'|^2+r'^2}$. Hence the following identity holds

$$\begin{aligned} a+{\bar{b}}\frac{{\bar{z}}'}{|z'|^2+r'^2}+b\frac{z}{|z'|^2+r'^2}+d\frac{z'{\bar{z}}'}{(|z'|^2+r'^2)^2}+d\frac{r'^2}{(|z'|^2+r'^2)^2}=0. \end{aligned}$$

Then it is not hard to see that $\psi (\sigma \cdot f))=\sigma \cdot \psi (f)$ for $\sigma =\left( {\begin{matrix}0&{}-1\\ 1&{}0\end{matrix}}\right)$. $\square$

3 Orthogonal group of sign (3, 1) and cycles

In this section we recall some basic aspects on orthogonal groups of signature (3,1) and cycles in this case from [3, Sect.  4].

Let $F={\mathbb {Q}}(\sqrt{d})$ $(d<0)$ be an imaginary quadratic field of class number 1. Denote by ${\mathcal {O}}$ by its ring of integers. For an ideal ${\mathfrak {n}}\subset {\mathcal {O}}$ put

$$\begin{aligned} \Gamma _0({\mathfrak {n}})=\left\{ \left( \begin{matrix}a&{}b\\ c&{}d\end{matrix}\right) \in {{\,\textrm{SL}\,}}_2({\mathcal {O}}):c\in {\mathfrak {n}}\right\} . \end{aligned}$$

Assume that the four-dimensional space V over ${\mathbb {Q}}$ is given by the hermitian matrices

$$\begin{aligned} V=\{{\textbf{x}}\in M_2(F):{\textbf{x}}^t=\overline{{\textbf{x}}}\}, \end{aligned}$$

with quadratic form

$$\begin{aligned} {\textbf{x}}\mapsto -\det ({\textbf{x}}) \end{aligned}$$

and corresponding bilinear form

$$\begin{aligned} ({\textbf{x}},{\textbf{y}})\mapsto -\frac{1}{2}{{\,\textrm{tr}\,}}({\textbf{x}}{\textbf{y}}^*), \end{aligned}$$

where

$$\begin{aligned} \left( \begin{matrix}a&{}b\\ c&{}d\end{matrix}\right) ^*=\left( \begin{matrix}d&{}-b\\ -c&{}a\end{matrix}\right) . \end{aligned}$$

(2)

Note that this bilinear form is preserved under the action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ where its action is given in (1); that is, for $g\in {{\,\textrm{GL}\,}}_2({\mathbb {C}})$,

$$\begin{aligned} (g\cdot {\textbf{x}},g\cdot {\textbf{y}})&=(|\det (g)|^{-1}g{\textbf{x}}{\bar{g}}^t,|\det (g)|^{-1}g{\textbf{y}}{\bar{g}}^t)\nonumber \\&=-\frac{1}{2}{{\,\textrm{tr}\,}}(|\det (g)|^{-1}g{\textbf{x}}{\bar{g}}^t|\det (g)|({\bar{g}}^t)^{-1}{\textbf{y}}^{-1}g^{-1}\det (|\det (g)|^{-1}g{\textbf{y}}{\bar{g}}^t))\nonumber \\&=-\frac{1}{2}{{\,\textrm{tr}\,}}(g{\textbf{x}}{\textbf{y}}^{-1}\det ({\textbf{y}})g^{-1})=({\textbf{x}},{\textbf{y}}). \end{aligned}$$

(3)

We fix an orthogonal basis of $V({\mathbb {Q}})$ given by $e_1=\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right)$, $e_2=\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right)$, $e_3=\left( {\begin{matrix}0&{}\sqrt{d}\\ -\sqrt{d}&{}0\end{matrix}}\right)$ and $e_4=\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right) =Z_0$ such that the discriminant of V is d. The basis of $Z_0^\perp$ can be identified with $\{ e_1,e_2,e_3\}$.

The symmetric space in this case can be realized as

$$\begin{aligned} D=\{Z\in V({\mathbb {R}}):(Z,Z_0)<0\}, \end{aligned}$$

which is isomorphic to hyperbolic 3-space ${\mathbb {H}}_3$. The isomorphism can be given by

$$\begin{aligned} \mu :z+rj\in {\mathbb {H}}_3\longmapsto \frac{1}{r}\left( \begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}\right) . \end{aligned}$$

(4)

The ${{\,\textrm{GL}\,}}_2$-action on the Hermitian form defined as in (1) induces that on ${\mathbb {H}}_3$ in the following. For $g=\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right)$ we have $g\cdot \frac{1}{r}\left( {\begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}}\right) =\frac{1}{r'}\left( {\begin{matrix}|z'|^2+r'^2&{}z'\\ {\bar{z}}'&{}1\end{matrix}}\right)$; expand the LHS,

$$\begin{aligned} g\cdot \frac{1}{r}\left( {\begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}}\right)&=r^{-1}|\det (g)|^{-1}\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \left( {\begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}}\right) \left( {\begin{matrix}{\bar{\alpha }}&{}{\bar{\gamma }}\\ {\bar{\beta }}&{}{\bar{\delta }}\end{matrix}}\right) \\&=r^{-1}|\det (g)|^{-1}\left( {\begin{matrix}\alpha |z|^2+\alpha r^2+\beta {\bar{z}}&{}\alpha z+\beta \\ \gamma |z|^2+\gamma r^2+\delta {\bar{z}}&{}\gamma z+\delta \end{matrix}}\right) \left( {\begin{matrix}{\bar{\alpha }}&{}{\bar{\gamma }}\\ {\bar{\beta }}&{}{\bar{\delta }}\end{matrix}}\right) \\&=r^{-1}|\det (g)|^{-1}\left( {\begin{matrix}\alpha {\bar{\alpha }}|z|^2+\alpha {\bar{\alpha }}r^2+{\bar{\alpha }}\beta {\bar{z}}+\alpha {\bar{\beta }}z+\beta {\bar{\beta }}&{}\alpha {\bar{\gamma }}|z|^2+\alpha {\bar{\gamma }}r^2+\beta {\bar{\gamma }}{\bar{z}}+\alpha {\bar{\delta }}z+\beta {\bar{\delta }}\\ {\bar{\alpha }}\gamma |z|^2+{\bar{\alpha }}\gamma r^2+{\bar{\alpha }}\delta {\bar{z}}+{\bar{\beta }}\gamma z+{\bar{\beta }}\delta &{}\gamma {\bar{\gamma }}|z|^2+\gamma {\bar{\gamma }} r^2+{\bar{\gamma }}\delta {\bar{z}}+\gamma {\bar{\delta }}z+\delta {\bar{\delta }}\end{matrix}}\right) , \end{aligned}$$

and then

$$\begin{aligned} z^{'}=\frac{(\alpha z+\beta )({\bar{\gamma }}{\bar{z}}+{\bar{\delta }})+\alpha {\bar{\gamma }}r^2}{|\gamma z+\delta |^2+|\gamma |^2r^2},r^{'}=\frac{|\alpha \delta -\beta \gamma |r}{|\gamma z+\delta |^2+|\gamma |^2r^2}. \end{aligned}$$

By (4), we can define the action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ on ${\mathbb {H}}_3$ to be as

$$\begin{aligned} \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \cdot (z,r)=\left( \frac{(\alpha z+\beta )({\bar{\gamma }}{\bar{z}}+{\bar{\delta }})+\alpha {\bar{\gamma }}r^2}{|\gamma z+\delta |^2+|\gamma |^2r^2},\quad \frac{|\alpha \delta -\beta \gamma |r}{|\gamma z+\delta |^2+|\gamma |^2r^2}\right) . \end{aligned}$$

(5)

Proposition 3.1

The above map $\mu$ as in (4) intertwines the ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$-action on $V({\mathbb {R}})$ and ${\mathbb {H}}_3$; that is $\mu (g\cdot (z,r))=g\cdot \mu (z,r)$ for $g\in {{\,\textrm{GL}\,}}_2({\mathbb {C}})$.

Proof

See [26, Proposition 4.2.1]. $\square$

The set $\text {Iso}(V)$ of all isotropic lines (1-dimensional ${\textbf{x}}\in V$ such that $q({\textbf{x}})=0$) in $V({\mathbb {Q}})$ can be identified with ${\mathbb {P}}^1(F)=F\cup \infty$ ($\infty =[1:0]$). Assume that the cusp $\infty$ corresponds to the isotropic line spanned by $u_\infty =\left( {\begin{matrix}1&{}0\\ 0&{}0\end{matrix}}\right)$. Given an element $g=\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \in {{\,\textrm{SL}\,}}_2(F)$ transforming the cusp $\infty$ to another cusp $\kappa =[\alpha :\gamma ]$, we can see that

$$\begin{aligned} g\cdot u_\infty =\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}1&{}0\\ 0&{}0\end{matrix}\right) \left( \begin{matrix}{\bar{\alpha }}&{}{\bar{\gamma }}\\ {\bar{\beta }}&{}{\bar{\delta }}\end{matrix}\right) =\left( \begin{matrix}\alpha {\bar{\alpha }}&{}\alpha {\bar{\gamma }}\\ {\bar{\alpha }}\gamma &{}\gamma {\bar{\gamma }}\end{matrix}\right) . \end{aligned}$$

Hence we can identify the cusp with the isotropic line by means of the map

$$\begin{aligned} \nu :[a:b]\longmapsto \text {span}\left( \begin{matrix}a{\bar{a}}&{}a{\bar{b}}\\ {\bar{a}}b&{}b{\bar{b}}\end{matrix}\right) \in \text {Iso}(V). \end{aligned}$$

(6)

Proposition 3.2

The above map $\nu$ satisfies

$$\begin{aligned} \nu (g\cdot [a:b])=g\cdot \nu ([a:b]) \end{aligned}$$

for $g\in {{\,\textrm{GL}\,}}_2(F)$ and $[a:b]\in {\mathbb {P}}^1(F)$.

Proof

See [26, Proposition 4.2.2]. $\square$

Let $U\subset V$ be a ${\mathbb {Q}}$-subspace with $\dim _{\mathbb {Q}}U=2$ such that $(\,\ )|_U$ is positive definite; say $U=\langle {\textbf{x}}_1,{\textbf{x}}_2\rangle$. Define the special cycle as

$$\begin{aligned} D_U=\{Z\in D:Z\perp U\}. \end{aligned}$$

Let $\Gamma$ be a torsion-free congruence subgroup of ${{\,\textrm{GL}\,}}(V)$ preserving the quadratic form on V and let $\Gamma _U$ be the stabilizer of U in $\Gamma \cap {{\,\textrm{SO}\,}}_0(3,1)({\mathbb {R}})$, where ${{\,\textrm{SO}\,}}_0$ is the identity component of ${{\,\textrm{SO}\,}}$. We denote the image of the quotient $\Gamma _U\backslash D_U$ in $\Gamma \backslash D$ by $C_U$. The stabilizer $\Gamma _U$ is either trivial (if the orthogonal complement $U^\perp \subset V$ is split over ${\mathbb {Q}}$) or infinite cyclic (if $U^\perp$ is non-split over ${\mathbb {Q}}$) (see [11, Lemma 4.2]). If $\Gamma _U$ is infinite, then $C_U$ is a closed geodesic in $\Gamma \backslash D$, while $C_U$ is infinite if $\Gamma _U$ is trivial (see [3, Sect. 4.3]).

Lemma 3.3

For $U=\langle {\textbf{x}}_1,{\textbf{x}}_2\rangle \subset V$ as above. Then the following two statements are equivalent:

• (1) $U^\perp$ is split over ${\mathbb {Q}}$,

• (2) $disc (U)\in -d({\mathbb {Q}}^\times )^2$.

Proof

For an arbitrary subspace U of a non-degenerate quadratic space V we have $\text{ dim }(V)=\text{ dim }(U)+\text{ dim }(U^\perp )$. Thus $U^\perp$ is also 2-dimensional. By assumption $U^\perp$ is a hyperbolic plane. By Witt’s Theorem (a 2-dimensional quadratic space over a field F is a hyperbolic plane if and only if its discriminant lies in $-(F^\times )^2$) we have ${\text{ disc }(U^\perp )\in -({\mathbb {Q}}^\times )^2}$. Thus $\text{ disc }(U)\in -d({\mathbb {Q}}^\times )^2$ as $\text{ disc }(V)=\text{ disc }(U)\text{ disc }(U^\perp )$.

Conversely suppose ${\text{ disc }(U)\in -d({\mathbb {Q}}^\times )^2}$. Again by $\text{ disc }(V)=\text{ disc }(U)\text{ disc }(U^\perp )$, we have $\text{ disc }(U^\perp )\in -({\mathbb {Q}}^\times )^2$ implying that $U^\perp$ is split over ${\mathbb {Q}}$. $\square$

We orient $D_U$ by requiring that a tangent vector $v\in T_Z(D_U)\simeq Z^\perp \cap U^\perp$ followed by $Z^\perp \cap U$ gives a properly oriented basis of $T_Z(D)\simeq Z^\perp$. Then $\langle Z^\perp \cap U^\perp ,Z^\perp \cap U,Z\rangle$ has the same orientation as $\langle e_1,e_2,e_3,e_4\rangle$, i.e. the determinant of the base change is positive.

For $\beta =\beta ^t\in M_2({\mathbb {Q}})$ a positive definite symmetric matrix, let

$$\begin{aligned} \Omega _\beta =\left\{ ({\textbf{x}}_1,{\textbf{x}}_2)\in V^2({\mathbb {Q}}):(({\textbf{x}}_i,{\textbf{x}}_i))=\beta \right\} . \end{aligned}$$

Consider the subspace $U({\textbf{x}}_1,{\textbf{x}}_2):=\langle {\textbf{x}}_1,{\textbf{x}}_2\rangle \subset V$. For a fixed cusp $\kappa _i$ corresponding to the isotropic line $l_{\kappa _i}$, we write

$$\begin{aligned} \Omega _{\beta ,\kappa _i}=\{({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _\beta :U({\textbf{x}}_1,{\textbf{x}}_2)\perp l_{\kappa _i}\}. \end{aligned}$$

From now on, fix a $\beta$ such that ${{\,\textrm{disc}\,}}(U)\in -d({\mathbb {Q}}^\times )^2$, i.e. $\det \beta \in -d({\mathbb {Q}}^\times )^2$. Let $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _\beta$ and $U=U({\textbf{x}}_1,{\textbf{x}}_2)=\langle {\textbf{x}}_1,{\textbf{x}}_2\rangle$. Given a vector ${\textbf{x}}\in U$, by Lemma 3.3, it is orthogonal to two isotropic lines $l_{\kappa _1}$ and $l_{\kappa _2}$ generated by $u_{\kappa _1}$ and $u_{\kappa _2}$ respectively associated to two cusps $\kappa _1$ and $\kappa _2$. Again, if these two cusps are not equivalent with respect to $\Gamma$, we can give a positive orientation to U to distinguish the cusps in the sense that the new base $\langle u_{\kappa _1},{\textbf{x}}_1,{\textbf{x}}_2,u_{\kappa _2}\rangle$ preserves the orientation of $\langle e_1,e_2,e_3,e_4\rangle$. For a fixed cusp $\kappa _i$ corresponding to the isotropic line $l_{\kappa _i}$, we write

$$\begin{aligned} \Omega _{\beta ,\kappa _i,+}\!=\!\{({\textbf{x}}_1,\!{\textbf{x}}_2)\!\in \!\Omega _{\beta ,\kappa _i}\!:\!\langle \! u_{\kappa _i},\!u_{\kappa _j}\rangle \!\perp \! U({\textbf{x}}_1,\!{\textbf{x}}_2),\!\langle u_{\kappa _i},\!{\textbf{x}}_1,\!{\textbf{x}}_2,\!u_{\kappa _j}\rangle \!\text { has a positive orientation}\}. \end{aligned}$$

It should be mentioned here that $\langle u_{\kappa _i},{\textbf{x}}_1,{\textbf{x}}_2,u_{\kappa _j}\rangle$ and $\langle u_{\kappa _i},-{\textbf{x}}_1,-{\textbf{x}}_2,u_{\kappa _j}\rangle$ have the same orientation which means that we need to count $({\textbf{x}}_1,{\textbf{x}}_2)$ and $(-{\textbf{x}}_1,-{\textbf{x}}_2)$ simultaneously in $\Omega _{\beta ,\kappa _i,+}$. Alternatively, the following Lemma 3.7 describes the orientations associated to two pairs $({\textbf{x}}_1,{\textbf{x}}_2)$ and $(-{\textbf{x}}_1,-{\textbf{x}}_2)$ in $\Omega _{\beta ,\infty ,+}$ in a different way. Note that the stabilizer $\Gamma _{\kappa _i}\subset \Gamma _0({\mathfrak {n}})$ of the cusp $\kappa _i$ acts on $\Omega _{\beta ,\kappa _i,+}$ as ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ preserves bilinear forms and the orientation.

Proposition 3.4

For $\det \beta \in -d({\mathbb {Q}}^\times )^2$, we have

$$\begin{aligned} \Gamma \backslash \Omega _{\beta }=\sum _{\kappa _i\in \Gamma \backslash {\mathbb {P}}^1(F)}\Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}. \end{aligned}$$

Proof

Given a representative $[({\textbf{x}}_1,{\textbf{x}}_2)]$ in $\Gamma \backslash \Omega _\beta$ such that $U({\textbf{x}}_1,{\textbf{x}}_2)\perp u_{\kappa _i}$, we consider its $\Gamma$-orbit $\Gamma \cdot ({\textbf{x}}_1,{\textbf{x}}_2)$. The corresponding $D_U$ for $U=\langle {\Gamma \cdot ({\textbf{x}}_1,{\textbf{x}}_2)}\rangle$ has the image $C_{({\textbf{x}}_1,{\textbf{x}}_2)}$ in $\Gamma \backslash {\mathbb {H}}_3$ under the natural projection ${\mathbb {H}}_3\rightarrow \Gamma \backslash {\mathbb {H}}_3$. For a $\gamma \in \Gamma$ we have $U_{\Gamma \cdot ({\textbf{x}}_1,{\textbf{x}}_2)}\perp \gamma \cdot u_{\kappa _i}$. By Proposition 3.1, we know that $\gamma \cdot u_{\kappa _i}=u_{\gamma \cdot \kappa _i}$. It follows that $\gamma \cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ lies in $\Omega _{\beta ,\gamma \cdot \kappa _i,+}$. Thus, modulo the $\Gamma$-action, we have a well-defined map:

$$\begin{aligned} \iota :\Gamma \backslash \Omega _{\beta }\longrightarrow \coprod _{\kappa _i\in \Gamma \backslash {\mathbb {P}}^1(F)}\Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}. \end{aligned}$$

If two pairs $({\textbf{x}}_1,{\textbf{x}}_2)$ and $({\textbf{y}}_1,{\textbf{y}}_2)$ are not $\Gamma$-equivalent then they are not $\Gamma _{\kappa _i}$-equivalent since $\Gamma _{\kappa _i}\subset \Gamma$. Hence this map is injective.

We will show that the inverse map $\iota ^{-1}$ is injective in the following. For ${\textbf{x}}=\left( {\begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}}\right)$, we calculate its orthogonal complement in ${\mathbb {H}}_3$, due to the isomorphism (4),

$$\begin{aligned} {\textbf{x}}^\perp \cap {\mathbb {H}}_3=\{z+rj\in {\mathbb {H}}_3:d(|z|^2+r^2)-b{\bar{z}}-{\bar{b}}z+a=0\}=\psi ({\textbf{x}}) \end{aligned}$$

where $\psi$ is defined as in Definition 2.3. Observe that ${\textbf{x}}_1^\perp \cap {\textbf{x}}_2^\perp \cap {\mathbb {H}}_3=\psi ({\textbf{x}}_1)\cap \psi ({\textbf{x}}_2)$ of which one boundary point on the complex plane is $\kappa _i$. Suppose that two pairs $({\textbf{x}}_1,{\textbf{x}}_2)$ and $({\textbf{y}}_1,{\textbf{y}}_2)$ are not $\Gamma _{\kappa _i}$-equivalent in $\Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}$. Note that $\psi ({\textbf{x}}_1)\cap \psi ({\textbf{x}}_2)$ and $\psi ({\textbf{y}}_1)\cap \psi ({\textbf{y}}_2)$ have a boundary point in common, the cusp $\kappa _i$. Assume that there exists an element $\gamma \in \Gamma$ such that $\gamma \cdot ({\textbf{x}}_1,{\textbf{x}}_2)=({\textbf{y}}_1,{\textbf{y}}_2)$. Then, by Proposition 2.5, we have $\gamma \cdot \psi ({\textbf{x}}_1)=\psi ({\textbf{y}}_1)$ and $\gamma \cdot \psi ({\textbf{x}}_2)=\psi ({\textbf{y}}_2)$. It is easy to observe that

$$\begin{aligned} \gamma \cdot (\psi ({\textbf{x}}_1)\cap \psi ({\textbf{x}}_2))=\gamma \cdot \psi ({\textbf{x}}_1)\cap \gamma \cdot \psi ({\textbf{x}}_2)=\psi ({\textbf{y}}_1)\cap \psi ({\textbf{y}}_2). \end{aligned}$$

It follows that $\gamma$ must be in $\Gamma _{\kappa _i}$, which is a contradiction to that $({\textbf{x}}_1,{\textbf{x}}_2)$ and $({\textbf{y}}_1,{\textbf{y}}_2)$ are not $\Gamma _{\kappa _i}$-equivalent. So such a $\gamma$ does not exist. We have proven the injectivity of $\iota ^{-1}$. $\square$

Set $\det (\beta )\in -d({\mathbb {Q}}^\times )^2$. It is easy to observe that, for the cusp $\infty$,

$$\begin{aligned} \Omega _{\beta ,\infty }=\left\{ \left( \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ,\left( \begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right) \in \Omega _\beta :\det (\beta )\in -d({\mathbb {Q}}^\times )^2,a_1,a_2\in {\mathbb {Q}},b_1,b_2\in F\right\} . \end{aligned}$$

Setting $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\infty }=\left( \left( {\begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}}\right) ,\left( {\begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}}\right) \right)$, the associated Gram matrix is of form

$$\begin{aligned} \beta =\left( \begin{matrix}({\textbf{x}}_1,{\textbf{x}}_1)&{}({\textbf{x}}_1,{\textbf{x}}_2)\\ ({\textbf{x}}_1,{\textbf{x}}_2)&{}({\textbf{x}}_2,{\textbf{x}}_2)\end{matrix}\right) =\left( \begin{matrix}b_1{\bar{b}}_1&{}\frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)\\ \frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)&{}b_2{\bar{b}}_2\end{matrix}\right) . \end{aligned}$$

of which the determinant is

$$\begin{aligned} \det (\beta )={{\,\textrm{disc}\,}}(U({\textbf{x}}_1,{\textbf{x}}_2))=-\frac{1}{4}(b_1{\bar{b}}_2-{\bar{b}}_1b_2)^2. \end{aligned}$$

We are not interested in the case when $b_1{\bar{b}}_2\in {\mathbb {Q}}$ since then $\det (\beta )=0$.

Let $U=U({\textbf{x}}_1,{\textbf{x}}_2)$ for $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\infty }$. We will calculate its corresponding special cycle $D_U$ in the following. Given a point $z+rj\in {\mathbb {H}}_3$ identified with $\frac{1}{r}\left( \begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}\right)$, we compute

$$\begin{aligned} \frac{1}{r}\left( \begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}\right) \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ^*=\frac{1}{r}\left( \begin{matrix}|z|^2+r^2&{}z\\ {\bar{z}}&{}1\end{matrix}\right) \left( \begin{matrix}0&{}-b_1\\ -{\bar{b}}_1&{}a_1\end{matrix}\right) =\frac{1}{r}\left( \begin{matrix}-{\bar{b}}_1z&{}*\\ *&{}-b_1{\bar{z}}+a_1\end{matrix}\right) . \end{aligned}$$

Thus we have

$$\begin{aligned} {\textbf{x}}_1^\perp =\{z+rj\in {\mathbb {H}}_3:a_1-b_1{\bar{z}}-{\bar{b}}_1z=0\}, \end{aligned}$$

and similarly,

$$\begin{aligned} {\textbf{x}}_2^\perp =\{z+rj\in {\mathbb {H}}_3:a_2-b_2{\bar{z}}-{\bar{b}}_2z=0\}. \end{aligned}$$

Then, solving above equations, we can deduce that the special cycle $D_U$ consists of the infinite geodesic line joining two cusps $\infty$ and

$$\begin{aligned} z_U=\frac{a_2b_1-a_1b_2}{b_1{\bar{b}}_2-{\bar{b}}_1b_2}. \end{aligned}$$

(7)

Lemma 3.5

Suppose that ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ with $\omega$ is either $\sqrt{d}$ or $\frac{1+\sqrt{d}}{2}$ and denote the stabilizer of the cusp $\infty$ by $\Gamma _\infty =\left\{ \left( \begin{matrix}1&{}\alpha \\ 0&{}1\end{matrix}\right) :\alpha \in {\mathcal {O}}\right\}$. Denote

$$\begin{aligned} L_{\infty ,\dag }=\left\{ \left( \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ,\left( \begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right) :a_1,a_2\in {\mathbb {Z}},b_1,b_2\in {\mathcal {O}}, \text { the condition }\dag \text { holds }\right\} \end{aligned}$$

where the condition $\dag$ is given by

$$\begin{aligned} \left( \begin{matrix}b_1\\ b_2\end{matrix}\right) =m\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}1\\ \omega \end{matrix}\right) \end{aligned}$$

(8)

with $\alpha ,\beta ,\gamma ,\delta \in {\mathbb {Z}}$, $\alpha \delta -\beta \gamma =\pm 1$ and $m\in F$. Then the cusp $z_U$ associated to $\Gamma _\infty \backslash L_{\infty ,\dag }$ runs through all the representatives in $({\overline{m}}\sqrt{d_F}|)^{-1}{\mathcal {O}}/{\mathcal {O}}$.

Proof

Write $U=U({\textbf{x}}_1,{\textbf{x}}_2)$. The $\Gamma _\infty$-action on $({\textbf{x}}_1,{\textbf{x}}_2)$ is given explicitly by

$$\begin{aligned} \begin{pmatrix}1&{}\alpha \\ 0&{}1\end{pmatrix}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( \begin{matrix}a_1+\alpha {\bar{b}}_1+{\bar{\alpha }}b_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ,\left( \begin{matrix}a_2+\alpha {\bar{b}}_2+{\bar{\alpha }}b_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right) . \end{aligned}$$

Under the $\Gamma _\infty$-action, the cusp $z_U$ becomes $z_U^{'}$; that is

$$\begin{aligned} z_U^{'}=&\frac{(a_2+\alpha {\bar{b}}_2+{\bar{\alpha }}b_2)b_1-(a_1+\alpha {\bar{b}}_1+{\bar{\alpha }}b_1)b_2}{b_1{\bar{b}}_2-{\bar{b}}_1b_2}\\ =&\frac{a_2b_1+\alpha {\bar{b}}_2b_1-a_1b_2-\alpha {\bar{b}}_1b_2}{b_1{\bar{b}}_2-{\bar{b}}_1b_2}=z_U+\alpha . \end{aligned}$$

By our assumption, the cusp $z_U$ can be rewritten as

$$\begin{aligned} z_U&=\frac{ma_2(\alpha +\beta \omega )-ma_1(\gamma +\delta \omega )}{-m{\overline{m}}(\alpha +\beta \omega )(\gamma +\delta {\bar{\omega }})+m{\overline{m}}(\alpha +\beta {\bar{\omega }})(\gamma +\delta \omega )}\\&=\frac{a_2(\alpha +\beta \omega )-a_1(\gamma +\delta \omega )}{-{\overline{m}}(\alpha \delta -\beta \gamma )({\bar{\omega }}-\omega )}=\frac{a_2(\alpha +\beta \omega )-a_1(\gamma +\delta \omega )}{{\overline{m}}\sqrt{d_F}} \end{aligned}$$

of which the numerator ranges over the whole ${\mathcal {O}}$.

Thus, modulo the $\Gamma _\infty$-action, the corresponding cusp $z_U$ runs through all the representatives in $({\overline{m}}\sqrt{d_F})^{-1}{\mathcal {O}}/{\mathcal {O}}$. $\square$

Remark 3.6

Let m be a square-free product of split or inert primes.

• (1) Let $d\equiv 1{{\,\textrm{mod}\,}}4$ and then $d_F=d$. The above $z_U$ ranges over $(m\sqrt{d})^{-1}{\mathcal {O}}$. Writing ${\mathfrak {f}}=(m\sqrt{d}){\mathcal {O}}$, we have

  $$\begin{aligned} z_U{\mathfrak {f}}=\left( \frac{a_2b_1-a_1b_2}{b_1{\bar{b}}_2-{\bar{b}}_1b_2}\right) {\mathfrak {f}}=\left( \frac{a_2b_1-a_1b_2}{m}\right) {\mathcal {O}}. \end{aligned}$$

• (2) Let $d\equiv 2,3{{\,\textrm{mod}\,}}4$ and then $d_F=4d$. Note that in this case prime 2 is ramified in $F={\mathbb {Q}}(\sqrt{d})$. Rewrite (8) above as

  $$\begin{aligned} \left( \begin{matrix}b_1\\ b_2\end{matrix}\right) =\frac{1}{2}m\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}1\\ \omega \end{matrix}\right) \end{aligned}$$

  with $\alpha ,\beta ,\gamma ,\delta \in {\mathbb {Z}}$, $\alpha \delta -\beta \gamma =\pm 1$. Then the above $z_U$ ranges over $(m\sqrt{d})^{-1}{\mathcal {O}}$. Writing ${\mathfrak {f}}=(m\sqrt{d}){\mathcal {O}}$, we have

  $$\begin{aligned} z_U{\mathfrak {f}}=\left( \frac{a_2b_1-a_1b_2}{b_1{\bar{b}}_2-{\bar{b}}_1b_2}\right) {\mathfrak {f}}=\left( \frac{2(a_2b_1-a_1b_2)}{m}\right) {\mathcal {O}}. \end{aligned}$$

In Sect. 4 we will define the Schwartz function evaluated at $\frac{a_2b_1-a_1b_2}{m}$ or $\frac{2(a_2b_1-a_1b_2)}{m}$ depending on d.

Let $U=\langle {\textbf{x}}_1,{\textbf{x}}_2\rangle =\left\langle \left( {\begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}}\right\rangle ,\left( {\begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}}\right) \right\rangle$ where $a_1,a_2\in {\mathbb {Q}}$ and $b_1,b_2\in F^\times$. We have seen that $D_U$ consists of the infinite geodesic line joining the cusps $\infty$ and $z_U$ as in (7). Choose a point $Z=z_U+rj$ on $D_U$ and then the orientation of $T_Z(D_U)$ depends on the sign of $\text {Im}(b_1{\bar{b}}_2)$ (assuming $\text {Im}(b_1{\bar{b}}_2)\ne 0$) by the following lemma.

Lemma 3.7

Let $U,D_U,Z$ be as above. Then the sign of $\text {Im}(b_1{\bar{b}}_2)$ (assuming $\text {Im}(b_1{\bar{b}}_2)\ne 0$) determines the orientation of $T_Z(D_U)$.

Proof

Here we sketch the proof and for more details see [26, Lemma 4.2.7].

Let $Z=z_U+rj$ be a point on $D_U$which can be identified with $\frac{1}{r}\left( \begin{matrix}z_U{\bar{z}}_U+r^2&{}z_U\\ {\bar{z}}_U&{}1\end{matrix}\right)$. Suppose that $\left( \begin{matrix}\alpha &{}\beta \\ {\bar{\beta }}&{}\delta \end{matrix}\right) \in Z^\perp$ and we compute, recalling $*$ action in (2)

$$\begin{aligned} \left( \begin{matrix}\alpha &{}\beta \\ {\bar{\beta }}&{}\delta \end{matrix}\right) \left( \begin{matrix}z_U{\bar{z}}_U+r^2&{}z_U\\ {\bar{z}}_U&{}1\end{matrix}\right) ^*=\left( \begin{matrix}\alpha -\beta {\bar{z}}_U&{}*\\ *&{}-{\bar{\beta }}z_U+\delta (z_U{\bar{z}}_U+r^2)\end{matrix}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} Z^\perp =\left\{ \left( \begin{matrix}\alpha &{}\beta \\ {\bar{\beta }}&{}\delta \end{matrix}\right) :\alpha -\beta {\bar{z}}_U-{\bar{\beta }}z_U+\delta (z_U{\bar{z}}_U+r^2)=0\right\} . \end{aligned}$$

We describe the subspace $Z^\perp \cap U$ as

$$\begin{aligned} Z^\perp \cap U=\left\{ \left( \begin{matrix}\alpha &{}\beta \\ {\bar{\beta }}&{}0\end{matrix}\right) :\alpha -\beta {\bar{z}}_U-{\bar{\beta }}z_U=0,\alpha \in {\mathbb {R}},\beta \in {\mathbb {C}}\right\} \\ =\left\{ \left( \begin{matrix}\beta {\bar{z}}_U+{\bar{\beta }}z_U&{}\beta \\ {\bar{\beta }}&{}0\end{matrix}\right) \right\} \end{aligned}$$

where $\beta \in {{\,\textrm{Span}\,}}\{b_1,b_2\}$. We see that

$$\begin{aligned} \begin{aligned} Z^\perp \cap U=\left\langle \frac{1}{2}a_1(e_1+e_4)+\frac{1}{2}(b_1+{\bar{b}}_1)e_2-\frac{1}{2}i(b_1-{\bar{b}}_1)e_3,\right. \\ \left. \frac{1}{2}a_2(e_1+e_4)+\frac{1}{2}(b_2+{\bar{b}}_2)e_2-\frac{1}{2}i(b_2-{\bar{b}}_2)e_3\right\rangle \end{aligned} \end{aligned}$$

is identical to $U=\left\langle \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right\rangle ,\left( \begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right\rangle$.

Similarly, we have that

$$\begin{aligned}{} & {} Z^\perp \cap U^\perp \\{} & {} \quad =\left\langle \frac{\varepsilon }{2}(z_U{\bar{z}}_U-r^2-1)e_1+\frac{1}{2}(z_U+{\bar{z}}_U)e_2-\frac{1}{2}i(z_U-{\bar{z}}_U)+e_3\frac{\varepsilon }{2}(z_U{\bar{z}}_U-r^2+1)e_4\right\rangle \end{aligned}$$

where $\varepsilon =\pm 1$ describes the orientation of $T_Z(D_U)\simeq Z^\perp \cap U^\perp$. If the cycle $D_U$ is directed from $z_U$ to $\infty$ then we take $\varepsilon =-1$. For a different direction of $D_U$ we take $\varepsilon =1$.

Then we consider the base change, which describes the orientation related to introducing $\Omega _{\beta ,\kappa _i,+}$, given by

$$\begin{aligned} \left( \begin{matrix}Z^\perp \cap U^\perp \\ Z^\perp \cap U\\ Z\end{matrix}\right) =M\left( \begin{matrix}e_1\\ e_2\\ e_3\\ e_4\end{matrix}\right) \end{aligned}$$

where the determinnant of M can ba calculated as

$$\begin{aligned} \det M=-\varepsilon r\frac{1}{2}i(b_1{\bar{b}}_2-{\bar{b}}_1b_2)=\frac{1}{2}\varepsilon r\text {Im}({b_1{\bar{b}}_2})>0. \end{aligned}$$

It is obvious that the sign of $\text {Im}({b_1{\bar{b}}_2})$ determines the orientation $\varepsilon$ of $T_Z(D_U)$. $\square$

4 Schwartz function

In this paper we need to consider the pair ${{\,\textrm{Sp}\,}}_4\times {{\,\textrm{SO}\,}}(3,1)$ to construct the theta liftings of weight 2 Bianchi modular forms. In the following Subsects. 4.1, 4.2 and 4.3, we define local Schwartz functions at split primes dividing m, inert primes dividing m and ramified primes away from 2 respectively. In Sect. 5 we will construct the theta lifting of a weight 2 Bianchi modular form of $\Gamma _0({\mathfrak {n}})$ with square-free ${\mathfrak {n}}$ coprime to $(m|d_F|)$. To avoid the vanishing of our theta lifting, in Subsect. 4.4 we define the local Schwartz function at each place dividing $N({\mathfrak {n}})$ (norm of ${\mathfrak {n}}$) and ramified prime 2, to be different to the characteristic function of integral lattice. In Subsect. 4.5 we consider all other finite places.

Let $F={\mathbb {Q}}(\sqrt{d})$ be an imaginary quadratic field and denote by ${\mathcal {O}}$ its ring of integers. Choose $m\in {\mathbb {Z}}$ as a square-free product of inert or split primes and put ${\mathfrak {m}}=m{\mathcal {O}}$. Let $\chi _{\mathfrak {m}}$ be a finite order Hecke character of conductor ${\mathfrak {m}}\sqrt{d}$. Denote by ${\widetilde{\chi }}_{\mathfrak {m}}$ the induced idelic one and by ${\widetilde{\chi }}_{{\mathfrak {m}},v}$ its local component. In this section we will define a Schwartz function $\varphi ^{\chi _{\mathfrak {m}}}$ related to this character $\chi _{\mathfrak {m}}$.

We first describe how to localise the quadratic space in the following proposition. In Sect. 3 we have chosen the rational quadratic space of dimension 4 such that $(V({\mathbb {Q}}),q)\simeq ({\mathcal {H}}(F),-\text{ det})$. Thus, to extend it to the 4-dimensional quadratic space over p-adic numbers ${\mathbb {Q}}_p$, we can consider ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p$. Following [20, p.273], there are two four dimensional quadratic spaces over ${\mathbb {Q}}_p$ with discriminant ${\mathfrak {d}}\in {\mathbb {Q}}_p^\times /({\mathbb {Q}}_p^\times )^2$ up to isometry. If ${\mathfrak {d}}=1$, it is isometric to $M_{2\times 2}({\mathbb {Q}}_p)$ equipped with the determinant; if ${\mathfrak {d}}\ne 1$, it is isometric to

$$\begin{aligned} V_1({\mathbb {Q}}_p)=\left\{ \left( \begin{matrix}e&{}f\sqrt{{\mathfrak {d}}}\\ g\sqrt{{\mathfrak {d}}}&{}{\overline{e}}\end{matrix}\right) :f,g\in {\mathbb {Q}}_p,e\in {\mathbb {Q}}_p(\sqrt{{\mathfrak {d}}})\right\} \subset M_2\left( {\mathbb {Q}}_p(\sqrt{{\mathfrak {d}}})\right) \end{aligned}$$

equipped with the determinant.

Proposition 4.1

For a prime p, the four dimensional quadratic space over ${\mathbb {Q}}_p$ is isometric to either $(V_1({\mathbb {Q}}_p),\det )$ when p is inert or ramified in $F/{\mathbb {Q}}$, or $(M_2({\mathbb {Q}}_p),\det )$ when p splits in $F/{\mathbb {Q}}$.

Proof

Given a diagonal quadratic form $Q=\sum _{i=1}^n a_ix_i^2$ with $a_i\in {\mathbb {Q}}^\times _p$, we define the Hasse invariant as $c_p(Q)=c(Q)=\prod _{i<j}(a_i,a_j)_p=\pm 1$ where $(\,\ )$ denotes the Hilbert symbol. The non-degenerate quadratic spaces over ${\mathbb {Q}}_p$ ($p<\infty$) are in 1-1 correspondence with the triples $(n,{\mathfrak {d}},c)$, where n is the dimension, ${\mathfrak {d}}$ is the discriminant, and c is the Hasse invariant [4, Theorem 1.1, Chapter 4].

Let p be inert in $F/{\mathbb {Q}}$ which implies that $\sqrt{d}\notin {\mathbb {Q}}_p$ and that $F\otimes {\mathbb {Q}}_p={\mathbb {Q}}_p(\sqrt{d})$. Then we have

$$\begin{aligned} {\mathcal {H}}(F)\otimes {\mathbb {Q}}_p=\left\{ \left( \begin{matrix}a&{}b\\ {\overline{b}}&{}d\end{matrix}\right) :a,d\in {\mathbb {Q}}_p,b\in {\mathbb {Q}}_p(\sqrt{d})\right\} ={\mathcal {H}}(F\otimes {\mathbb {Q}}_p), \end{aligned}$$

where $^{-}$ denotes the non-trivial action in ${\text {Gal}}({\mathbb {Q}}_p(\sqrt{d})/{\mathbb {Q}}_p)$. Equipping ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p$ with the quadratic form being $-\det$ and choosing an orthogonal basis $e_1=\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$, $e_2=\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right)$, $e_3=\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right)$ and $e_4=\left( {\begin{matrix}0&{}\sqrt{d}\\ -\sqrt{d}&{}0\end{matrix}}\right)$, we have an associated diagonal form ${Q=-x_1^2+x_2^2+x_3^2-dx_4^2}$. It follows that ${\mathfrak {d}}=d$ and $c=(-1,-d)_p=1$ since $p\not \mid d$. Similarly, for $V_1({\mathbb {Q}}_p)$ with the discriminant ${\mathfrak {d}}=d$, choosing an orthogonal basis $e_1^{'}=\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$, ${e_2^{'}=\left( {\begin{matrix}\sqrt{d}&{}0\\ 0&{}-\sqrt{d}\end{matrix}}\right) }$, $e_3^{'}=\left( {\begin{matrix}0&{}\sqrt{d}\\ \sqrt{d}&{}0\end{matrix}}\right)$ and $e_4^{'}=\left( {\begin{matrix}0&{}\sqrt{d}\\ -\sqrt{d}&{}0\end{matrix}}\right)$ in $V_1({\mathbb {Q}}_p)$ above, we have a diagonal form $Q^{'}=x_1^2-dx_2^2-dx_3^2+dx_4^2$. Then ${\mathfrak {d}}^{'}=d^3$ and $c^{'}=(1,-d)_p^2(1,d)_p(-d,-d)_p(-d,d)_p^2=1$ since $p\not \mid d$. Thus, we can deduce that $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_p,-\det )\simeq (V_1({\mathbb {Q}}_p),\det )$ if p is inert in $F/{\mathbb {Q}}$.

Let p split in $F/{\mathbb {Q}}$ such that $(p)={\mathfrak {p}}\overline{{\mathfrak {p}}}$. Then d has a square root $\alpha$ in the ring ${\mathbb {Z}}_p$ of p-adic integers by Hensel’s lemma. It is known that $F\otimes {\mathbb {Q}}_p=F_{\mathfrak {p}}\times F_{\overline{{\mathfrak {p}}}}$ where $F_{{\mathfrak {p}}},F_{\overline{{\mathfrak {p}}}}$ are both isomorphic to ${\mathbb {Q}}_p$. Consider the map ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p\rightarrow M_2({\mathbb {Q}}_p)$ via $\left( {\begin{matrix}a&{}b\\ {\overline{b}}&{}d\end{matrix}}\right) \otimes x\mapsto \left( {\begin{matrix}a_px&{}b_{{\mathfrak {p}}}x\\ b_{\overline{{\mathfrak {p}}}}x&{}d_px\end{matrix}}\right)$ where the subscripts $p, {\mathfrak {p}}, \bar{{\mathfrak {p}}}$ denote images under ${\mathbb {Q}}\hookrightarrow {\mathbb {Q}}_p$, $F\hookrightarrow F_{{\mathfrak {p}}}$ and $F\hookrightarrow F_{\overline{{\mathfrak {p}}}}$ respectively. Note that ${\overline{b}}_{{\mathfrak {p}}}, b_{\overline{{\mathfrak {p}}}}$ have the same image in ${\mathbb {Q}}_p$. It is not hard to see the map ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p\rightarrow M_2({\mathbb {Q}}_p)$ is surjective: for any element $\left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right) \cdot \lambda$ with $\lambda \in {\mathbb {Q}}_p$ we can find its preimage $\left( {\begin{matrix}0&{}\sqrt{d}\\ -\sqrt{d}&{}0\end{matrix}}\right) \otimes \alpha ^{-1}\lambda$ in ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p$; for $\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right) \cdot \lambda$, $\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right) \cdot \lambda$ and $\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right) \cdot \lambda$, we can find their preimages $\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right) \otimes \lambda$, $\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right) \otimes \lambda$ and $\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right) \otimes \lambda$ respectively in ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p$. Then ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p\simeq M_2({\mathbb {Q}}_p)$ as they are both 4-dimensional over ${\mathbb {Q}}_p$. In fact $M_2({\mathbb {Q}}_p)$ equipped with the determinant is one isometric class of four dimensional quadratic spaces of discriminant 1 [20, p.273]. Again, equipping ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p$ with minus determinant and choosing an orthogonal basis $e_1=\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$, $e_2=\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right)$, $e_3=\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right)$ and $e_4=\left( {\begin{matrix}0&{}\sqrt{d}\\ -\sqrt{d}&{}0\end{matrix}}\right)$, we have an associated diagonal form $Q=-x_1^2+x_2^2+x_3^2-dx_4^2$. It follows that ${\mathfrak {d}}=d$ (square in ${\mathbb {Z}}_p$) and $c=(-1,-d)_p=1$. Choosing an orthogonal basis $e_1^{'}=\left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$, $e_2^{'}=\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right)$, $e_3^{'}=\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right)$ and $e_4^{'}=\left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right)$ in $M_2({\mathbb {Q}}_p)$ we have ${\mathfrak {d}}^{'}=1$ and $c^{'}=1$. Thus we can deduce that $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_p,-\det )\simeq (M_2({\mathbb {Q}}_p),\det )$ if p splits in $F/{\mathbb {Q}}$.

Suppose that p is ramified in $F/{\mathbb {Q}}_p$ and then we have $F\otimes {\mathbb {Q}}_p={\mathbb {Q}}_p(\sqrt{d})$. As in the inert case, ${\mathcal {H}}(F)\otimes {\mathbb {Q}}_p={\mathcal {H}}(F\otimes {\mathbb {Q}}_p)$. Corresponding to $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_p,-\det )$ the Hasse invariant $c=(-1,-d)_p=(-1)^{\frac{p-1}{2}}$. For $(V_1({\mathbb {Q}}_p),\det )$ we calculate

$$\begin{aligned} c^{'}=(1,d)_p(-d,-d)_p=(1,-d)_p(1,-1)_p(1,-d)_p(-d,-d)_p=(-d,-d)_p=(-1)^{\frac{p-1}{2}} \end{aligned}$$

where the last equality holds as d is square-free and divisible by q. Thus, if p is ramified we have $({\mathcal {H}}(F\otimes {\mathbb {Q}}_p),-\det )\simeq (V_1({\mathbb {Q}}_p),\det )$. $\square$

Fix the additive character $\psi =\prod _q\psi _q$ of ${\mathbb {A}}_{\textbf{Q}}$ given by $\psi _\infty (x)=\exp (2\pi ix)$ and, for every rational prime q, $\psi _q(x)=\exp (2\pi i\text {Fr}_q(q^{-1}x))$ for $x\in {\mathbb {Q}}_q$, where $\text {Fr}_q(x)$ denotes the fractional part of x. In this case we have $\ker (\psi _q)=q{\mathbb {Z}}_q$. Note that computing the congruence subgroup of ${{\,\textrm{Sp}\,}}_4$ under which $\varphi _q$ is invariant is related to $\ker (\psi _q)$.

Let $\chi _V(x)=(x,\det V):{\mathbb {Q}}^\times \backslash {\mathbb {A}}^\times \mapsto \{\pm 1\}$ be the quadratic Hecke character associated to V. For ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)\in V({\mathbb {Q}})^2$, represent the associated Gram matrix by

$$\begin{aligned} ({\textbf{X}},{\textbf{X}})=(({\textbf{x}}_i,{\textbf{x}}_j))\in \text {Sym}_2({\mathbb {Q}}), \end{aligned}$$

where $(\,\ )$ is the symmetric bilinear form on $V({\mathbb {Q}})^2$. Let $\omega =\omega _{\psi _q}$ denote the Weil representation on the Schwartz space $S(V^2)$ characterized (for every q) by the following actions of ${{\,\textrm{Sp}\,}}_4\times {{\,\textrm{SO}\,}}(3,1)$ locally on $\varphi _q\in S((V\otimes {\mathbb {Q}}_q)^2)$:

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{X}})&=\varphi _q(h^{-1}{\textbf{X}}), \end{aligned}$$

(9)

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}b\\ 0&{}1\end{matrix}\right) ,1\right) \varphi _q({\textbf{X}})&=\psi _q\left( \frac{1}{2}{{\,\textrm{tr}\,}}(b({\textbf{X}},{\textbf{X}}))\right) \varphi _q({\textbf{X}}), \end{aligned}$$

(10)

$$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) ,1\right) \varphi _q({\textbf{X}})&=\chi _{V,q}(\det ( a))|\det (a)|_q^{2}\varphi _q({\textbf{X}}a), \end{aligned}$$

(11)

$$\begin{aligned} \omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) ,1\right) \varphi _q({\textbf{X}})&=\gamma {\hat{\varphi }}_q({\textbf{X}}). \end{aligned}$$

(12)

Here the Fourier transform is defined by

$$\begin{aligned} {\hat{\varphi }}_q({\textbf{X}})=\int _{(V\otimes {\mathbb {Q}}_q)^2}\varphi _q({\textbf{Y}})\psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))d{\textbf{Y}} \end{aligned}$$

and $\gamma$ is a certain complex number of absolute value 1.

4.1 At split prime dividing m

Let q|m be a split prime such that $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$. According to Proposition 4.1, there is an isomorphism $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_q,-\det )\simeq (V_1({\mathbb {Q}}_q),\det )$ for q split in $F/{\mathbb {Q}}$ given by

$$\begin{aligned} \left( \begin{matrix}a&{}b\\ {\bar{b}}&{}d\end{matrix}\right) \longmapsto \left( \begin{matrix}a&{}b\\ c&{}d\end{matrix}\right) \quad \end{aligned}$$

where c is the image of ${\bar{b}}\in F$ under $F\hookrightarrow F_{{\mathfrak {q}}}\simeq {\mathbb {Q}}_q$.

Definition 4.2

• (1) Suppose that $d\equiv 1{{\,\textrm{mod}\,}}4$. The local Schwartz function $\varphi _q^{\chi _{\mathfrak {m}}}$ at q is vanishing unless

  $$\begin{aligned} a_i\in {\mathbb {Z}}_q,b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}},c_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}},d_i\in {\mathbb {Z}}_q,a_2d_1-a_1d_2\in q{\mathbb {Z}}_q, \end{aligned}$$

  in which case

  $$\begin{aligned}&\varphi _{q}^{\chi _{\mathfrak {m}}}\left( \left( \begin{matrix}a_1&{}b_1\\ c_1&{}d_1\end{matrix}\right) ,\left( \begin{matrix}a_2&{}b_2\\ c_2&{}d_2\end{matrix}\right) \right) \\ =&\left\{ \begin{array}{cl}({\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}{\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}})\left( \frac{a_2b_1-a_1b_2}{m}+\frac{c_2d_1-c_1d_2}{m}\right) ,&{}\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{c_2d_1-c_1d_2}{m}\in {\mathcal {O}}_{\mathfrak {q}}^\times \times {\mathcal {O}}_{\bar{{\mathfrak {q}}}}^\times ,\\ 0,&{}\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{c_2d_1-c_1d_2}{m}\in {\mathfrak {q}} {\mathcal {O}}_{\mathfrak {q}} \text{ or } \bar{{\mathfrak {q}} }{\mathcal {O}}_{\bar{{\mathfrak {q}}}}.\end{array}\right. \end{aligned}$$

  where $({\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}{\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}})(-)={\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}(-){\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}}(-)$. Note that $b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}},c_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ is equivalent to $b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}\times \bar{{\mathfrak {q}}}{\mathcal {O}}_{\bar{{\mathfrak {q}}}}$.

• (2) Suppose that $d\equiv 2,3{{\,\textrm{mod}\,}}4$. Replace above $\frac{a_2b_1-a_1b_2}{m}$ by $\frac{2(a_2b_1-a_1b_2)}{m}$ and $\frac{c_2d_1-c_1d_2}{m}$ by $\frac{2(c_2d_1-c_1d_2)}{m}$ as discussed in Remark 3.6

In the following we will check the invariance properties of this local Schwartz function under some congruence subgroups of ${{\,\textrm{Sp}\,}}_4$ and ${{\,\textrm{SO}\,}}(3,1)$ in details in case of $d\equiv 1{{\,\textrm{mod}\,}}4$ and the other case can be treated similarly. We need to calculate the transformation properties (9), (10), (11) and (12). For simplicity we write $\varphi _q=\varphi ^{\chi _{\mathfrak {m}}}_q$ and $\chi =\chi _{\mathfrak {m}}$ in the following computation.

Set ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}a_1&{}b_1\\ c_1&{}d_1\end{matrix}}\right) ,\left( {\begin{matrix}a_2&{}b_2\\ c_2&{}d_2\end{matrix}}\right) \right)$. For $a_i\in {\mathbb {Z}}_q$, $b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, $c_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, $d_i\in {\mathbb {Z}}_q$, it is not difficult to observe that

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) \varphi _q({\textbf{X}})=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text { for }u\in M_2(q{\mathbb {Z}}_q) \end{aligned}$$

(13)

as $\psi _q\left( \frac{1}{2}{{\,\textrm{tr}\,}}(u({\textbf{X}},{\textbf{X}}))\right)$ is trivial for such $({\textbf{x}}_1,{\textbf{x}}_2)$ and u.

Set ${\textbf{Y}}=({\textbf{y}}_1,{\textbf{y}}_2)=\left( \left( {\begin{matrix}\alpha _1&{}\beta _1\\ \gamma _1&{}\delta _1\end{matrix}}\right) ,\left( {\begin{matrix}\alpha _2&{}\beta _2\\ \gamma _2&{}\delta _2\end{matrix}}\right) \right)$. Inspired by Prasanna’s computations in the proof of [19, Proposition 3.4], we will calculate the Fourier transform

$$\begin{aligned} {\hat{\varphi }}_q({\textbf{Y}})=\int \limits _{\begin{array}{c} a_i\in {\mathbb {Z}}_q,b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}\\ c_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}},d_i\in {\mathbb {Z}}_q \end{array}}\psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))\varphi _q({\textbf{X}})d{\textbf{X}}\quad \text { for }i=1,2, \end{aligned}$$

where

$$\begin{aligned} {{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}})=-\frac{1}{2}(a_1\delta _1-b_1\gamma _1-c_1\beta _1+d_1\alpha _1+a_2\delta _2-b_2\gamma _2-c_2\beta _2+d_2\alpha _2). \end{aligned}$$

Denote by $\lambda _{\sqrt{d}}$ the image of $\sqrt{d}$ in ${\mathcal {O}}_{\mathfrak {q}}$. By the above definition, $\varphi _q$ is invariant under the transformations $a_i\mapsto a_i+q$, $b_i\mapsto b_i+q^2$, $b_i\mapsto b_i+q^2\lambda _{\sqrt{d}}$ (or $b_i\mapsto b_i+q^2\frac{1+\lambda _{\sqrt{d}}}{2}$) and $d_i\mapsto d_i+q$. Sending $a_i\mapsto a_i+q$, we will have $\psi _q(-q\delta _i)$, factored out of the above integral, which has to be trivial for the non-vanishing of ${\hat{\varphi }}_q$. So for ${\hat{\varphi }}_q({\textbf{Y}})$ non-vanishing we need $\delta _i\in {\mathbb {Z}}_q$. Sending $b_1\mapsto b_1+q^2$, we get $\psi _q(q^2(\gamma _1+\beta _1))$. For $b_1\mapsto b_1+q^2\lambda _{\sqrt{d}}$ and $b_1\mapsto b_1+q^2\frac{1+\lambda _{\sqrt{d}}}{2}$, we get $\psi _q(q^2(\gamma _1-\beta _1)\lambda _{\sqrt{d}})$ and $\psi _q(\frac{1}{2}q^2(\gamma _1+\beta _1)+\frac{1}{2}q^2(\gamma _1-\beta _1)\lambda _{\sqrt{d}})$ respectively. For ${\hat{\varphi }}_q({\textbf{Y}})$ non-vanishing we need $\gamma _1+\beta _1,\gamma _1-\beta _1\in q^{-1}{\mathbb {Z}}_q$ which implies $\beta _1,\gamma _1\in q^{-1}{\mathbb {Z}}_q$. Repeating the same argument we can deduce that for the non-vanishing of ${\hat{\varphi }}_q$ the following conditions must be satisfied, for $i=1,2$,

$$\begin{aligned} \alpha _i\in {\mathbb {Z}}_q, \beta _i\in q^{-1} {\mathbb {Z}}_q, \gamma _i\in q^{-1}{\mathbb {Z}}_q,\delta _i\in {\mathbb {Z}}_q. \end{aligned}$$

It follows that $\omega \left( \left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right) \right) \varphi _q({\textbf{Y}})$ is vanishing unless $\alpha _i\in {\mathbb {Z}}_q$, $\beta _i\in q^{-1}{\mathbb {Z}}_q$, $\gamma _i\in q^{-1}{\mathbb {Z}}_q$ and $\delta _i\in {\mathbb {Z}}_q$.

Recall from (10) that

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) {\hat{\varphi }}_q({\textbf{Y}})=\psi _q\left( \frac{1}{2}{{\,\textrm{tr}\,}}(u({\textbf{Y}},{\textbf{Y}}))\right) {\hat{\varphi }}_q({\textbf{Y}}). \end{aligned}$$

Set $\alpha _i\in {\mathbb {Z}}_q, \beta _i\in q^{-1} {\mathbb {Z}}_q, \gamma _i\in q^{-1}{\mathbb {Z}}_q,\delta _i\in {\mathbb {Z}}_q$. Then when $u\in M_2(q^3{\mathbb {Z}}_q)$ we can see that $\psi _q\left( \frac{1}{2}{{\,\textrm{tr}\,}}(u({\textbf{Y}},{\textbf{Y}}))\right)$ is trivial. Thus we can deduce that

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{Y}})=\omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{Y}})\quad \text { for }u\in M_2(q^3{\mathbb {Z}}_q),\ u=u^t \end{aligned}$$

which implies

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}0\\ u&{}1\end{matrix}\right) \right) \varphi _q({\textbf{Y}})=\varphi _q({\textbf{Y}})\quad \text { for }u\in M_2(q^3{\mathbb {Z}}_q),\ u=u^t. \end{aligned}$$

(14)

For $a=\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \in {{\,\textrm{GL}\,}}_2({\mathbb {Z}}_q)$, we compute

$$\begin{aligned} {\textbf{X}}a&=({\textbf{x}}_1,{\textbf{x}}_2)\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \\&=\left( \left( \begin{matrix}\alpha a_1+\gamma a_2&{}\alpha b_1+\gamma b_2\\ \alpha c_1+\gamma c_2&{}\alpha d_1+\gamma d_2\end{matrix}\right) ,\left( \begin{matrix}\beta a_1+\delta a_2&{}\beta b_1+\delta b_2\\ \beta c_1+\delta c_2&{}\beta d_1+\delta d_2\end{matrix}\right) \right) =:\left( \left( \begin{matrix}a_1^{'}&{}b^{'}_1\\ c^{'}_1&{}d^{'}_1\end{matrix}\right) ,\left( \begin{matrix}a_1^{'}&{}b^{'}_1\\ c^{'}_1&{}d^{'}_1\end{matrix}\right) \right) \end{aligned}$$

due to which we obtain that

$$\begin{aligned} \frac{a_2^{'}b_1^{'}-a_1^{'}b_2^{'}}{m}&=\frac{(\beta a_1+\delta a_2)(\alpha b_1+\gamma b_2)-(\alpha a_1+\gamma a_2)(\beta b_1+\delta b_2)}{m}\\ {}&=\frac{(\alpha \delta -\beta \gamma )a_2b_1-(\alpha \delta -\beta \gamma )a_1b_2}{m}=\det (a)\cdot \frac{a_2b_1-a_1b_2}{m} \end{aligned}$$

and similarly that $\frac{c_2^{'}d_1^{'}-c_1^{'}d_2^{'}}{m}=\det (a)\cdot \frac{c_2d_1-c_1d_2}{m}$. Then from (11) we see that if $\det (a)\in {\mathbb {Z}}_q^\times$ then

$$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) \right) \varphi _q({\textbf{X}})=\chi _{V,q}(\det (a))|\det (a)|^2_q({\widetilde{\chi }}_{\mathfrak {q}}{\widetilde{\chi }}_{\bar{{\mathfrak {q}}}})(\det (a))\varphi _q({\textbf{X}}). \end{aligned}$$

(15)

Combining (13), (14) and (15), we have proved the following lemma:

Lemma 4.3

We have

$$\begin{aligned} \omega (k_1)\varphi _q=\chi _{V,q}(\det (A))|\det (A)|^2_q({\widetilde{\chi }}_{\mathfrak {q}}{\widetilde{\chi }}_{\bar{{\mathfrak {q}}}})(\det (A))\varphi _q \end{aligned}$$

for

$$\begin{aligned} k_1=\left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in \left\{ \left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q):B\in M_2(q{\mathbb {Z}}_q),C\in M_2(q^3{\mathbb {Z}}_q)\right\} . \end{aligned}$$

Proof

The assertion follows from the Iwahori decomposition of ${{\,\textrm{Sp}\,}}_4$. $\square$

We next discuss the action of ${{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q)$ on $\varphi _q$ characterised by

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)\quad \text { for }h\in {{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q). \end{aligned}$$

Recall from [26, Sect. 1.2], for odd split prime q we have the exact sequence

$$\begin{aligned} 1\rightarrow \{\pm 1\}\rightarrow {{\,\textrm{SL}\,}}_2({\mathbb {Q}}_q\times {\mathbb {Q}}_q)\overset{\Lambda }{\longrightarrow }{{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q)\rightarrow {\mathbb {Q}}_q^\times /{\mathbb {Q}}_q^{\times 2}\rightarrow 1 \end{aligned}$$

from which we can deduce the isomorphism ${{\,\textrm{PSL}\,}}_2({\mathbb {Q}}_q\times {\mathbb {Q}}_q)\simeq {{\,\textrm{SO}\,}}^+(3,1)({\mathbb {Q}}_q)$ where ${{\,\textrm{SO}\,}}^+:=\text {Im}(\Lambda )$. In this case of split q, we want to check congruence subgroups of ${{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q\times {\mathbb {Z}}_q)$ under which the Schwartz function $\varphi _q$ is invariant. Recall from [20, Sect. 2] that $h^{-1}{\textbf{x}}_i:=h_1^{-1}{\textbf{x}}_i{}^t(h_2^{-1})^*$ for $h=(h_1,h_2)\in {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)\times {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)={{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q\times {\mathbb {Z}}_q)$.

Lemma 4.4

For $h=(h_1,h_2)\in {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)\times {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)$ satisfying

$$\begin{aligned} h_i\in {\overline{\Gamma }}(q)=\left\{ \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \in {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q):\pm \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \equiv \pm \left( \begin{matrix}1&{}0\\ 0&{}1\end{matrix}\right) {\text {mod}} q \right\} , \end{aligned}$$

we have that

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

(16)

Proof

Set

$$\begin{aligned} h_j^{-1}=\left( \begin{matrix}\alpha _j&{}\beta _j\\ \gamma _j&{}\delta _j\end{matrix}\right) \text { and }{\textbf{x}}_i=\left( \begin{matrix}a_i&{}b_i\\ c_i&{}d_i\end{matrix}\right) \quad \text { for }i,j\in \{1,2\} \end{aligned}$$

with $\alpha _j,\delta _j\equiv 1{{\,\textrm{mod}\,}}q$ and $\beta _j,\gamma _j\equiv 0{{\,\textrm{mod}\,}}q$.

First we assume that $a_i,d_i\in {\mathbb {Z}}_q$ and $b_i,c_i\in q{\mathbb {Z}}_q$ so that $\varphi _q$ is non-vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$.

We compute

$$\begin{aligned}&\left( \begin{matrix}a^{'}_i&{}b^{'}_i\\ c^{'}_i&{}d^{'}_i\end{matrix}\right) := h_1^{-1}{\textbf{x}}_i{}^t(h_2^{-1})^*\\ \\&\quad =\left( \begin{matrix}\delta _2(\alpha _1 a_i+\beta _1 c_i)-\beta _2(\alpha _1 b_i+\beta _1 d_i)&{}-\gamma _2 (\alpha _1 a_i+\beta _1 c_i)+\alpha _2(\alpha _1 b_i+\beta _1 d_i)\\ \delta _2 (\gamma _1 a_i+\delta _1 c_i)-\beta _2 (\gamma _1 b_i+\delta _1 d_i)&{}-\gamma _2 (\gamma _1 a_i+\delta _1 c_i)+\alpha _2 (\gamma _1 b_i+\delta _1 d_i)\end{matrix}\right) . \end{aligned}$$

It is not hard to observe that$b^{'}_i,c^{'}_i\in q{\mathbb {Z}}_q$ as $b_i,c_i,\beta _j,\gamma _j\in q{\mathbb {Z}}_q$, and

$$\begin{aligned} a^{'}_2d^{'}_1-a^{'}_1d^{'}_2\equiv \alpha _1\alpha _2\delta _1\delta _2(a_2d_1-a_1d_2)\equiv a_2d_1-a_1d_2\mod q. \end{aligned}$$

Modulo $q^2$, we have

$$\begin{aligned}&\!\!\!\!\!\!\! a^{'}_2b^{'}_1-a^{'}_1b^{'}_2 \\&\quad \equiv \delta _2 \alpha _1 a_2 (-\gamma _2\alpha _1 a_1+\alpha _2\alpha _1 b_1+\alpha _2\beta _1 d_1)-\delta _2 \alpha _1 a_1 (-\gamma _2\alpha _1 a_2+\alpha _2\alpha _1 b_2+\alpha _2\beta _1 d_2)\\&\quad \equiv \alpha _1^2\alpha _2\delta _2(a_2b_1-a_1b_2)+\alpha _1\alpha _2\beta _1\delta _2(a_2d_1-a_1d_2). \end{aligned}$$

It follows that

$$\begin{aligned} \frac{a^{'}_2b^{'}_1-a^{'}_1b^{'}_2}{m}\equiv \frac{a_2b_1-a_1b_2}{m}\mod q. \end{aligned}$$

Similarly, we obtain that modulo $q^2$

$$\begin{aligned} c^{'}_2d^{'}_1-c^{'}_1d^{'}_2\equiv&\alpha _2\delta _1 d_1 (\delta _2\gamma _1 a_2+\delta _2\delta _1 c_2-\beta _2\delta _1 d_2)-\alpha _2\delta _1 d_2 (\delta _2\gamma _1 a_1+\delta _2\delta _1 c_1-\beta _2\delta _1 d_1)\\ \equiv&\alpha _2\delta _1^2\delta _2(c_2d_1-c_1d_2)+\alpha _2\gamma _1\delta _1\delta _2(a_2d_1-a_1d_2). \end{aligned}$$

and

$$\begin{aligned} \frac{c^{'}_2d^{'}_1-c^{'}_1d^{'}_2}{m}\equiv \frac{c_2d_1-c_1d_2}{m} \mod q. \end{aligned}$$

Therefore, when $\varphi _q$ is non-vanishing, we can deduce that

$$\begin{aligned} \omega (1,h)\varphi _q(({\textbf{x}}_1,{\textbf{x}}_2))=\varphi _q(({\textbf{x}}_1,{\textbf{x}}_2))\quad \text { for }h=(h_1,h_2)\in {\overline{\Gamma }}(q)\times {\overline{\Gamma }}(q). \end{aligned}$$

When $\varphi _q$ is vanishing, we consider that $b_1\in {\mathbb {Z}}_q^\times$ and other cases that $b_2$, $d_1$ or $d_2$ in ${\mathbb {Z}}_q^\times$ can be treated similarly. For $h^{-1}=\left( \left( {\begin{matrix}\alpha _1&{}\beta _1\\ \gamma _1&{}\delta _1\end{matrix}}\right) ,\left( {\begin{matrix}\alpha _2&{}\beta _2\\ \gamma _2&{}\delta _2\end{matrix}}\right) \right) \in {\overline{\Gamma }}(q)$, it is observed that

$$\begin{aligned} b^{'}_1=-\gamma _2 (\alpha _1 a_1+\beta _1 c_1)+\alpha _2(\alpha _1 b_1+\beta _1 d_1)\in {\mathbb {Z}}_q^\times \end{aligned}$$

which makes $\varphi _q$ vanish on $(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)$. Now we have proven this lemma. $\square$

4.2 At inert prime dividing m

Let q be an inert prime dividing ${\mathfrak {m}}$ such that $(q)={\mathfrak {q}}$. According to Proposition 4.1, there is an isomorphism $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_q,-\det )\simeq (V_1({\mathbb {Q}}_q\sqrt{d}),\det )$ for q inert in $F/{\mathbb {Q}}$ given by

$$\begin{aligned} \left( \begin{matrix}a&{}b\\ {\bar{b}}&{}c\end{matrix}\right) \longmapsto \left( \begin{matrix}b&{}a\sqrt{d}\\ c\sqrt{d}&{}{\bar{b}}\end{matrix}\right) \quad \text { for }a,c\in {\mathbb {Q}}_q, b\in {\mathbb {Q}}_q(\sqrt{d}) \end{aligned}$$

where $^{-}$ on the right hand side denotes the non-trivial action in $\text {Gal}({\mathbb {Q}}_q(\sqrt{d})/{\mathbb {Q}}_q)$.

Definition 4.5

• (1) Suppose that $d\equiv 1{{\,\textrm{mod}\,}}4$. The local Schwartz function $\varphi _q^{\chi _{\mathfrak {m}}}$ at q is vanishing unless, for $i=1,2$,

  $$\begin{aligned} a_i\in {\mathbb {Z}}_q,b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}},c_i\in {\mathbb {Z}}_q\text { and }a_2c_1-a_1c_2\in q{\mathbb {Z}}_q, \end{aligned}$$

  in which case

  $$\begin{aligned}&\varphi _{q}^{\chi _{\mathfrak {m}}}\left( \left( \begin{matrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{matrix}\right) ,\left( \begin{matrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{matrix}\right) \right) \\ =&\left\{ \begin{array}{cl}{\widetilde{\chi }}_{{\mathfrak {m}},\mathfrak {{\mathfrak {q}}}}\left( \frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\right) ,&{}\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\in {\mathcal {O}}_{\mathfrak {q}}^\times ,\\ 0,&{}\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}.\end{array}\right. \end{aligned}$$

• (2) Suppose that $d\equiv 2,3{{\,\textrm{mod}\,}}4$. We replace above $\frac{a_2b_1-a_1b_2}{m}$ by $\frac{2(a_2b_1-a_1b_2)}{m}$ and $\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}$ by $\frac{2({\bar{b}}_2c_1-{\bar{b}}_1c_2)}{m}$ as discussed in Remark 3.6.

In the following we will check the invariance properties of this local Schwartz function with respect to ${{\,\textrm{Sp}\,}}_4\times {{\,\textrm{SO}\,}}(3,1)$ in detail in case of $d\equiv 1{{\,\textrm{mod}\,}}4$ and the other case can be treated similarly. For simplicity we write $\varphi _{q}=\varphi _{q}^{\chi _{\mathfrak {m}}}$ and $\chi =\chi _{\mathfrak {m}}$.

Set ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{matrix}}\right) ,\left( {\begin{matrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{matrix}}\right) \right)$. For $a_i\in {\mathbb {Z}}_q$,$b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ and $c_i\in {\mathbb {Z}}_q$, it is easy to observe that

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) \varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text { for }u\in M_2(q{\mathbb {Z}}_q). \end{aligned}$$

(17)

Set ${\textbf{Y}}=({\textbf{y}}_1,{\textbf{y}}_2)=\left( \left( {\begin{matrix}\beta _1&{}\alpha _1\sqrt{d}\\ \gamma _1\sqrt{d}&{}{\bar{\beta }}_1\end{matrix}}\right) ,\left( {\begin{matrix}\beta _2&{}\alpha _2\sqrt{d}\\ \gamma _2\sqrt{d}&{}{\bar{\beta }}_2\end{matrix}}\right) \right)$. Consider the Fourier transform

$$\begin{aligned} {\hat{\varphi }}_q({\textbf{Y}})=\int \psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))\varphi _q({\textbf{X}})d{\textbf{X}} \end{aligned}$$

where

$$\begin{aligned} {{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}})=-\frac{1}{2}(b_1{\bar{\beta }}_1+{\bar{b}}_1\beta _1-a_1\gamma _1d-\alpha _1c_1d+b_2{\bar{\beta }}_2+{\bar{b}}_2\beta _2-a_2\gamma _2d-\alpha _2c_2d). \end{aligned}$$

By the definition, $\varphi _q$ is invariant under the transformations $a_i\mapsto a_i+q$, $b_i\mapsto b_i+q^2$, $b_i\mapsto b_i+q^2\sqrt{d}$ (or $b_i\mapsto b_i+q^2\cdot \frac{1+\sqrt{d}}{2}$) and $c_i\mapsto c_i+q$. Repeating arguments in the previous subsection, we can observe that the Fourier transform ${\hat{\varphi }}({\textbf{y}}_1,{\textbf{y}}_2)$ is vanishing unless, for $i=1,2$,

$$\begin{aligned} \alpha _i\in {\mathbb {Z}}_q;\beta ,{\bar{\beta }}\in {\mathfrak {q}}^{-1}{\mathcal {O}}_{\mathfrak {q}}(\text {as } \beta _i+{\bar{\beta }}_i\in q^{-1}{\mathbb {Z}}_q,\beta _i-{\bar{\beta }}_i\in q^{-1}\sqrt{d}{\mathbb {Z}}_q);\gamma _i\in {\mathbb {Z}}_q. \end{aligned}$$

It follows that, for $u\in M_2(q^3{\mathbb {Z}}_q)$ such that $u=u^t$,

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2) \end{aligned}$$

which implies

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}0\\ u&{}1\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\varphi _q({\textbf{y}}_1,{\textbf{y}}_2)\quad \text { for }u\in M_2(q^3{\mathbb {Z}}_q),\ u=u^t. \end{aligned}$$

(18)

For $a\in {{\,\textrm{GL}\,}}_2({\mathbb {Z}}_q)$, write ${\textbf{X}}a:=\left( \left( {\begin{matrix}b^{'}_1&{}a^{'}_1\sqrt{d}\\ c^{'}_1\sqrt{d}&{}{\bar{b}}^{'}_1\end{matrix}}\right) ,\left( {\begin{matrix}b^{'}_2&{}a^{'}_2\sqrt{d}\\ c^{'}_2\sqrt{d}&{}{\bar{b}}^{'}_2\end{matrix}}\right) \right)$. We have $a^{'}_2b^{'}_1-a^{'}_1b^{'}_2=\det (a)(a_2b_1-a_1b_2)$, ${\bar{b}}^{'}_2c^{'}_1-{\bar{b}}^{'}_1c^{'}_2=\det (a)({\bar{b}}_2c_1-{\bar{b}}_1c_2)$ and $a^{'}_2c^{'}_1-a^{'}_1c^{'}_2=\det (a)(a_2c_1-a_1c_2)$. So for $\det (a)\in {\mathbb {Z}}_q^\times$, we obtain

$$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) ,1\right) \varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\chi _{V,q}(\det (a))|\det (a)|^2_q{\widetilde{\chi }}_{\mathfrak {q}}(\det (a))\varphi _q({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

(19)

Combining (17),(18) and (19), we can deduce the following lemma:

Lemma 4.6

We have

$$\begin{aligned} \omega (k_2)\varphi _q=\chi _{V,q}(\det (A))|\det (A)|^2_q{\widetilde{\chi }}_{\mathfrak {q}}(\det (A)\varphi _q \end{aligned}$$

for

$$\begin{aligned} k_2=\left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in \left\{ \left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q):B\in M_2(q{\mathbb {Z}}_q), C\in M_2(q^3{\mathbb {Z}}_q)\right\} . \end{aligned}$$

We next discuss the action of ${{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q)$ on $\varphi _q$ characterised by

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)\quad \text { for }h\in {{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q). \end{aligned}$$

Due to the exceptional isomorphism ${{\,\textrm{PSL}\,}}_2({\mathbb {Q}}_q(\sqrt{d}))\simeq {{\,\textrm{SO}\,}}^+(3,1)({\mathbb {Q}}_q)$ as in [26, Sect. 1.2], in this case we check the invariance property under some congruence subgroups of ${{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}})$. Here we have that $h^{-1}{\textbf{x}}_i:=h^{-1}{\textbf{x}}_i({\bar{h}}^{-1})^*$ for $i=1,2$ where $^{-}$ denotes the non-trivial action in $\text {Gal}({\mathbb {Q}}_q(\sqrt{d})/{\mathbb {Q}}_q)$ (see [20, Sect. 2]).

Lemma 4.7

For

$$\begin{aligned} h\in {\overline{\Gamma }}(q)=\left\{ \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \in {{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}}):\pm \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \equiv \pm \left( \begin{matrix}1&{}0\\ 0&{}1\end{matrix}\right) {{\,\textrm{mod}\,}}q \right\} , \end{aligned}$$

we have that

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

(20)

Proof

Set

$$\begin{aligned} h^{-1}=\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \text { and }{\textbf{x}}_i=\left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right) . \end{aligned}$$

with $\alpha ,\delta \equiv 1$ mod q and $\beta ,\gamma \equiv 0$ mod q.

First we assume $a_i\in {\mathbb {Z}}_q$, ${\bar{b}}_i\in {\mathfrak {q}} {\mathcal {O}}_{{\mathfrak {q}}}$,$c_i\in {\mathbb {Z}}_q$ so that $\varphi _q$ is non-vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$. Writing $h^{-1}{\textbf{x}}_i=\left( \begin{matrix}b^{'}_i&{}a^{'}_i\sqrt{d}\\ c^{'}_i\sqrt{d}&{}{\bar{b}}^{'}_i\end{matrix}\right)$, we compute

$$\begin{aligned} \left( \begin{matrix}b^{'}_i&{}a^{'}_i\sqrt{d}\\ c^{'}_i\sqrt{d}&{}{\bar{b}}^{'}_i\end{matrix}\right) =&\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right) \left( \begin{matrix}{\bar{\delta }}&{}-{\bar{\beta }}\\ -{\bar{\gamma }}&{}{\bar{\alpha }}\end{matrix}\right) \\ =&\left( \begin{matrix}\alpha b_i+\beta c_i\sqrt{d}&{}\alpha a_i\sqrt{d}+\beta {\bar{b}}_i\\ \gamma b_i+\delta c_i\sqrt{d}&{}\gamma a_i\sqrt{d}+\delta {\bar{b}}_i\end{matrix}\right) \left( \begin{matrix}{\bar{\delta }}&{}-{\bar{\beta }}\\ -{\bar{\gamma }}&{}{\bar{\alpha }}\end{matrix}\right) \\ =&\left( \begin{matrix}{\bar{\delta }}(\alpha b_i+\beta c_i\sqrt{d})-{\bar{\gamma }}(\alpha a_i\sqrt{d}+\beta {\bar{b}}_i)&{}-{\bar{\beta }}(\alpha b_i+\beta c_i\sqrt{d})+{\bar{\alpha }}(\alpha a_i\sqrt{d}+\beta {\bar{b}}_i)\\ {\bar{\delta }}(\gamma b_i+\delta c_i\sqrt{d})-{\bar{\gamma }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)&{}-{\bar{\beta }}(\gamma b_i+\delta c_i\sqrt{d})+{\bar{\alpha }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)\end{matrix}\right) . \end{aligned}$$

It is not hard to observe that $a^{'}_i\in {\mathbb {Z}}_q$,$b^{'}_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, $c^{'}_i\in {\mathbb {Z}}_q$ and that $a^{'}_2c^{'}_1-a^{'}_1c^{'}_2\equiv a_2c_1-a_1c_2$ mod q. Then we expand

$$\begin{aligned}&a^{'}_2b^{'}_1-a^{'}_1b^{'}_2\\ =&\left( \frac{{\bar{\alpha }}\beta {\bar{b}}_2-{\bar{\beta }}\alpha b_2}{\sqrt{d}}-{\bar{\beta }}\beta c_2+{\bar{\alpha }}\alpha a_2\right) ({\bar{\delta }}(\alpha b_1+\beta c_1\sqrt{d})-{\bar{\gamma }}(\alpha a_1\sqrt{d}+\beta {\bar{b}}_1))\\&-\left( \frac{{\bar{\alpha }}\beta {\bar{b}}_1-{\bar{\beta }}\alpha b_1}{\sqrt{d}}-{\bar{\beta }}\beta c_1+{\bar{\alpha }}\alpha a_1\right) ({\bar{\delta }}(\alpha b_2+\beta c_2\sqrt{d})-{\bar{\gamma }}(\alpha a_2\sqrt{d}+\beta {\bar{b}}_2)), \end{aligned}$$

and, modulo $q^2$, we get

$$\begin{aligned} a^{'}_2b^{'}_1-a^{'}_1b^{'}_2\equiv&\alpha ^2{\bar{\alpha }}{\bar{\delta }}(a_2b_1-a_1b_2)+\alpha {\bar{\alpha }}\beta {\bar{\delta }}\sqrt{d}(a_2c_1-a_1c_2). \end{aligned}$$

Similarly, we have, modulo $q^2$,

$$\begin{aligned}&{\bar{b}}^{'}_2c^{'}_1-{\bar{b}}^{'}_1c^{'}_2\\ \equiv&\delta {\bar{\delta }}c_1(-{\bar{\beta }}\delta c_2\sqrt{d}+{\bar{\alpha }}(\gamma a_2\sqrt{d}+\delta {\bar{b}}_2))-\delta {\bar{\delta }}c_2(-{\bar{\beta }}\delta c_1\sqrt{d}+{\bar{\alpha }}(\gamma a_1\sqrt{d}+\delta {\bar{b}}_1))\\ \equiv&\delta ^2{\bar{\alpha }}{\bar{\delta }}({\bar{b}}_2c_1-{\bar{b}}_1c_2)+{\bar{\alpha }}\gamma \delta {\bar{\delta }}\sqrt{d}(a_2c_1-a_1c_2). \end{aligned}$$

Then we can deduce that

$$\begin{aligned} \frac{a^{'}_2b^{'}_1-a^{'}_1b^{'}_2}{m}+\frac{{\bar{b}}^{'}_2c^{'}_1-{\bar{b}}^{'}_1c^{'}_2}{m}\equiv \frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\mod q \end{aligned}$$

which implies that

$$\begin{aligned} \varphi _q\left( \frac{a^{'}_2b^{'}_1-a^{'}_1b^{'}_2}{m}+\frac{{\bar{b}}^{'}_2c^{'}_1-{\bar{b}}^{'}_1c^{'}_2}{m}\right) =\varphi _q\left( \frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\right) . \end{aligned}$$

Next we assume $b_1\in {\mathcal {O}}_{\mathfrak {q}}^\times$ so that $\varphi _q$ is vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$. It follows that ${\bar{\delta }}\alpha b_1\in {\mathcal {O}}_{\mathfrak {q}}^\times$ and then $b^{'}_1\in {\mathcal {O}}_{\mathfrak {q}}^\times$ which makes $\varphi _q$ is vanishing on $(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)$. Other cases that ${\bar{b}}^{'}_1,b^{'}_2,{\bar{b}}^{'}_2\in {\mathcal {O}}^\times _{\mathfrak {q}}$ can be treated in the same way and recall that $a^{'}_2c^{'}_1-a^{'}_1c^{'}_2\equiv a_2c_1-a_1c_2$ mod q. Hence, if $\varphi _q$ is vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$, so is that on $(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)$.

Now we have proven this lemma. $\square$

4.3 At ramified prime away from 2

Let q be a ramified prime away from 2 such that $(q)={\mathfrak {q}}^2$. According to Proposition 4.1, there is an isomorphism $({\mathcal {H}}(F)\otimes {\mathbb {Q}}_q,-\det )\simeq (V_1({\mathbb {Q}}_q),\det )$ for q ramified in $F/{\mathbb {Q}}$ given by

$$\begin{aligned} \left( \begin{matrix}a&{}b\\ {\bar{b}}&{}c\end{matrix}\right) \longmapsto \left( \begin{matrix}b&{}a\sqrt{d}\\ c\sqrt{d}&{}{\bar{b}}\end{matrix}\right) \quad \text { for }a,c\in {\mathbb {Q}}_q, b\in {\mathbb {Q}}_q(\sqrt{d}) \end{aligned}$$

where $^{-}$ on the right hand side denotes the non-trivial action in $\text {Gal}({\mathbb {Q}}_q(\sqrt{d})/{\mathbb {Q}}_q)$. Note that when $d\equiv 2,3$ (mod 4) the prime 2 is ramified and at the ramified 2 the local Schwartz function is defined in the next subsection.

Definition 4.8

• (1) Suppose that $d\equiv 1{{\,\textrm{mod}\,}}4$. The local Schwartz function $\varphi _q^{\chi _{\mathfrak {m}}}$ at q is vanishing unless, for $i=1,2$,

  $$\begin{aligned} a_i\in {\mathbb {Z}}_q,c_i\in {\mathbb {Z}}_q,b_i\in {\mathcal {O}}_{\mathfrak {q}},b_1{\bar{b}}_2-{\bar{b}}_1b_2\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}, \end{aligned}$$

  in which case

  $$\begin{aligned}&\varphi _{q}^{\chi _{\mathfrak {m}}}\left( \left( \begin{matrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{matrix}\right) ,\left( \begin{matrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{matrix}\right) \right) \\&\quad \left\{ \begin{array}{cl}{\widetilde{\chi }}_{{\mathfrak {m}},\mathfrak {{\mathfrak {q}}}}\left( \frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\right) ,&{}\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\in {\mathcal {O}}_{\mathfrak {q}}^\times ,\\ &{} \qquad 0,\text { if }\frac{a_2b_1-a_1b_2}{m}+\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}.\end{array}\right. \end{aligned}$$

• (2) Suppose that $d\equiv 2,3{{\,\textrm{mod}\,}}4$. We replace above $\frac{a_2b_1-a_1b_2}{m}$ by $\frac{2(a_2b_1-a_1b_2)}{m}$ and $\frac{{\bar{b}}_2c_1-{\bar{b}}_1c_2}{m}$ by $\frac{2({\bar{b}}_2c_1-{\bar{b}}_1c_2)}{m}$ as discussed in Remark 3.6.

In the following we will check the invariance properties of this local Schwartz function with respect to ${{\,\textrm{Sp}\,}}_4\times {{\,\textrm{SO}\,}}(3,1)$ in case of $d\equiv 1{{\,\textrm{mod}\,}}4$ and the other case can be treated similarly. For simplicity we write $\varphi _{q}=\varphi _{q}^{\chi _{\mathfrak {m}}}$ and $\chi =\chi _{\mathfrak {m}}$.

Set ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{matrix}}\right) ,\left( {\begin{matrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{matrix}}\right) \right)$. For $a_i,c_i\in {\mathbb {Z}}_q$ and $b_i,{\bar{b}}_i\in {\mathcal {O}}_{\mathfrak {q}}$, it is easy to observe that

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) \varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text { for }u\in M_2(q{\mathbb {Z}}_q). \end{aligned}$$

(21)

Set ${\textbf{Y}}=({\textbf{y}}_1,{\textbf{y}}_2)=\left( \left( {\begin{matrix}\beta _1&{}\alpha _1\sqrt{d}\\ \gamma _1\sqrt{d}&{}{\bar{\beta }}_1\end{matrix}}\right) ,\left( {\begin{matrix}\beta _2&{}\alpha _2\sqrt{d}\\ \gamma _2\sqrt{d}&{}{\bar{\beta }}_2\end{matrix}}\right) \right)$. Consider the Fourier transform

$$\begin{aligned} {\hat{\varphi }}_q({\textbf{Y}})=\int \psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))\varphi _q({\textbf{X}})d{\textbf{X}} \end{aligned}$$

where

$$\begin{aligned} {{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}})=-\frac{1}{2}(b_1{\bar{\beta }}_1+{\bar{b}}_1\beta _1-a_1\gamma _1d-\alpha _1c_1d+b_2{\bar{\beta }}_2+{\bar{b}}_2\beta _2-a_2\gamma _2d-\alpha _2c_2d). \end{aligned}$$

By the definition, $\varphi _q$ is invariant under the transformations $a_i\mapsto a_i+q$, $b_i\mapsto b_i+q$, $b_i\mapsto b_i+\sqrt{d}$ and $c_i\mapsto c_i+q$. Repeating arguments in the previous subsection, we can observe that the Fourier transform ${\hat{\varphi }}({\textbf{y}}_1,{\textbf{y}}_2)$ is vanishing unless, for $i=1,2$,

$$\begin{aligned} \alpha _i, \gamma _i\in q^{-1}{\mathbb {Z}}_q\text { and }\beta _i,{\bar{\beta }}_i\in {\mathcal {O}}_{\mathfrak {q}}(\text {as }\beta _i+{\bar{\beta }}_i\in {\mathbb {Z}}_q,\beta _i-{\bar{\beta }}_i\in \sqrt{d}{\mathbb {Z}}_q). \end{aligned}$$

It follows that, for $u\in M_2(q^2{\mathbb {Z}}_q)$ such that $u=u^t$,

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2) \end{aligned}$$

which implies

$$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}0\\ u&{}1\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\varphi _q({\textbf{y}}_1,{\textbf{y}}_2)\quad \text { for }u\in M_2(q^2{\mathbb {Z}}_q),\ u=u^t. \end{aligned}$$

(22)

For $a\in {{\,\textrm{GL}\,}}_2({\mathbb {Z}}_q)$ with $\det (a)\in {\mathbb {Z}}_q^\times$, we also have

$$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) ,1\right) \varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\chi _{V,q}(\det (a))|\det (a)|^2_q{\widetilde{\chi }}_{\mathfrak {q}}(\det (a))\varphi _q({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

(23)

Again, combining (21),(22) and (23), we can deduce the following lemma:

Lemma 4.9

We have

$$\begin{aligned} \omega (k_3)\varphi _q=\chi _{V,q}(\det (A))|\det (A)|^2_q{\widetilde{\chi }}_{\mathfrak {q}}(\det (A))\varphi _q \end{aligned}$$

for

$$\begin{aligned} k_3=\left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in \left\{ \left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q):B\in M_2(q{\mathbb {Z}}_q),C\in M_2(q^2{\mathbb {Z}}_q)\right\} . \end{aligned}$$

We next discuss the action of ${{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q)$ on $\varphi _q$ characterised by

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)\quad \text { for }h\in {{\,\textrm{SO}\,}}(3,1)({\mathbb {Q}}_q). \end{aligned}$$

In this case, we check the invariance property under congruence subgroups of ${{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}})$ and have that $h^{-1}{\textbf{x}}_i:=h^{-1}{\textbf{x}}_i({\bar{h}}^{-1})^*$ for $i=1,2$ where $^{-}$ denotes the non-trivial action in $\text {Gal}({\mathbb {Q}}_q(\sqrt{d})/{\mathbb {Q}}_q)$ (see [20, Sect. 2]).

Lemma 4.10

For

$$\begin{aligned} h\in {\overline{\Gamma }}({\mathfrak {q}})=\left\{ \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \in {{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}}):\pm \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \equiv \pm \left( \begin{matrix}1&{}0\\ 0&{}1\end{matrix}\right) {{\,\textrm{mod}\,}}{\mathfrak {q}} \right\} , \end{aligned}$$

we have that

$$\begin{aligned} \omega (1,h)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

(24)

Proof

Set

$$\begin{aligned} h^{-1}=\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \text { and }{\textbf{x}}_i=\left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right) . \end{aligned}$$

with $\alpha ,\delta \equiv 1$ mod ${\mathfrak {q}}$ and $\beta ,\gamma \equiv 0$ mod ${\mathfrak {q}}$.

First we assume $a_i,c_i\in {\mathbb {Z}}_q$ and $b_i,{\bar{b}}_i\in {\mathcal {O}}_{\mathfrak {q}}$ so that $\varphi _q$ is non-vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$. Writing $h^{-1}{\textbf{x}}_i=\left( \begin{matrix}b^{'}_i&{}a^{'}_i\sqrt{d}\\ c^{'}_i\sqrt{d}&{}{\bar{b}}^{'}_i\end{matrix}\right)$, we compute

$$\begin{aligned} \left( \begin{matrix}b^{'}_i&{}a^{'}_i\sqrt{d}\\ c^{'}_i\sqrt{d}&{}{\bar{b}}^{'}_i\end{matrix}\right) =&\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right) \left( \begin{matrix}{\bar{\delta }}&{}-{\bar{\beta }}\\ -{\bar{\gamma }}&{}{\bar{\alpha }}\end{matrix}\right) \\ =&\left( \begin{matrix}\alpha b_i+\beta c_i\sqrt{d}&{}\alpha a_i\sqrt{d}+\beta {\bar{b}}_i\\ \gamma b_i+\delta c_i\sqrt{d}&{}\gamma a_i\sqrt{d}+\delta {\bar{b}}_i\end{matrix}\right) \left( \begin{matrix}{\bar{\delta }}&{}-{\bar{\beta }}\\ -{\bar{\gamma }}&{}{\bar{\alpha }}\end{matrix}\right) \\ =&\left( \begin{matrix}{\bar{\delta }}(\alpha b_i+\beta c_i\sqrt{d})-{\bar{\gamma }}(\alpha a_i\sqrt{d}+\beta {\bar{b}}_i)&{}-{\bar{\beta }}(\alpha b_i+\beta c_i\sqrt{d})+{\bar{\alpha }}(\alpha a_i\sqrt{d}+\beta {\bar{b}}_i)\\ {\bar{\delta }}(\gamma b_i+\delta c_i\sqrt{d})-{\bar{\gamma }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)&{}-{\bar{\beta }}(\gamma b_i+\delta c_i\sqrt{d})+{\bar{\alpha }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)\end{matrix}\right) . \end{aligned}$$

It is not hard to observe that $a^{'}_i,c^{'}_i\in {\mathbb {Z}}_q,b^{'}_i,{\bar{b}}^{'}_i\in {\mathcal {O}}_{\mathfrak {q}}$ and $b^{'}_1{\bar{b}}^{'}_2-{\bar{b}}^{'}_1b^{'}_2\equiv b_1{\bar{b}}_2-{\bar{b}}_1b_2$ mod ${\mathfrak {q}}$. Modulo ${\mathfrak {q}}$, we have

$$\begin{aligned}&a^{'}_2b^{'}_1-a^{'}_1b^{'}_2\\ \equiv&{\bar{\delta }}\alpha b_1\left( \frac{{\bar{\alpha }}\beta {\bar{b}}_2-{\bar{\beta }}\alpha b_2}{\sqrt{d}}+{\bar{\alpha }}\alpha a_2\right) -{\bar{\delta }}\alpha b_2\left( \frac{{\bar{\alpha }}\beta {\bar{b}}_1-{\bar{\beta }}\alpha b_1}{\sqrt{d}}+{\bar{\alpha }}\alpha a_1\right) \\ \equiv&\alpha ^2{\bar{\alpha }}{\bar{\delta }}(a_2b_1-a_1b_2)-\alpha {\bar{\alpha }}\beta {\bar{\delta }}/\sqrt{d}(b_1{\bar{b}}_2-{\bar{b}}_1b_2), \end{aligned}$$

and

$$\begin{aligned}&c^{'}_2{\bar{b}}^{'}_1-c^{'}_1{\bar{b}}^{'}_2\\ \equiv&{\bar{\alpha }}\delta {\bar{b}}_1\left( \frac{{\bar{\delta }}\gamma b_2-\delta {\bar{\gamma }}{\bar{b}}_2}{\sqrt{d}}+{\bar{\delta }}\delta c_2\right) -{\bar{\alpha }}\delta {\bar{b}}_ 2\left( \frac{{\bar{\delta }}\gamma b_1-\delta {\bar{\gamma }}{\bar{b}}_1}{\sqrt{d}}+{\bar{\delta }}\delta c_1\right) \\ \equiv&\delta ^2{\bar{\alpha }}{\bar{\delta }}(c_2{\bar{b}}_1-c_1{\bar{b}}_2)+{\bar{\alpha }}\gamma \delta {\bar{\delta }}/\sqrt{d}(b_1{\bar{b}}_2-{\bar{b}}_1b_2). \end{aligned}$$

So, modulo ${\mathfrak {q}}$, we get

$$\begin{aligned} \frac{a^{'}_2b^{'}_1-a^{'}_1b^{'}_2}{m}\equiv \alpha ^2{\bar{\alpha }}{\bar{\delta }}\frac{a_2b_1-a_1b_2}{m}\quad \text { and }\quad \frac{c^{'}_2{\bar{b}}^{'}_1-c^{'}_1{\bar{b}}^{'}_2}{m}\equiv \delta ^2{\bar{\alpha }}{\bar{\delta }}\frac{c_2{\bar{b}}_1-c_1{\bar{b}}_2}{m}. \end{aligned}$$

It follows that

$$\begin{aligned} \varphi _q\left( \frac{a^{'}_2b^{'}_1-a^{'}_1b^{'}_2}{m}+\frac{c^{'}_2{\bar{b}}^{'}_1-c^{'}_1{\bar{b}}^{'}_2}{m}\right) =\varphi _q\left( \frac{a_2b_1-a_1b_2}{m}+\frac{c_2{\bar{b}}_1-c_1{\bar{b}}_2}{m}\right) . \end{aligned}$$

Next we assume $b_1\in {\mathfrak {q}}^{-1}{\mathcal {O}}^\times _{\mathfrak {q}}$ so that $\varphi _q$ is vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$. It follows that ${\bar{\delta }}\alpha b_1\in {\mathfrak {q}}^{-1}{\mathcal {O}}^\times _{\mathfrak {q}}$ and then $b^{'}_1\in {\mathfrak {q}}^{-1}{\mathcal {O}}^\times _{\mathfrak {q}}$ which makes $\varphi _q$ be vanishing on $(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)$. Other cases for $a_i,c_i,{\bar{b}}_1, b_2,{\bar{b}}_2$ can be treated in the same way. Hence, if $\varphi _q$ is vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$, so is that on $(h^{-1}{\textbf{x}}_1,h^{-1}{\textbf{x}}_2)$.

Now we have proven this lemma. $\square$

4.4 At places dividing $N({\mathfrak {n}})$ and ramified 2

In this subsection we consider the local Schwartz function at finite places dividing $N({\mathfrak {n}})$ and at ramified prime 2 (when $d\equiv 2,3{{\,\textrm{mod}\,}}4$). For a place q and an integral lattice X on V, we put $X_q=X\otimes _{\mathbb {Z}}{\mathbb {Z}}_q$.

Definition 4.11

• (1) Let $q|N({\mathfrak {n}})$ be split with $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$.

  • Suppose that $({\mathfrak {n}},(q))={\mathfrak {q}}$. Define the local Schwartz function $\varphi _{q}$ to be the characteristic function of

    $$\begin{aligned} \left\{ \left( \begin{pmatrix}a_1&{}b_1\\ c_1&{}d_1\end{pmatrix},\begin{pmatrix}a_2&{}b_2\\ c_2&{}d_2\end{pmatrix}\right) \in X_q^2:b_1c_2+c_1b_2\in {\mathcal {O}}_{\mathfrak {q}}^\times ,d_i\in q{\mathbb {Z}}_q\right\} . \end{aligned}$$

  • Suppose that $({\mathfrak {n}},(q))=\bar{{\mathfrak {q}}}$. Define $\varphi _{q}$ to be the characteristic function of

    $$\begin{aligned} \left\{ \left( \begin{pmatrix}a_1&{}b_1\\ c_1&{}d_1\end{pmatrix},\begin{pmatrix}a_2&{}b_2\\ c_2&{}d_2\end{pmatrix}\right) \in X_q^2:b_1c_2+c_1b_2\in {\mathcal {O}}_{\bar{{\mathfrak {q}}}}^\times ,d_i\in q{\mathbb {Z}}_q\right\} . \end{aligned}$$

  • Suppose that $({\mathfrak {n}},(q))=(q)$. Define $\varphi _{q}$ to be the characteristic function of

    $$\begin{aligned} \left\{ \left( \begin{pmatrix}a_1&{}b_1\\ c_1&{}d_1\end{pmatrix},\begin{pmatrix}a_2&{}b_2\\ c_2&{}d_2\end{pmatrix}\right) \in X_q^2:b_1c_2+c_1b_2\in {\mathcal {O}}_{\mathfrak {q}}^\times \times {\mathcal {O}}_{\bar{{\mathfrak {q}}}}^\times ,d_i\in q{\mathbb {Z}}_q\right\} . \end{aligned}$$

• (2) At inert place $q|{\mathfrak {n}}$ with $(q)={\mathfrak {q}}$, we define $\varphi _{q}$ to be the characteristic function of

  $$\begin{aligned} \left\{ \left( \begin{pmatrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{pmatrix},\begin{pmatrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{pmatrix}\right) \in X_q^2:b_1{\bar{b}}_2+{\bar{b}}_1b_2\in {\mathcal {O}}_{\mathfrak {q}}^\times ,c_i\in q{\mathbb {Z}}_q\right\} \end{aligned}$$

• (3) If 2 is ramified with $(2)={\mathfrak {q}}_2^2$, we define $\varphi _2$ to be the characteristic function of

  $$\begin{aligned} \left\{ \left( \begin{pmatrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{pmatrix},\begin{pmatrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{pmatrix}\right) :a_i,c_i\in {\mathbb {Z}}_2, b_i\in \frac{1}{2}{\mathcal {O}}_{{\mathfrak {q}}_2},b_1{\bar{b}}_2+{\bar{b}}_1b_2\in \frac{1}{2}{\mathcal {O}}_{{\mathfrak {q}}_2}^\times \right\} . \end{aligned}$$

Note that if we take the local Schwartz function at finite places dividing $N({\mathfrak {n}})$ and ramified 2 as the characteristic function of integral lattice, the resulting theta lifting would be vanishing.

In the following we will check the invariance properties of this local Schwartz function with respect to ${{\,\textrm{Sp}\,}}_4\times {{\,\textrm{SO}\,}}(3,1)$.

Lemma 4.12

• (1) For $\varphi _q$ as in above Definition 4.11 (1) and (2), We have

  $$\begin{aligned} \omega (k_4)\varphi _q=\varphi _q \end{aligned}$$

  for

  $$\begin{aligned} k_4\in \left\{ \begin{pmatrix}A&{}B\\ C&{}D\end{pmatrix}\in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q):A\in \left( \begin{matrix}{\mathbb {Z}}_q&{}q{\mathbb {Z}}_q\\ q{\mathbb {Z}}_q&{}{\mathbb {Z}}_q\end{matrix}\right) , B\in M_2(q{\mathbb {Z}}_q),C\in M_2(q{\mathbb {Z}}_q)\right\} . \end{aligned}$$

• (2) At ramified 2, we have

  $$\begin{aligned} \omega (k_5)\varphi _2=\varphi _2 \end{aligned}$$

  for

  $$\begin{aligned} k_5\in \left\{ \begin{pmatrix}A&{}B\\ C&{}D\end{pmatrix}\in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q):A\in \left( \begin{matrix}{\mathbb {Z}}_2&{}2{\mathbb {Z}}_2\\ 2{\mathbb {Z}}_2&{}{\mathbb {Z}}_2\end{matrix}\right) , B\in M_2(2^4{\mathbb {Z}}_2)\right\} . \end{aligned}$$

Proof

• (1) We prove this lemma in details only for split q with $({\mathfrak {n}},(q))={\mathfrak {q}}$ and other cases can be treated similarly. Set ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}a_1&{}b_1\\ c_1&{}d_1\end{matrix}}\right) ,\left( {\begin{matrix}a_2&{}b_2\\ c_2&{}d_2\end{matrix}}\right) \right) \in X_q^2$. It is not difficult to observe that

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) \varphi _q({\textbf{X}})=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text { for }u\in M_2(q{\mathbb {Z}}_q). \end{aligned}$$

  Set ${\textbf{Y}}=({\textbf{y}}_1,{\textbf{y}}_2)=\left( \left( {\begin{matrix}\alpha _1&{}\beta _1\\ \gamma _1&{}\delta _1\end{matrix}}\right) ,\left( {\begin{matrix}\alpha _2&{}\beta _2\\ \gamma _2&{}\delta _2\end{matrix}}\right) \right)$. Consider the Fourier transform

  $$\begin{aligned} {\hat{\varphi }}_q({\textbf{Y}})=\int _{X_{q,v}^2}\psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))\varphi _q({\textbf{X}})d{\textbf{X}}\quad \text { for }i=1,2, \end{aligned}$$

  where

  $$\begin{aligned} {{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}})=-\frac{1}{2}(a_1\delta _1-b_1\gamma _1-c_1\beta _1+d_1\alpha _1+a_2\delta _2-b_2\gamma _2-c_2\beta _2+d_2\alpha _2). \end{aligned}$$

  By the above definition, $\varphi _q$ is invariant under the transformations $a_i\mapsto a_i+{\mathbb {Z}}_q$, $b_i\mapsto b_i+q$, $b_i\mapsto b_i+q\lambda _{\sqrt{d}}$ (or $b_i\mapsto b_i+q\frac{1+\lambda _{\sqrt{d}}}{2}$) and $d_i\mapsto d_i+q$. Note that $b_1c_2+c_1b_2\in {\mathcal {O}}_{\mathfrak {q}}^\times$ is not preserved under $b_i\mapsto b_i+{\mathfrak {q}}$ or $b_i\mapsto b_i+\bar{{\mathfrak {q}}}$. Then we can deduce that $\omega \left( \left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right) \right) \varphi _q({\textbf{Y}})$ is vanishing unless $\alpha _i\in {\mathbb {Z}}_q$, $\beta _i\in {\mathbb {Z}}_q$, $\gamma _i\in {\mathbb {Z}}_q$ and $\delta _i\in q{\mathbb {Z}}_q$. It follows that

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{Y}})=\omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{Y}})\quad \text { for }u\in M_2(q{\mathbb {Z}}_q),\ u=u^t \end{aligned}$$

  which implies

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}0\\ u&{}1\end{matrix}\right) \right) \varphi _q({\textbf{Y}})=\varphi _q({\textbf{Y}})\quad \text { for }u\in M_2(q{\mathbb {Z}}_q),\ u=u^t. \end{aligned}$$

  For $a=\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \in {{\,\textrm{GL}\,}}_2({\mathbb {Z}}_q)$, set $\left( \left( {\begin{matrix}a^{'}_1&{}b^{'}_1\\ c^{'}_1&{}d^{'}_1\end{matrix}}\right) ,\left( {\begin{matrix}a^{'}_2&{}b^{'}_2\\ c^{'}_2&{}d^{'}_2\end{matrix}}\right) \right) :=({\textbf{x}}_1,{\textbf{x}}_2)a$. It is clear that

  $$\begin{aligned} d^{'}_1=\alpha d_1+\beta d_2\text { and }d^{'}_2=\gamma d_1+\delta d_2 \end{aligned}$$

  lie in $q{\mathbb {Z}}_q$. Also we have

  $$\begin{aligned} b^{'}_1c^{'}_2+c^{'}_1b^{'}_2=(\alpha b_1+\gamma b_2)(\beta c_1+\delta c_2)+(\beta b_1+\delta b_2)(\alpha c_1+\gamma c_2). \end{aligned}$$

  If $\beta ,\gamma \equiv 0{{\,\textrm{mod}\,}}q$ and $\det (a)\in {\mathbb {Z}}_q^\times$, then $b^{'}_1c^{'}_2+c^{'}_1b^{'}_2\in {\mathcal {O}}_{\mathfrak {q}}^\times$. We can deduce that

  $$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) \right) \varphi _q({\textbf{X}})=\varphi _q({\textbf{X}})\quad \text { for }a\in \begin{pmatrix}{\mathbb {Z}}_q&{}q{\mathbb {Z}}_q\\ q{\mathbb {Z}}_q&{}{\mathbb {Z}}_q\end{pmatrix}\text { and }\det (a)\in {\mathbb {Z}}_q^\times . \end{aligned}$$

• (2) Set ${\textbf{X}}=({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}b_1&{}a_1\sqrt{d}\\ c_1\sqrt{d}&{}{\bar{b}}_1\end{matrix}}\right) ,\left( {\begin{matrix}b_2&{}a_2\sqrt{d}\\ c_2\sqrt{d}&{}{\bar{b}}_2\end{matrix}}\right) \right)$. For $a_i,c_i\in {\mathbb {Z}}_2$ and $b_i\in \frac{1}{2}{\mathcal {O}}_{{\mathfrak {q}}_2}$, it is easy to observe that

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \right) \varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text { for }u\in M_2(2^4{\mathbb {Z}}_2). \end{aligned}$$

  Set ${\textbf{Y}}=({\textbf{y}}_1,{\textbf{y}}_2)=\left( \left( {\begin{matrix}\beta _1&{}\alpha _1\sqrt{d}\\ \gamma _1\sqrt{d}&{}{\bar{\beta }}_1\end{matrix}}\right) ,\left( {\begin{matrix}\beta _2&{}\alpha _2\sqrt{d}\\ \gamma _2\sqrt{d}&{}{\bar{\beta }}_2\end{matrix}}\right) \right)$. Consider the Fourier transform

  $$\begin{aligned} {\hat{\varphi }}_q({\textbf{Y}})=\int \psi _q({{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}}))\varphi _q({\textbf{X}})d{\textbf{X}} \end{aligned}$$

  where

  $$\begin{aligned} {{\,\textrm{tr}\,}}({\textbf{X}},{\textbf{Y}})=-\frac{1}{2}(b_1{\bar{\beta }}_1+{\bar{b}}_1\beta _1-a_1\gamma _1d-\alpha _1c_1d+b_2{\bar{\beta }}_2+{\bar{b}}_2\beta _2-a_2\gamma _2d-\alpha _2c_2d). \end{aligned}$$

  By the definition, $\varphi _q$ is invariant under the transformations $a_i\mapsto a_i+{\mathbb {Z}}_2$, $b_i\mapsto b_i+{\mathbb {Z}}_2$, $b_i\mapsto b_i+{\mathbb {Z}}_2\sqrt{d}$ and $c_i\mapsto c_i+{\mathbb {Z}}_2$. Repeating arguments in the previous subsection, we can observe that the Fourier transform ${\hat{\varphi }}({\textbf{y}}_1,{\textbf{y}}_2)$ is vanishing unless, for $i=1,2$,

  $$\begin{aligned} \alpha _i, \gamma _i\in 4{\mathbb {Z}}_2\text { and }\beta _i\in 4{\mathcal {O}}_{{\mathfrak {q}}_2}(\text {as }\beta _i+{\bar{\beta }}_i\in 4{\mathbb {Z}}_q,\beta _i-{\bar{\beta }}_i\in 4\sqrt{d}{\mathbb {Z}}_q). \end{aligned}$$

  It follows that, for $u\in M_2({\mathbb {Z}}_2)$ such that $u=u^t$,

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}u\\ 0&{}1\end{matrix}\right) \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\omega \left( \left( \begin{matrix}0&{}1\\ -1&{}0\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2) \end{aligned}$$

  which implies

  $$\begin{aligned} \omega \left( \left( \begin{matrix}1&{}0\\ u&{}1\end{matrix}\right) \right) \varphi _q({\textbf{y}}_1,{\textbf{y}}_2)=\varphi _q({\textbf{y}}_1,{\textbf{y}}_2)\quad \text { for }u\in M_2({\mathbb {Z}}_2),\ u=u^t. \end{aligned}$$

  For $a=\left( {\begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}}\right) \in {{\,\textrm{GL}\,}}_2({\mathbb {Z}}_q)$, set $\left( \left( {\begin{matrix}b^{'}_1&{}a^{'}_1\sqrt{d}\\ c^{'}_1\sqrt{d}&{}{\bar{b}}^{'}_1\end{matrix}}\right) ,\left( {\begin{matrix}b^{'}_2&{}a^{'}_2\sqrt{d}\\ c^{'}_2\sqrt{d}&{}{\bar{b}}^{'}_2\end{matrix}}\right) \right) :=({\textbf{x}}_1,{\textbf{x}}_2)a$. We have

  $$\begin{aligned} b^{'}_1=\alpha b_1+\gamma b_2 \text{ and } b^{'}_2=\beta b_1+\delta b_2 \end{aligned}$$

  and then

  $$\begin{aligned} b^{'}_1{\bar{b}}^{'}_2+{\bar{b}}^{'}_1b^{'}_2=(\alpha b_1+\gamma b_2)(\beta {\bar{b}}_1+\delta {\bar{b}}_2)+(\alpha {\bar{b}}_1+\gamma {\bar{b}}_2)(\beta b_1+\delta b_2). \end{aligned}$$

  If $\beta ,\gamma \in 2{\mathbb {Z}}_2$ and $\alpha \gamma \in {\mathbb {Z}}_2^\times$, we have

  $$\begin{aligned} \omega \left( \left( \begin{matrix}a&{}0\\ 0&{}^ta^{-1}\end{matrix}\right) ,1\right) \varphi _2({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _2({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

$\square$

Lemma 4.13

• (1) Let $q|N({\mathfrak {n}})$ split with $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$.

  • Suppose that $({\mathfrak {n}},(q))={\mathfrak {q}}$. We have

    $$\begin{aligned} \omega (1,h_1)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2) \end{aligned}$$

    for $h_1=h_{1,1}\times h_{1,2}$ with

    $$\begin{aligned} h_{1,i}\in {\overline{\Gamma }}({\mathfrak {q}})=\left\{ \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \in {{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}}):\pm \left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \equiv \pm \left( \begin{matrix}1&{}0\\ 0&{}1\end{matrix}\right) {{\,\textrm{mod}\,}}{\mathfrak {q}}\right\} . \end{aligned}$$

  • If $({\mathfrak {n}},(q))=\bar{{\mathfrak {q}}}$, we have

    $$\begin{aligned} \omega (1,h_2)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text{ for } h_2\in {\overline{\Gamma }}(\bar{{\mathfrak {q}}}). \end{aligned}$$

  • If $({\mathfrak {n}},(q))=(q)$, we have

    $$\begin{aligned} \omega (1,h_3)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text{ for } h_3\in {\overline{\Gamma }}(q). \end{aligned}$$

• (2) For inert $q|N({\mathfrak {n}})$ with $(q)={\mathfrak {q}}$, we have

  $$\begin{aligned} \omega (1,h_4)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text{ for } h_4\in {\overline{\Gamma }}({\mathfrak {q}}). \end{aligned}$$

• (3) For 2 ramified with $(2)={\mathfrak {q}}_2^2$, we have

  $$\begin{aligned} \omega (1,h_5)\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _q({\textbf{x}}_1,{\textbf{x}}_2)\quad \text{ for } h_5\in {\overline{\Gamma }}({\mathfrak {q}}_2). \end{aligned}$$

Proof

In part (1), we prove the first statement and other cases can be treated similarly. Let $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$. Set $h^{-1}_{j}=\left( \begin{matrix}\alpha _j&{}\beta _j\\ \gamma _j&{}\delta _j\end{matrix}\right)$ and ${\textbf{x}}_i=\left( \begin{matrix}a_i&{}b_i\\ c_i&{}d_i\end{matrix}\right)$ with $\alpha _j,\delta _j\equiv 1{{\,\textrm{mod}\,}}{\mathfrak {q}}$ and $\beta _j,\gamma _j\equiv 0{{\,\textrm{mod}\,}}{\mathfrak {q}}$. We compute

$$\begin{aligned} \left( \begin{matrix}a^{'}_i&{}b^{'}_i\\ c^{'}_i&{}d^{'}_i\end{matrix}\right) :=&h_1^{-1}{\textbf{x}}_i{}^t(h_2^{-1})^*\\ =&\left( \begin{matrix}\delta _2(\alpha _1 a_i+\beta _1 c_i)-\beta _2(\alpha _1 b_i+\beta _1 d_i)&{}-\gamma _2 (\alpha _1 a_i+\beta _1 c_i)+\alpha _2(\alpha _1 b_i+\beta _1 d_i)\\ \delta _2 (\gamma _1 a_i+\delta _1 c_i)-\beta _2 (\gamma _1 b_i+\delta _1 d_i)&{}-\gamma _2 (\gamma _1 a_i+\delta _1 c_i)+\alpha _2 (\gamma _1 b_i+\delta _1 d_i)\end{matrix}\right) \\ \equiv&\begin{pmatrix}*&{}b_i\\ c_i&{}d_i\end{pmatrix}\mod \mathfrak {q}. \end{aligned}$$

So the conditions on $b_1c_2+b_2c_1$ and $d^{'}_i$ for $\varphi _q$ non-vanishing are preserved.

Let q be an inert prime. Set $h^{-1}=\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right)$ and ${\textbf{x}}_i=\left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right)$ with $\alpha ,\delta \equiv 1$ mod q and $\beta ,\gamma \equiv 0$ mod q. It suffices to show that

$$\begin{aligned} \left( \begin{matrix}b^{'}_i&{}a^{'}_i\sqrt{d}\\ c^{'}_i\sqrt{d}&{}{\bar{b}}^{'}_i\end{matrix}\right) =&\left( \begin{matrix}\alpha &{}\beta \\ \gamma &{}\delta \end{matrix}\right) \left( \begin{matrix}b_i&{}a_i\sqrt{d}\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{matrix}\right) \left( \begin{matrix}{\bar{\delta }}&{}-{\bar{\beta }}\\ -{\bar{\gamma }}&{}{\bar{\alpha }}\end{matrix}\right) \\ =&\left( \begin{matrix}{\bar{\delta }}(\alpha b_i+\beta c_i\sqrt{d})-{\bar{\gamma }}(\alpha a_i\sqrt{d}+\beta {\bar{b}}_i)&{}*\\ {\bar{\delta }}(\gamma b_i+\delta c_i\sqrt{d})-{\bar{\gamma }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)&{}-{\bar{\beta }}(\gamma b_i+\delta c_i\sqrt{d})+{\bar{\alpha }}(\gamma a_i\sqrt{d}+\delta {\bar{b}}_i)\end{matrix}\right) \\ \equiv&\begin{pmatrix}b_i&{}*\\ c_i\sqrt{d}&{}{\bar{b}}_i\end{pmatrix} \mod q. \end{aligned}$$

It is clear that if $\varphi _q$ is vanishing on $({\textbf{x}}_1,{\textbf{x}}_2)$, then so is $\omega (1,h_j)\varphi _q$ on $({\textbf{x}}_1,{\textbf{x}}_2)$ in the same way as dicussed in previous subsections. Similarly at ramified 2 we obtain the same result. $\square$

4.5 At other finite places

We consider non-archimedean places away from $m|d_F|N({\mathfrak {n}})$. For such a place q and an integral lattice X on V, we put $X_q=X\otimes _{\mathbb {Z}}{\mathbb {Z}}_q$. Define its dual lattice

$$\begin{aligned} X_q^\sharp =\{{\textbf{x}}\in V\otimes {\mathbb {Q}}_q:({\textbf{x}},{\textbf{y}})\in {\mathbb {Z}}_q\ \forall {\textbf{y}}\in X_q\} \end{aligned}$$

and let $(q^{-l_q})$ be the ${\mathbb {Z}}_q$-module generated by $\{({\textbf{x}},{\textbf{x}}):{\textbf{x}}\in X_q^\sharp \}$. In [3, Lemma 27], it is shown that $l_q=0$ at these places. At each place q, we define the local Schwartz function $\varphi _q$ to be the characteristic function of $X_q^2$. Note that $\varphi _q$ is invariant under ${{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)\times {{\,\textrm{PSL}\,}}_2({\mathbb {Z}}_q)$ for split q and ${{\,\textrm{PSL}\,}}_2({\mathcal {O}}_{\mathfrak {q}})$ for inert or ramified q due to $l_q=0$ (see [3, Sect. 5.2]).

Lemma 4.14

( [25, Lemma 2.1]) At non-archimedean $q\not \mid m|d_F|N({\mathfrak {n}})$ we have

$$\begin{aligned} \omega (\sigma )\varphi _q=\chi _{V,q}(\det A)\varphi _q\quad \text {for }\sigma =\left( \begin{matrix}A&{}B\\ C&{}D\end{matrix}\right) \in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}_q). \end{aligned}$$

5 Theta lift and Fourier coefficient

Let $F={\mathbb {Q}}(\sqrt{d})$ be an imaginary quadratic field of class number 1. Denote its ring of integers by ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ and the discriminant by $d_F$. Let m be a product of distinct inert or split primes as introduced in Sect. 4 and choose a quadratic Hecke character $\chi _{\mathfrak {m}}$ (${\mathfrak {m}}=m{\mathcal {O}}$) of conductor ${\mathfrak {f}}=\sqrt{d}{\mathfrak {m}}$. Let ${\mathfrak {n}}$ be square-free and coprime to $(m|d_F|)$. Suppose that ${\mathcal {F}}=({\mathcal {F}}_0,{\mathcal {F}}_1,{\mathcal {F}}_2):{\mathbb {H}}_3\rightarrow {\mathbb {C}}^3$ is a weight 2 cusp form for $\Gamma _0({\mathfrak {n}})$ with the corresponding $\Gamma _0({\mathfrak {n}})$-invariant differential $\eta _{\mathcal {F}}$ of the form $-{\mathcal {F}}_0\frac{dz}{r}+{\mathcal {F}}_1\frac{dr}{r}+{\mathcal {F}}_2\frac{d{\bar{z}}}{r}$ on ${\mathbb {H}}_3$. More detailed discussion on Bianchi modular form can be found in [26, Chapter 1].

Remark 5.1

• (1) Suppose that $d\equiv 1{{\,\textrm{mod}\,}}4$ with $d_F=d$. We choose the local Schwartz function $\varphi _q^{\chi _{\mathfrak {m}}}$ at each place q dividing $m|d_F|$ as defined in Definition 4.2, 4.5 and 4.8. At each place q dividing $N({\mathfrak {n}})$, the local Schwartz function $\varphi ^{\mathfrak {n}}_q$ is chosen to be as in Definition 4.11. For all other finite places we take the local Schwartz function as in Sect. 4.5.

• (2) Suppose that $d\equiv 2,3{{\,\textrm{mod}\,}}4$ with $d_F=4d$. We choose the local Schwartz function $\varphi _q^{\chi _{\mathfrak {m}}}$ at each place q dividing m as defined in Definition 4.2 and 4.5, and that at ramified place away from 2 as in Definition 4.8. At each place q dividing $2N({\mathfrak {n}})$, the local Schwartz functions $\varphi ^{\mathfrak {n}}_q$ and $\varphi _2$ are chosen to be as in Definition 4.11. For all other places we take the local Schwartz function as in Sect. 4.5.

It has been shown in Lemma 4.4, 4.7, 4.10 and 4.13 that the local Schwartz function $\varphi _v$ at each place v dividing $m|d_F|N({\mathfrak {n}})$ is invariant under the action of the principal congruence subgroup ${\overline{\Gamma }}({\mathfrak {q}}_v)\subset {{\,\textrm{SO}\,}}^+(3,1)(V({\mathbb {Q}}_v))$. We now consider a ${\overline{\Gamma }}_0({\mathfrak {q}}_v)$-invariant local Schwartz function $\varphi _v^{\text {new}}$ at these places defined by

$$\begin{aligned} \varphi _v^{\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)=\sum _{[\gamma ]\in {\overline{\Gamma }}_0({\mathfrak {q}}_v)/{\overline{\Gamma }}({\mathfrak {q}}_v)}\omega (1,\gamma )\varphi _v({\textbf{x}}_1,{\textbf{x}}_2) \end{aligned}$$

where the sum is taken over all the representatives of ${\overline{\Gamma }}_0({\mathfrak {q}}_v)/{\overline{\Gamma }}({\mathfrak {q}}_v)$. With this new local Schwartz function we know that $\varphi ^{\text {new}}_f$ is invariant under ${\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}})$ when $d\equiv 1$ mod 4 or ${\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}}{\mathfrak {q}}_2)$ when $d\equiv 2,3$ mod 4.

Give the Schwartz form $\varphi _2\in S(V({\mathbb {R}})^2)\otimes \Omega ^2(D)$ as constructed in [13, Sect. 5], [9, Sect. 4] and [10, Sect. 5], and the above finite Schwartz function $\varphi _f^{{{\,\textrm{new}\,}}}$ on $V({\mathbb {A}}_f)^2$, we now consider a Schwartz form

$$\begin{aligned} \varphi ({\textbf{X}},z):=\varphi _2\otimes \varphi _f^{{{\,\textrm{new}\,}}}\in S(V({\mathbb {A}})^2)\otimes \Omega ^2(D)\quad \text { for }{\textbf{X}}\in V({\mathbb {A}})^2,z\in D. \end{aligned}$$

Then we consider the theta series in this case is given by

$$\begin{aligned} \theta (g^{'},\varphi _f^{{{\,\textrm{new}\,}}},z):=\sum _{{\textbf{X}}\in V({\mathbb {Q}})^2}\omega (g^{'})\varphi ({\textbf{X}},z)\quad \text { for }g^{'}\in {{\,\textrm{Sp}\,}}_4({\mathbb {A}}) \end{aligned}$$

which defines a closed differential 2-form on ${\overline{\Gamma }}_0(2mN)\backslash D$.

Following [13, Theorem 1], the theta lifting of ${\mathcal {F}}$, which is a holomophic Siegel modular form of weight 2, is given by

$$\begin{aligned} \Theta _{\varphi }(\eta _{\mathcal {F}})(g^{'}):=\int _{\Gamma \backslash D}\eta _{\mathcal {F}}(z)\wedge \theta (g^{'},\varphi _f^{{{\,\textrm{new}\,}}},z) \end{aligned}$$

where $\Gamma ={\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}})\cap \Gamma _0({\mathfrak {n}})={\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}})$ when $d\equiv 1$ mod 4 or $\Gamma ={\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}}{\mathfrak {q}}_2)\cap \Gamma _0({\mathfrak {n}})={\overline{\Gamma }}_0({\mathfrak {f}}{\mathfrak {n}}{\mathfrak {q}}_2)$ when $d\equiv 2,3$ mod 4. Moreover, the Fourier coefficients are given as periods of $\eta _{{\mathcal {F}}}$ over certain special cycles $C_\beta$ in $\Gamma \backslash D$ attached to positive definite $\beta \in {{\,\textrm{Sym}\,}}_2({\mathbb {Q}})$, i.e.,

$$\begin{aligned} \Theta (\eta _{{\mathcal {F}}})(g^{'})=\sum _{\beta >0}a_\beta (\eta _{{\mathcal {F}}})e^{2\pi i{{\,\textrm{tr}\,}}(\beta g^{'})} \end{aligned}$$

By Lemma 4.3, 4.6, 4.9 and 4.14, we can determine that it has level

$$\begin{aligned} {\mathcal {L}}_{m,{\mathfrak {n}}}=\left\{ \begin{pmatrix}A&{}B\\ C&{}D\end{pmatrix}\in {{\,\textrm{Sp}\,}}_4({\mathbb {Z}}):A\in \begin{pmatrix}{\mathbb {Z}}&{}n_1N({\mathfrak {n}}){\mathbb {Z}}\\ n_1N({\mathfrak {n}}){\mathbb {Z}}&{}{\mathbb {Z}}\end{pmatrix},\right. \\ \left. B\in M_2(n_2m|d|N({\mathfrak {n}}){\mathbb {Z}}),C\in M_2(m^3d^2N({\mathfrak {n}}){\mathbb {Z}})\right\} \end{aligned}$$

with

$$\begin{aligned} \left\{ \begin{array}{cl}n_1=n_2=1,&{}\text { if }d\equiv 1{{\,\textrm{mod}\,}}4\\ n_1=2,n_2=2^4,&{}\text { if }d\equiv 2,3{{\,\textrm{mod}\,}}4.\end{array}\right. \end{aligned}$$

Recall

$$\begin{aligned} \Omega _\beta =\left\{ ({\textbf{x}}_1,{\textbf{x}}_2)\in V({\mathbb {Q}})^2:\begin{pmatrix}({\textbf{x}}_1,{\textbf{x}}_1)&{}({\textbf{x}}_1,{\textbf{x}}_2)\\ ({\textbf{x}}_1,{\textbf{x}}_2)&{}({\textbf{x}}_2,{\textbf{x}}_2)\end{pmatrix}=\beta \right\} . \end{aligned}$$

By [3, Theorem 9], the Fourier coefficient of the theta lifting $\Theta _{\varphi }(\eta _{\mathcal {F}})$ at $\beta >0$ is given by

$$\begin{aligned} C_{\Theta _{\varphi }(\eta _{\mathcal {F}}),\beta }=&\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma \backslash \Omega _\beta }\varphi _f^{\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)\int _{C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\eta _{\mathcal {F}}\nonumber \\ =&\sum _{[\kappa _i]\in \Gamma \backslash {\mathbb {P}}^1(F)}\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}}\varphi _f^{\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)\int _{C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\eta _{\mathcal {F}} \end{aligned}$$

(25)

where the second equality is the consequence of Proposition 3.4. For simplicity we will denote $C_{\Theta _{\varphi ^{\text {new}}}(\eta _{\mathcal {F}}),\beta }:=I=\sum _{[\kappa _i]\in \Gamma \backslash {\mathbb {P}}^1(F)}I_{\kappa _i}$ where

$$\begin{aligned} I_{\kappa _i}=\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}}\varphi _f^{\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)\int _{C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\eta _{\mathcal {F}}. \end{aligned}$$

(26)

We will first express $I_\infty$ in terms of the twisted L-value $L({\mathcal {F}},\chi _{\mathfrak {m}},1)$ in Subsect. 5.1 and then use Atkin-Lehner operators to calculate $I_{\kappa _i}$ for $\kappa _i\ne \infty$ in Subsect. 5.2.

Remark 5.2

We will describe how to choose the Gram matrix $\beta$ for which we will show that $C_{\Theta _{\varphi }(\eta _{\mathcal {F}}),\beta }$ is non-vanishing.

• Let $\det \beta \in -d({\mathbb {Q}}^\times )^2$. Then for $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _\beta$, $U({\textbf{x}}_1,{\textbf{x}}_2)^\perp$ is split over ${\mathbb {Q}}$ due to Proposition 3.3 and $U({\textbf{x}}_1,{\textbf{x}}_2)^\perp$ has signature (1,1). The stabilizer $\Gamma _U\subset \Gamma$ of $U=U({\textbf{x}}_1,{\textbf{x}}_2)$ is trivial if $U^\perp$ is split over ${\mathbb {Q}}$, see [11, Lemma 4.2].

• For $({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( {\begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}}\right) ,\left( {\begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}}\right) \right) \in \Omega _{\beta ,\infty ,+}$, we have

  $$\begin{aligned} \beta =\begin{pmatrix}\beta _{11}&{}\beta _{12}\\ \beta _{21}&{}\beta _{22}\end{pmatrix}=\begin{pmatrix}b_1{\bar{b}}_1&{}\frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)\\ \frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)&{}b_2{\bar{b}}_2\end{pmatrix}. \end{aligned}$$

  We want this pair to satisfy the condition $\dagger$ as in Remark 3.6. This will allow us to apply Lemma 3.5 to deduce that the corresponding cusp $z_{U({\textbf{x}}_1,{\textbf{x}}_2)}$ runs through all the representatives in ${\mathfrak {f}}^{-1}/{\mathcal {O}}$. For the non-vanishing of $\varphi ^{\chi _{\mathfrak {m}}}_q$ at split or inert q dividing ${\mathfrak {f}}$, we only count $({\textbf{x}}_1,{\textbf{x}}_2)$ such that $b_i$ ($i=1,2$) is divisible by m. Via imposing conditions on $\beta$ itself, we can achieve that for any pair $({\textbf{x}}_1,{\textbf{x}}_2)$ in $\Omega _{\beta ,\infty ,+}$ the assumption $\dagger$ holds. Explicit examples of $\beta$ will be given at the end of the Subsect. 5.1, see from Example 5.6 to 5.11.

Assume that $\beta$ is given as in above Remark 5.2. For $({\textbf{x}}_1,{\textbf{x}}_2)$ in $\Omega _{\beta ,\infty ,+}$, we have

$$\begin{aligned} \begin{pmatrix}b_1\\ b_2\end{pmatrix}=m\begin{pmatrix}x&{}y\\ z&{}w\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\text { if }d\equiv 1 \end{aligned}$$

(27)

or

$$\begin{aligned} \begin{pmatrix}b_1\\ b_2\end{pmatrix}=\frac{1}{2}m\begin{pmatrix}x&{}y\\ z&{}w\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\text { if }d\equiv 2,3 \end{aligned}$$

(28)

with $x,y,z,w\in {\mathbb {Z}}$ and $xw-yz=\pm 1$. We want to find out if there is another pair $({\textbf{y}}_1,{\textbf{y}}_2)$ in $\Omega _{\beta ,\infty ,+}$ such that it gives rise to the same cycle $D_U$ as that generated by $({\textbf{x}}_1,{\textbf{x}}_2)$.

Assume that such a pair $({\textbf{y}}_1,{\textbf{y}}_2)$ exists in $\Omega _{\beta ,\infty ,+}$. For $U({\textbf{x}}_1,{\textbf{x}}_2)=U({\textbf{y}}_1,{\textbf{y}}_2)$ we consider an element $\sigma =\left( {\begin{matrix}a&{}b\\ c&{}d\end{matrix}}\right) \in {{\,\textrm{GL}\,}}_2({\mathbb {Q}})$ such that

$$\begin{aligned} \left( \begin{matrix}a&{}b\\ c&{}d\end{matrix}\right) \left( \begin{matrix}{\textbf{x}}_1\\ {\textbf{x}}_2\end{matrix}\right) =\left( \begin{matrix}{\textbf{y}}_1\\ {\textbf{y}}_2\end{matrix}\right) . \end{aligned}$$

(29)

To make $\langle u_{\infty },{\textbf{x}}_1,{\textbf{x}}_2,u_{\kappa _j}\rangle$ and $\langle u_{\infty },{\textbf{y}}_1,{\textbf{y}}_2,u_{\kappa _j}\rangle$ represent the same orientation, we need $\sigma \in {{\,\textrm{GL}\,}}_2^+({\mathbb {Q}})$. Additionally the Gram matrix $\beta$’s corresponding to $({\textbf{x}}_1,{\textbf{x}}_2)$ and $({\textbf{y}}_1,{\textbf{y}}_2)$ must be identical.

Expressing $({\textbf{y}}_i,{\textbf{y}}_j)$ in terms of $({\textbf{x}}_i,{\textbf{x}}_j)$ and using bilinearity, we have

$$\begin{aligned} ({\textbf{y}}_1,{\textbf{y}}_1)= & {} a^2({\textbf{x}}_1,{\textbf{x}}_1)+2ab({\textbf{x}}_1,{\textbf{x}}_2)+b^2({\textbf{x}}_2,{\textbf{x}}_2),\\ ({\textbf{y}}_2,{\textbf{y}}_2)= & {} c^2({\textbf{x}}_1,{\textbf{x}}_1)+2cd({\textbf{x}}_1,{\textbf{x}}_2)+d^2({\textbf{x}}_2,{\textbf{x}}_2),\\ ({\textbf{y}}_1,{\textbf{y}}_2)= & {} ac({\textbf{x}}_1,{\textbf{x}}_1)+(ad+bc)({\textbf{x}}_1,{\textbf{x}}_2)+bd({\textbf{x}}_2,{\textbf{x}}_2). \end{aligned}$$

Consider that $\det \beta$ is preserved; more explicitly,

$$\begin{aligned} \det \beta =&\det (({\textbf{y}}_i,{\textbf{y}}_j))=({\textbf{y}}_1,{\textbf{y}}_1)({\textbf{y}}_2,{\textbf{y}}_2)-({\textbf{y}}_1,{\textbf{y}}_2)^2\\ =&a^2c^2({\textbf{x}}_1,{\textbf{x}}_1)^2+b^2d^2({\textbf{x}}_2,{\textbf{x}}_2)^2+4abcd({\textbf{x}}_1,{\textbf{x}}_2)^2+(a^2d^2+b^2c^2)({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_2,{\textbf{x}}_2)\\&+2ac(ad+bc)({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_1,{\textbf{x}}_2)+2bd(ad+bc)({\textbf{x}}_1,{\textbf{x}}_2)({\textbf{x}}_2,{\textbf{x}}_2)\\&-a^2c^2({\textbf{x}}_1,{\textbf{x}}_1)^2-b^2d^2({\textbf{x}}_2,{\textbf{x}}_2)^2-(ad+bc)^2({\textbf{x}}_1,{\textbf{x}}_2)^2-2abcd({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_2,{\textbf{x}}_2)\\&-2ac(ad+bc)({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_1,{\textbf{x}}_2)-2bd(ad+bc)({\textbf{x}}_1,{\textbf{x}}_2)({\textbf{x}}_2,{\textbf{x}}_2)\\ =&(\det \sigma )^2\det (({\textbf{x}}_i,{\textbf{x}}_j))=\det (({\textbf{x}}_i,{\textbf{x}}_j)). \end{aligned}$$

Since $\sigma$ has positive determinant we know that $\sigma \in {{\,\textrm{SL}\,}}_2({\mathbb {Q}})$.

Moreover, to preserve $\beta$ the following identities must hold:

$$\begin{aligned} ({\textbf{x}}_1,{\textbf{x}}_1)= & {} a^2({\textbf{x}}_1,{\textbf{x}}_1)+2ab({\textbf{x}}_1,{\textbf{x}}_2)+b^2({\textbf{x}}_2,{\textbf{x}}_2), \end{aligned}$$

(30)

$$\begin{aligned} ({\textbf{x}}_2,{\textbf{x}}_2)= & {} c^2({\textbf{x}}_1,{\textbf{x}}_1)+2cd({\textbf{x}}_1,{\textbf{x}}_2)+d^2({\textbf{x}}_2,{\textbf{x}}_2), \end{aligned}$$

(31)

$$\begin{aligned} ({\textbf{x}}_1,{\textbf{x}}_2)= & {} ac({\textbf{x}}_1,{\textbf{x}}_1)+(ad+bc)({\textbf{x}}_1,{\textbf{x}}_2)+bd({\textbf{x}}_2,{\textbf{x}}_2). \end{aligned}$$

(32)

As $\det \sigma =ad-bc=1$, we can rewrite (32) as

$$\begin{aligned} ac({\textbf{x}}_1,{\textbf{x}}_1)+2bc({\textbf{x}}_1,{\textbf{x}}_2)+bd({\textbf{x}}_2,{\textbf{x}}_2)=0. \end{aligned}$$

(33)

We will describe $\sigma$ in different cases in the following.

• (I) Let b be 0. From (30) we know that $a^2=1$, and from (32) we have that $({\textbf{x}}_1,{\textbf{x}}_2)=ac({\textbf{x}}_1,{\textbf{x}}_1)+({\textbf{x}}_1,{\textbf{x}}_2)$ which implies $c=0$. So $\sigma =\pm \left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$. In the same way, if $c=0$ then $\sigma =\pm \left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$.

• (II) Assume that $bc\ne 0$. Substituting $({\textbf{x}}_1,{\textbf{x}}_2)$ in (30), we have

  $$\begin{aligned} ({\textbf{x}}_1,{\textbf{x}}_1)=a^2({\textbf{x}}_1,{\textbf{x}}_1)+b^2({\textbf{x}}_2,{\textbf{x}}_2)-\frac{a}{c}(ac({\textbf{x}}_1,{\textbf{x}}_1)+bd({\textbf{x}}_2,{\textbf{x}}_2)) \end{aligned}$$

  which is simplified to be

  $$\begin{aligned} c({\textbf{x}}_1,{\textbf{x}}_1)+b({\textbf{x}}_2,{\textbf{x}}_2)=0. \end{aligned}$$

  (34)

  Combining (33) and (34) we have

  $$\begin{aligned} ac({\textbf{x}}_1,{\textbf{x}}_1)+2bc({\textbf{x}}_1,{\textbf{x}}_2)-cd({\textbf{x}}_1,{\textbf{x}}_1)=0 \end{aligned}$$

  and then $d=a+2b\frac{({\textbf{x}}_1,{\textbf{x}}_2)}{({\textbf{x}}_1,{\textbf{x}}_1)}$. As $ad-bd=1$, we have

  $$\begin{aligned} a^2+2ab\frac{({\textbf{x}}_1,{\textbf{x}}_2)}{({\textbf{x}}_1,{\textbf{x}}_1)}+b^2\frac{({\textbf{x}}_2,{\textbf{x}}_2)}{({\textbf{x}}_1,{\textbf{x}}_1)}=1. \end{aligned}$$

  (35)

  Set ${\textbf{x}}_1=\left( {\begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}d_1\end{matrix}}\right)$, ${\textbf{x}}_2=\left( {\begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}d_2\end{matrix}}\right)$, ${\textbf{y}}_1=\left( {\begin{matrix}a_1^{'}&{}b_1^{'}\\ {\bar{b}}_1^{'}&{}d_1^{'}\end{matrix}}\right)$ and ${\textbf{y}}_2=\left( {\begin{matrix}a_2^{'}&{}b_2^{'}\\ {\bar{b}}_2^{'}&{}d_2^{'}\end{matrix}}\right)$. Combining (27) and (29), we can rewrite

  $$\begin{aligned} b_1^{'}=ab_1+bb_2=m((ax+bz)+(ay+bw)\omega ) \end{aligned}$$

  or combining (28) and (29)

  $$\begin{aligned} b_1^{'}=ab_1+bb_2=\frac{1}{2}((ax+bz)+(ay+bw)\omega ). \end{aligned}$$

  Then we need $ax+bz=\mu \in {\mathbb {Z}}$ and $ay+bw=\nu \in {\mathbb {Z}}$. Solving these two equations we get

  $$\begin{aligned} a=\frac{\mu \delta -\nu \gamma }{\alpha \delta -\beta \gamma }\in {\mathbb {Z}}\quad \text { and }\quad b=\frac{\nu \alpha -\mu \beta }{\alpha \delta -\beta \gamma }\in {\mathbb {Z}}. \end{aligned}$$

  Similarly, we can deduce that $c,d\in {\mathbb {Z}}$ when treating $b^{'}_2$. Therefore, the linear transform $\sigma \in {{\,\textrm{SL}\,}}_2({\mathbb {Q}})$ on $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\kappa _i,+}$ generates the same cycle, but the Schwartz function on $({\textbf{y}}_1,{\textbf{y}}_2)$ vanishes if $\sigma \notin {{\,\textrm{SL}\,}}_2({\mathbb {Z}})$. For particular choices of $\beta$we get limited possibilities of above $\sigma$. We can rewrite (35) as

  $$\begin{aligned} \left( a+b\frac{({\textbf{x}}_1,{\textbf{x}}_2)}{({\textbf{x}}_1,{\textbf{x}}_1)}\right) ^2+b^2\frac{({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_2,{\textbf{x}}_2)-({\textbf{x}}_1,{\textbf{x}}_2)^2}{({\textbf{x}}_1,{\textbf{x}}_1)^2}=1. \end{aligned}$$

  (36)

  • (II.1) If $({\textbf{x}}_1,{\textbf{x}}_2)=0$ and $({\textbf{x}}_1,{\textbf{x}}_1)=({\textbf{x}}_2,{\textbf{x}}_2)$, then $a^2=0$ as $bc\ne 0$ by our assumption. So in this case $\sigma =\pm \left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right)$.

  • (II.2) If $({\textbf{x}}_1,{\textbf{x}}_2)=0$ and $({\textbf{x}}_1,{\textbf{x}}_1)<({\textbf{x}}_2,{\textbf{x}}_2)$, i.e. $\frac{\det \beta }{({\textbf{x}}_1,{\textbf{x}}_1)^2}>1$, then there is no such a $\sigma$ that $bc\ne 0$.

  • (II.3) If $({\textbf{x}}_1,{\textbf{x}}_2)\ne 0$ and $\frac{({\textbf{x}}_1,{\textbf{x}}_1)({\textbf{x}}_2,{\textbf{x}}_2)-({\textbf{x}}_1,{\textbf{x}}_2)^2}{({\textbf{x}}_1,{\textbf{x}}_1)^2}>1$, then b has to be 0 which is a contradiction to $bc\ne 0$.

Remark 5.3

The possibilities of $\sigma$ in (29) will determine the constant $\mu _\beta$ in Proposition 5.5. After the whole treatment of this section, we will see that this $\mu _\beta$ does not effect the non-vanishing of our theta liftings since it appears in the Fourier coefficient as a non-zero multiplier. In Example from 5.6 to 5.11, we will show how to get the exact values of $\mu _\beta$.

5.1 On cycles through $\infty$

We will first calculate the part $I_\infty$ corresponding to the cusp $\infty$ as in (26). We pick a fundamental domain for $\Gamma _\infty \backslash D_U$ and integrate with respect to the cycle. Since we are integrating along a vertical path with z-coordinate constant, we can ignore dz and $d{\bar{z}}$. We obtain

$$\begin{aligned} I_\infty =\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}}\varphi ^{{{\,\textrm{new}\,}}}_f({\textbf{x}}_1,{\textbf{x}}_2)\int _{(z,r)\in C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\frac{1}{2}{\mathcal {F}}_1(z,r)dr. \end{aligned}$$

(37)

Lemma 5.4

For $({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$, we have

$$\begin{aligned} \varphi _f^{{{\,\textrm{new}\,}}}({\textbf{x}}_1,{\textbf{x}}_2)=\lambda _{{\mathfrak {m}},{\mathfrak {n}}}\varphi _f({\textbf{x}}_1,{\textbf{x}}_2) \end{aligned}$$

where

$$\begin{aligned} \lambda _{{\mathfrak {m}},{\mathfrak {n}}}=\prod _{q_1|m|d|}\left[ {\overline{\Gamma }}_0(q_1):{\overline{\Gamma }}(q_1)\right] \prod _{{\mathfrak {q}}_2 \text{ above } \text{ ramified } 2}\left[ {\overline{\Gamma }}_0({\mathfrak {q}}_2):{\overline{\Gamma }}({\mathfrak {q}}_2)\right] \prod _{{\mathfrak {q}}_3|{\mathfrak {n}}}\left[ {\overline{\Gamma }}_0({\mathfrak {q}}_3):{\overline{\Gamma }}({\mathfrak {q}}_3)\right] . \end{aligned}$$

Proof

Note that any pair $({\textbf{x}}_1,{\textbf{x}}_2)$ in $\Omega _{\beta ,\infty }$ is of form $\left( \left( {\begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}}\right) ,\left( {\begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}}\right) \right)$. Recall from [26, Sect. 1.3], for $a\in {\mathcal {O}}^\times$ satisfying $((a),{\mathfrak {f}})=1$ we have

$$\begin{aligned} \prod _{v|{\mathfrak {f}}}{\widetilde{\chi }}_v(a_v)=\prod _{v\not \mid {\mathfrak {f}}}{\widetilde{\chi }}_v^{-1}(a_v)=\chi ^{-1}((a)). \end{aligned}$$

Then, for our choice of finite Schwartz function $\varphi _f$, we have

$$\begin{aligned} \begin{aligned} \varphi _f({\textbf{x}}_1,{\textbf{x}}_2)&=\prod _{v|{\mathfrak {f}}}{\widetilde{\chi }}_{{\mathfrak {m}},v}\left( \frac{a_2b_1-a_1b_2}{m}\right) \\ {}&=\prod _{v\not \mid {\mathfrak {f}}}{\widetilde{\chi }}_{{\mathfrak {m}},v}^{-1}\left( \frac{a_2b_1-a_1b_2}{m}\right) =\chi ^{-1}\left( \left( \frac{a_2b_1-a_1b_2}{m}\right) \right) \end{aligned} \end{aligned}$$

or

$$\begin{aligned} \varphi _f({\textbf{x}}_1,{\textbf{x}}_2)=\chi ^{-1}\left( \left( \frac{2(a_2b_1-a_1b_2)}{m}\right) \right) . \end{aligned}$$

Let q be the split prime dividing m. Consider the representative

$$\begin{aligned} \gamma =(\gamma _1,\gamma _2)=\left( \begin{pmatrix}x_{\gamma ,1}&{}y_{\gamma ,1}\\ 0&{}x_{\gamma ,1}^{-1}\end{pmatrix},\begin{pmatrix}x_{\gamma ,2}&{}y_{\gamma ,2}\\ 0&{}x_{\gamma ,2}^{-1}\end{pmatrix}\right) \end{aligned}$$

for ${\overline{\Gamma }}_0(q)/{\overline{\Gamma }}(q)$ with $[x_{\gamma ,j}]\in ({\mathcal {O}}/(q))^\times$ and $[y_{\gamma ,j}]\in {\mathcal {O}}/(q)$. By the computation in the proof of Lemma 4.4, we can observe that

$$\begin{aligned} \varphi _q^{\chi _{\mathfrak {m}},\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)&=\sum _{[\gamma ]\in {\overline{\Gamma }}_0(q)/{\overline{\Gamma }}(q)}({\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}{\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}})(x^{-2}_{\gamma ,1})({\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}{\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}})\left( \frac{a_2b_1-a_1b_2}{m}\right) \\&=\sum _{[\gamma ]\in {\overline{\Gamma }}_0(q)/{\overline{\Gamma }}(q)}({\widetilde{\chi }}_{{\mathfrak {m}},{\mathfrak {q}}}{\widetilde{\chi }}_{{\mathfrak {m}},\bar{{\mathfrak {q}}}})\left( \frac{a_2b_1-a_1b_2}{m}\right) \\&=[{\overline{\Gamma }}_0(q):{\overline{\Gamma }}(q)]\varphi _q^{\chi _{\mathfrak {m}}}({\textbf{x}}_1,{\textbf{x}}_2) \end{aligned}$$

Similarly we have for inert q|m

$$\begin{aligned} \varphi _q^{\chi _{\mathfrak {m}},\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)=[{\overline{\Gamma }}_0(q):{\overline{\Gamma }}(q)]\varphi _q^{\chi _{\mathfrak {m}}}({\textbf{x}}_1,{\textbf{x}}_2), \end{aligned}$$

and for ramified prime q with $(q)={\mathfrak {q}}^2$

$$\begin{aligned} \varphi _q^{\chi _{\mathfrak {m}},\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)=[{\overline{\Gamma }}_0({\mathfrak {q}}):{\overline{\Gamma }}({\mathfrak {q}})]\varphi _q^{\chi _{\mathfrak {m}}}({\textbf{x}}_1,{\textbf{x}}_2). \end{aligned}$$

At the place $q|N({\mathfrak {n}})$ we will show that

$$\begin{aligned} \varphi _{q}^{{\mathfrak {n}},{{\,\textrm{new}\,}}}=[{\overline{\Gamma }}_0(q):{\overline{\Gamma }}(q)]\varphi _{q}^{{\mathfrak {n}}}. \end{aligned}$$

For q split, set ${\textbf{x}}_i=\begin{pmatrix}a_i&{}b_i\\ c_i&{}0\end{pmatrix}$ and compute

$$\begin{aligned} w(1,\gamma )\cdot ({\textbf{x}}_1,{\textbf{x}}_2)=\left( \begin{pmatrix}*&{}x_{\gamma _1}^{-1}x_{\gamma _2}^{-1}b_1\\ x_{\gamma _1}x_{\gamma _2}c_1&{}0\end{pmatrix},\begin{pmatrix}*&{}x_{\gamma _1}^{-1}x_{\gamma _2}^{-1}b_2\\ x_{\gamma _1}x_{\gamma _2}c_2&{}0\end{pmatrix}\right) . \end{aligned}$$

It is not hard to observe that the condition on $b_1c_2+b_2c_1$ is preserved as $x_{\gamma ,j}\in {\mathbb {Z}}_q^\times$. It follow that

$$\begin{aligned} w(1,\gamma )\varphi _{q}({\textbf{x}}_1,{\textbf{x}}_2)=\varphi _{q}({\textbf{x}}_1,{\textbf{x}}_2) \end{aligned}$$

which implies the assertion. For q inert, set ${\textbf{x}}_i=\begin{pmatrix}b_i&{}a_i\sqrt{d}\\ 0&{}{\bar{b}}_i\end{pmatrix}$ and compute

$$\begin{aligned} \gamma ^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)=\left( \begin{pmatrix}x^{-1}_\gamma {\bar{x}}_\gamma b_1&{}*\\ 0&{}x_\gamma {\bar{x}}^{-1}_\gamma {\bar{b}}_1\end{pmatrix},\begin{pmatrix}x_\gamma ^{-1} {\bar{x}}_\gamma b_2&{}*\\ 0&{}x_\gamma {\bar{x}}^{-1}_\gamma {\bar{b}}_2\end{pmatrix}\right) . \end{aligned}$$

Again the condition on $b_1{\bar{b}}_2+{\bar{b}}_1b_2$ is preserved and so the assertion follows. The case at ramified 2 can be treated similarly.

Now we have proven the lemma. $\square$

Proposition 5.5

Assume that the Gram matrix $\beta$ is chosen so that the condition $\dagger$ in Lemma 3.5 is satisfied. Then we can calculate

$$\begin{aligned} I_\infty =\frac{\mu _\beta \lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{2A(1,1,\chi _{\mathfrak {m}},1)} \end{aligned}$$

(38)

where $\mu _\beta$ is a non-zero integer depending on $\beta$ as stated in Remark 5.3 and $A(1,1,\chi _{\mathfrak {m}},1)$ is given explicitly in [24, Theorem 1.8].

Proof

By the above lemma, we can express

$$\begin{aligned} I_\infty =&\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}}\varphi _f^{\text {new}}({\textbf{x}}_1,{\textbf{x}}_2)\int _{(z,r)\in C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\frac{1}{2}{\mathcal {F}}_1(z,r)dr\\ =&\lambda _{{\mathfrak {m}},{\mathfrak {n}}}\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}}\varphi _f({\textbf{x}}_1,{\textbf{x}}_2)\int _{(z,r)\in C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\frac{1}{2}{\mathcal {F}}_1(z,r)dr. \end{aligned}$$

Under our assumption on $\beta$, by Lemma 3.5 we have

$$\begin{aligned} I_\infty =\mu _\beta \lambda _{{\mathfrak {m}},{\mathfrak {n}}}\sum _{[z_{U}]\in {{\mathfrak {f}}^{-1}/{\mathcal {O}}},(z_U{\mathfrak {f}},{\mathfrak {f}})=1}\chi ^{-1}_{\mathfrak {m}}(z_{U}{\mathfrak {f}})\int _0^\infty \frac{1}{2}{\mathcal {F}}_1(z,r)dr\nonumber \end{aligned}$$

where $\mu _\beta$ is a non-zero integer depending on the possibilities of $\sigma$ as discussed in Remark 5.3. At last, by [24, Theorem 1.8] with $n=1$, we can compute

$$\begin{aligned} I_\infty =\mu _\beta \lambda _{{\mathfrak {m}},{\mathfrak {n}}}\sum _{[z_{U}]\in {{\mathfrak {f}}^{-1}/{\mathcal {O}}},(z_U{\mathfrak {f}},{\mathfrak {f}})=1}\chi ^{-1}_{\mathfrak {m}}(z_{U}{\mathfrak {f}})\int _0^\infty \frac{1}{2}{\mathcal {F}}_1(z,r)dr=\frac{\mu _\beta \lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{2A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

$\square$

For a diagonal Gram matrix $\beta$, the pair $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\infty ,+}$ has $b_1{\bar{b}}_2+{\bar{b}}_1b_2=0$. It follows that $\varphi _{q}^{{\mathfrak {n}}}$ is vanishing on such a pair $({\textbf{x}}_1,{\textbf{x}}_2)$. So, for the non-vanishing of $I_\infty$, the Gram matrix $\beta$ being diagonal is ruled out of our consideration. In the following we give some examples of $\beta$ satisfying the condition $\dagger$ (as promised in Remark 5.2) for which $I_\infty$ can be expressed in terms of $L({\mathcal {F}},\chi _{\mathfrak {m}},1)$.

Example 5.6

Let $F={\mathbb {Q}}(\sqrt{-3})$ with $d_F=-3$ and ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ with $\omega =\frac{1+\sqrt{-3}}{2}$. Suppose that

$$\begin{aligned} \beta =\begin{pmatrix}b_1{\bar{b}}_1&{}\frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)\\ \frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)&{}b_2{\bar{b}}_2\end{pmatrix}=\left( \begin{matrix}m^2&{}\frac{1}{2}m^2\\ \frac{1}{2}m^2&{}m^2\end{matrix}\right) . \end{aligned}$$

We have

$$\begin{aligned} I_\infty =\frac{4\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Proof

For the non-vanishing of $\varphi _q^{\chi _{\mathfrak {m}}}$, we need $m|b_i$. Solving $b_i{\bar{b}}_i=m^2$, we must take $b_i=\pm m$, $\pm m\omega$ or $\pm m{\bar{\omega }}$. Observing

$$\begin{aligned} ({\textbf{x}}_1,{\textbf{x}}_2)=\frac{1}{2}(b_1{\bar{b}}_2+{\bar{b}}_1b_2)=\frac{1}{2}m^2, \end{aligned}$$

we can determine $b_i$ with the condition $\dagger$ as in Proposition 3.5 satisfied:

$$\begin{aligned}&\left\{ \begin{array}{l}b_1=m\\ b_2=m\omega \end{array}\right. \sim \begin{pmatrix}b_1\\ b_2\end{pmatrix}=m\begin{pmatrix}1&{}0\\ 0&{}1\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\\&\left\{ \begin{array}{l}b_1=m\\ b_2=m{\bar{\omega }}\end{array}\right. \sim \begin{pmatrix}b_1\\ b_2\end{pmatrix}=m\begin{pmatrix}1&{}0\\ 1&{}-1\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\\&\left\{ \begin{array}{l}b_1=-m\\ b_2=-m\omega \end{array}\right. \sim \begin{pmatrix}b_1\\ b_2\end{pmatrix}=m\begin{pmatrix}-1&{}0\\ 0&{}-1\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\\&\left\{ \begin{array}{l}b_1=-m\\ b_2=-m{\bar{\omega }}\end{array}\right. \sim \begin{pmatrix}b_1\\ b_2\end{pmatrix}=m\begin{pmatrix}-1&{}0\\ -1&{}1\end{pmatrix}\begin{pmatrix}1\\ \omega \end{pmatrix}\\&\left\{ \begin{array}{l}b_1=m\omega \text { or }m{\bar{\omega }}\\ b_2=m,\end{array}\right. \left\{ \begin{array}{l}b_1=-m\omega \text { or }-m{\bar{\omega }}\\ b_2=-m.\end{array}\right. \end{aligned}$$

We have seen in Lemma 3.7 that the sign of $\text {Im}(b_1{\bar{b}}_2)$ determines the orientation $\varepsilon$ of $T_Z(D_U)$ via $\varepsilon \text {Im}(b_1{\bar{b}}_2)>0$. If the cycle $D_U$ integrated over is directed from the cusp on the complex plane to the cusp $\infty$, we need $\varepsilon <0$ which implies $\text {Im}(b_1{\bar{b}}_2)<0$. We will list all pairs in $\Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$ with $\text {Im}(b_1{\bar{b}}_2)<0$.

First we consider one pair

$$\begin{aligned} ({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})=\left( \left( \begin{matrix}a_{1,1}&{}m\\ m&{}0\end{matrix}\right) ,\left( \begin{matrix}a_{1,2}&{}m\omega \\ m{\bar{\omega }}&{}0\end{matrix}\right) \right) \end{aligned}$$

with $a_{1,1},a_{1,2}\in {\mathbb {Z}}$ which gives rise to the cycle $D_{U({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}$ directed from the cusp $z_{U({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}=\frac{a_{1,1}\omega -a_{1,2}}{m\sqrt{-3}}\in F$ to the cusp $\infty$. Rewriting (35) as $a^2-ab+b^2=1$, we have either $a^2=1,b^2=0$ or $a^2=0,b^2=1$ and then $\sigma =\pm \left( {\begin{matrix}1&{}0\\ 0&{}1\end{matrix}}\right)$ or $\pm \left( {\begin{matrix}0&{}1\\ -1&{}0\end{matrix}}\right)$. So the following four pairs give rise to the same cycle $D_{U({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}$:

$$\begin{aligned} ({\textbf{x}}_{1,1},{\textbf{x}}_{1,2}),(-{\textbf{x}}_{1,1},-{\textbf{x}}_{1,2}),({\textbf{x}}_{1,2},-{\textbf{x}}_{1,1}),(-{\textbf{x}}_{1,2},{\textbf{x}}_{1,1}). \end{aligned}$$

(39)

Lemma 3.5 tells us that for $({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$, $z_{U({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}$ ranges over ${\mathfrak {f}}^{-1}/{\mathcal {O}}$ with ${\mathfrak {f}}=\sqrt{-3}{\mathfrak {m}}$.

Suppose that

$$\begin{aligned} ({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})=\left( \left( \begin{matrix}a_{2,1}&{}-m\\ -m&{}0\end{matrix}\right) ,\left( \begin{matrix}a_{2,2}&{}-m{\bar{\omega }}\\ -m\omega &{}0\end{matrix}\right) \right) \end{aligned}$$

with $a_{2,1},a_{2,2}\in {\mathbb {Z}}$ is another pair in $\Omega _{\beta ,\infty ,+}$ which gives rise to the cycle $D_{U({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})}$ directed from the cusp $z_{U({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})}=\frac{a_{2,1}{\bar{\omega }}-a_{2,2}}{m\sqrt{-3}}\in F$ to the cusp $\infty$. Similarly we have following pairs

$$\begin{aligned} ({\textbf{x}}_{2,1},{\textbf{x}}_{2,2}),(-{\textbf{x}}_{2,1},-{\textbf{x}}_{2,2}),({\textbf{x}}_{2,2},-{\textbf{x}}_{2,1}),(-{\textbf{x}}_{2,2},{\textbf{x}}_{2,1}) \end{aligned}$$

(40)

giving rise to the same cycle $D_{U({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})}$. Also for $({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$ we have $z_{U({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}$ running through ${\mathfrak {f}}^{-1}/{\mathcal {O}}$ with ${\mathfrak {f}}=\sqrt{-3}{\mathfrak {m}}$.

It is obvious that the eight pairs in (39) and (40) are not $\Gamma _\infty$-equivalent since the $\Gamma _\infty$-action on the pair preserves off-diagonal entries of each component of the pair. Then we can split $I_\infty$ as

$$\begin{aligned} I_\infty =&I_{({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}+I_{(-{\textbf{x}}_{1,1},-{\textbf{x}}_{1,2})}+I_{({\textbf{x}}_{1,2},-{\textbf{x}}_{1,1})}+I_{(-{\textbf{x}}_{1,2},{\textbf{x}}_{1,1})}\\&+I_{({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})}+I_{(-{\textbf{x}}_{2,1},-{\textbf{x}}_{2,2})}+I_{({\textbf{x}}_{2,2},-{\textbf{x}}_{2,1})}+I_{(-{\textbf{x}}_{2,2},{\textbf{x}}_{2,1})}, \end{aligned}$$

where the subscript $(-,-)$ indicates the sum as in (37) over $[z_{U(-,-)}]\in {\mathfrak {f}}^{-1}/{\mathcal {O}}$. By [24, Theorem 1.8] with $n=1$, we can calculate

$$\begin{aligned} I_{({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}=I_{({\textbf{x}}_{2,1},{\textbf{x}}_{2,2})}=\frac{\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{2A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

So, in this case we have $\mu _\beta =8$ and then we can deduce that

$$\begin{aligned} I_\infty =8\cdot I_{({\textbf{x}}_{1,1},{\textbf{x}}_{1,2})}=\frac{4\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

$\square$

Detailed calculations in the following examples can be found in [26, Sect. 4.4.1].

Example 5.7

Let $F={\mathbb {Q}}(\sqrt{d})$ with $d\equiv 1$ (mod 4) and $d\ne -3$ in which case $d_F=d$ and ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ with $\omega =\frac{1+\sqrt{d}}{2}$. Suppose that

$$\begin{aligned} \beta =\left( \begin{matrix}m^2&{}\frac{1}{2}m^2\\ \frac{1}{2}m^2&{}\frac{1-d}{4}m^2\end{matrix}\right) . \end{aligned}$$

We have

$$\begin{aligned} I_\infty =\frac{2\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Example 5.8

Let $F={\mathbb {Q}}(\sqrt{d})$ with $d\equiv 2,3$ (mod 4) and $d\ne -1$ in which case $d_F=4d$ and ${\mathcal {O}}={\mathbb {Z}}[d]$. Let

$$\begin{aligned} \beta =\left( \begin{array}{ll}\frac{1}{4}m^2 &{}\frac{1}{4}nm^2\\ \frac{1}{4}nm^2&{}\frac{1}{4}(n^2-d)m^2\end{array}\right) \end{aligned}$$

with $n\in {\mathbb {Z}}$ coprime to $2{\mathfrak {n}}$ (for the non-vanishing of $\varphi _{q}^{{\mathfrak {n}}}$ and $\varphi _2$). We have

$$\begin{aligned} I_\infty =\frac{2\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Example 5.9

Let $F={\mathbb {Q}}(i)$ with $d_F=-4$ and ${\mathcal {O}}={\mathbb {Z}}[i]$. Set

$$\begin{aligned} \beta =\begin{pmatrix}\frac{1}{4}m^2&{}\frac{1}{4}nm^2\\ \frac{1}{4}nm^2&{}\frac{1}{4}(n^2+1)m^2\end{pmatrix} \end{aligned}$$

with $1<n\in {\mathbb {Z}}$ coprime to $2{\mathfrak {n}}$ (for the non-vanishing of $\varphi _q^{{\mathfrak {n}}}$ and $\varphi _2$ at ramified 2). We have

$$\begin{aligned} I_\infty =\frac{4\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Example 5.10

Let $F={\mathbb {Q}}(\sqrt{d})$ with $d\equiv 1$ (mod 4) and $d\ne -3$ in which case $d_F=d$ and ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ with $\omega =\frac{1+\sqrt{d}}{2}$. Set

$$\begin{aligned} \beta =\left( \begin{matrix}m^2&{}\frac{nm^2}{2}\\ \frac{nm^2}{2}&{}\frac{n^2-d}{4}m^2\end{matrix}\right) \end{aligned}$$

with $1<n\in {\mathbb {Z}}$ coprime to $q_{\mathfrak {n}}$. We have

$$\begin{aligned} I_\infty =\frac{2\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Example 5.11

Let $F={\mathbb {Q}}(\sqrt{-3})$ with $d_F=d=-3$ and ${\mathcal {O}}={\mathbb {Z}}[\omega ]$ with $\omega =\frac{1+\sqrt{-3}}{2}$. Set

$$\begin{aligned} \beta =\left( \begin{matrix}m^2&{}\frac{nm^2}{2}\\ \frac{nm^2}{2}&{}\frac{n^2-d}{4}m^2\end{matrix}\right) \end{aligned}$$

with odd $n\in {\mathbb {Z}}$ greater than 1 and coprime to ${\mathfrak {n}}$. We have

$$\begin{aligned} I_\infty =\frac{3\lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{A(1,1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

Remark

We can swap the diagonal entries of each $\beta$ in Example from 5.6 to 5.11 and obtain same results.

5.2 On other cycles

We introduce the Atkin-Lehner operator as defined in Lingham’s thesis [15, Sect. 5.3]. Lingham developed this for all odd class numbers while we shall only use results for class number 1 since then we can follow Asai’s treatment of cusps (see [1, Sect. 1.1]) in the case of principal ideal domain. For ${\mathfrak {m}}$ in ${\mathcal {O}}$ dividing ${\mathfrak {n}}$ such that ${\mathfrak {m}}$ and $\frac{{\mathfrak {n}}}{{\mathfrak {m}}}$ are coprime, take

$$\begin{aligned} W_{\mathfrak {m}}=\left( \begin{matrix}x&{}y\\ z&{}w\end{matrix}\right) \end{aligned}$$

(41)

where $x\in {\mathfrak {m}}$, $y\in {\mathcal {O}}$, $z\in {\mathfrak {n}}$,$w\in {\mathfrak {m}}$ and $(xw-yz){\mathcal {O}}={\mathfrak {m}}$.

Proposition 5.12

• (1) For any ideal ${\mathfrak {m}}$ dividing ${\mathfrak {n}}$ such that ${\mathfrak {m}}$ and $\frac{{\mathfrak {n}}}{{\mathfrak {m}}}$ are coprime, we can find a matrix of the form (41).

• (2) $W_{\mathfrak {m}}$ is an involution (i.e. $W_{\mathfrak {m}}^2$ (modulo scalars) lies in $\Gamma _0({\mathfrak {n}})$), normalizes $\Gamma _0({\mathfrak {n}})$ and is independent of the particular choice of x, y, z, w.

Proof

See [15, Lemma 5.3.1 and Lemma 5.3.2]. $\square$

In particular if we take ${\mathfrak {m}}={\mathcal {O}}$ we get an element of $\Gamma _0({\mathfrak {n}})$ and if we take ${\mathfrak {m}}={\mathfrak {n}}$ we get the analogue of the classical Fricke involution. One can check that the Fricke involution can be formed as a product of Atkin-Lehner involutions, where ${\mathfrak {m}}$ runs over prime power divisors of ${\mathfrak {n}}$.

Lemma 5.13

Let $\alpha _1=\frac{p_1}{q_1}$, $\alpha _2=\frac{p_2}{q_2}$ be two cusps such that $\langle p_1,q_1\rangle =\langle p_2,q_2\rangle ={\mathcal {O}}$. Then the following are equivalent

• (1) $\alpha _2=M\alpha _1$ for some $M\in \Gamma _0({\mathfrak {n}})$;

• (2) $q_2s_1-q_1s_2\in q_1q_2{\mathcal {O}}+{\mathfrak {n}}$, where $s_i$ satisfies $p_is_i\equiv 1$ (mod $q_i$).

Proof

See [15, Lemma 1.5.1] for a more general version holding over any number field. $\square$

It follows that two cusps are equivalent relative to $\Gamma _0({\mathfrak {n}})$ if and only if the ideals generated by the denominators have the same ‘greatest common divisor’ with ${\mathfrak {n}}$, so each equivalence class of cusps is in one-to-one correspondence with each ordered decomposition ${\mathfrak {n}}={\mathfrak {M}}{\mathfrak {L}}$. Following Asai’s treatment (see [1, Sect. 1.1]) we say a cusp $\kappa _2/\kappa _1$ belongs to ${\mathfrak {L}}$-class if $\text {gcd}(\kappa _1{\mathcal {O}},{\mathfrak {n}})={\mathfrak {L}}$. For each decomposition ${\mathfrak {n}}={\mathfrak {M}}{\mathfrak {L}}$ with ${\mathfrak {M}}=M{\mathcal {O}}$ and any cusp $\kappa =\kappa _2/\kappa _1$ of ${\mathfrak {L}}$-class, we can take a typical matrix $W_\kappa$ which transforms $\kappa$ to $\infty$:

$$\begin{aligned} W_\kappa =\left( \begin{matrix}1&{}0\\ 0&{}M\end{matrix}\right) \alpha _\kappa \quad \text { with }\quad \alpha _\kappa =\left( \begin{matrix}M\lambda _1&{}\lambda _2\\ -\kappa _1&{}\kappa _2\end{matrix}\right) \in {{\,\textrm{SL}\,}}_2({\mathcal {O}}). \end{aligned}$$

(42)

As $\langle \kappa _1,\kappa _2\rangle =\langle \kappa _1,M\rangle ={\mathcal {O}}$ there exist $b,c\in {\mathcal {O}}$ such that $b\kappa _2\equiv 1$ (mod $\kappa _1$) and $cM\equiv 1$ (mod $\kappa _1$). Taking $\lambda _1=bc\in {\mathcal {O}}$ we observe that $\lambda _2=\frac{1-M\lambda _1\kappa _2}{\kappa _1}$ belongs to ${\mathcal {O}}$. So $W_\kappa$ is well-defined. It is not difficult to see that $W_\kappa$ is of type of Atkin-Lehner operator as defined in (41).

Fix a representative $\kappa _i=\kappa _{i,2}/\kappa _{i,1}\in {\mathbb {P}}^1(F)/\Gamma$ of each equivalence class of cusps corresponding to the ordered decomposition $m\sqrt{d_F}{\mathfrak {n}}={\mathfrak {M}}_i{\mathfrak {L}}_{i}$ with ${\mathfrak {M}}_i$ generated by $M_i$ and ${\mathfrak {L}}_{i}$ by $L_{i}$. Write as defined in (42)

$$\begin{aligned} W_{\kappa _i}=\left( \begin{matrix}1&{}0\\ 0&{}M_i\end{matrix}\right) \left( \begin{matrix}M_i\lambda _1&{}\lambda _2\\ -\kappa _{i,1}&{}\kappa _{i,2}\end{matrix}\right) \end{aligned}$$

which transforms $\kappa _i$ to $\infty$.

It is well known that the fractional linear transformation on the extended upper half space is composition of an even number of inversions (see e.g. [2, Sect. 2.3]). So the action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ on the subspace U preserves the orientation. By Proposition 3.2 we know that if $U({\textbf{x}}_1,{\textbf{x}}_2)\perp \nu (\infty )$ then $U(W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))\perp \nu (W_{\kappa _i}^{-1}\cdot \infty )$. We have proven that the bilinear form on a pair of vectors is preserved under the action of ${{\,\textrm{GL}\,}}_2({\mathbb {C}})$ in (3) and hence so is the Gram matrix $\beta$. Thus for $({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\infty ,+}$ we have $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)\in \Omega _{\beta ,\kappa _i,+}$. Then we obtain

$$\begin{aligned} I_{\kappa _i}&=\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}}\varphi ^{\text {new}}_f({\textbf{x}}_1,{\textbf{x}}_2)\int _{C_{U({\textbf{x}}_1,{\textbf{x}}_2)}}\eta _{\mathcal {F}}\\&=\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\infty }\backslash \Omega _{\beta ,\infty ,+}}\varphi ^{\text {new}}_f(|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))\int _{C_{U(|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))}}\eta _{\mathcal {F}}\\&=\sum _{({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\infty }\backslash \Omega _{\beta ,\infty ,+}}\varphi ^{\text {new}}_f(|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))\int _{C_{U(W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))}}\eta _{\mathcal {F}} \end{aligned}$$

where the last equality is the consequence of $U(|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))=U(W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))$.

Remark 5.14

We introduce the factor $|\det (W_{\kappa _i})|$ to make sure that for $({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\kappa _i}\backslash \Omega _{\beta ,\kappa _i,+}$, $|\det (W_{\kappa _i})|^{-1}\cdot W_{\kappa _i}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ lies in $V({\mathbb {Q}})^2$ (so in $\Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$) but not just in $V({\mathbb {R}})^2$.

Next we will analyze $\varphi _f^{\text {new}}(|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2))$ for $({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\infty }\backslash \Omega _{\beta ,\infty ,+}$. For simplicity we write $\chi =\chi _{\mathfrak {m}}$.

We begin the calculation in a slightly more general setting. Given $g=\left( \begin{matrix}x&{}y\\ z&{}w\end{matrix}\right)$ and $({\textbf{x}}_1,{\textbf{x}}_2)=\left( \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ,\left( \begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right)$, we compute

$$\begin{aligned}&\left( \left( \begin{matrix}a^{'}_1&{}b^{'}_1\\ {\bar{b}}^{'}_1&{}d^{'}_1\end{matrix}\right) ,\left( \begin{matrix}a^{'}_2&{}b^{'}_2\\ {\bar{b}}^{'}_2&{}d^{'}_2\end{matrix}\right) \right) :=|\det {g}|^{-1}\cdot g\cdot \left( \left( \begin{matrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{matrix}\right) ,\left( \begin{matrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{matrix}\right) \right) \nonumber \\ =&|\det (g)|^{-2}\left( \left( {\begin{matrix}a_1x{\bar{x}}+{\bar{b}}_1{\bar{x}}y+b_1x{\bar{y}}&{}a_1x{\bar{z}}+{\bar{b}}_1y{\bar{z}}+b_1x{\bar{w}}\\ a_1{\bar{x}}z+{\bar{b}}_1{\bar{x}}w+b_1{\bar{y}}z&{}a_1z{\bar{z}}+{\bar{b}}_1{\bar{z}}w+b_1z{\bar{w}}\end{matrix}}\right) ,\left( {\begin{matrix}a_2x{\bar{x}}+{\bar{b}}_2{\bar{x}}y+b_2x{\bar{y}}&{}a_2x{\bar{z}}+{\bar{b}}_2y{\bar{z}}+b_2x{\bar{w}}\\ a_2{\bar{x}}z+{\bar{b}}_2{\bar{x}}w+b_2{\bar{y}}z&{}a_2z{\bar{z}}+{\bar{b}}_2{\bar{z}}w+b_2z{\bar{w}}\end{matrix}}\right) \right) \end{aligned}$$

(43)

and then

$$\begin{aligned} (a^{'}_2b^{'}_1-a^{'}_1b^{'}_2)=&|\det (g)|^{-4}[(a_2x{\bar{x}}+{\bar{b}}_2{\bar{x}}y+b_2x{\bar{y}})(a_1x{\bar{z}}+{\bar{b}}_1y{\bar{z}}+b_1x{\bar{w}})\nonumber \\&-(a_1x{\bar{x}}+{\bar{b}}_1{\bar{x}}y+b_1x{\bar{y}})(a_2x{\bar{z}}+{\bar{b}}_2y{\bar{z}}+b_2x{\bar{w}})]\nonumber \\ =&|\det (g)|^{-4}[a_2b_1(x{\bar{x}}x{\bar{w}}-x{\bar{y}}x{\bar{z}})-a_1b_2(x{\bar{x}}x{\bar{w}}-x{\bar{y}}x{\bar{z}})\nonumber \\&+b_1{\bar{b}}_2(x{\bar{x}}y{\bar{w}}-x{\bar{y}}x{\bar{z}})+{\bar{b}}_1b_2(x{\bar{y}}y{\bar{z}}-{\bar{x}}yx{\bar{w}})]\nonumber \\ =&\det (g)^{-2}\det ({\bar{g}})^{-1}[(a_2b_1-a_1b_2)x^2+(b_1{\bar{b}}_2-{\bar{b}}_1b_2)xy], \end{aligned}$$

(44)

$$\begin{aligned} ({\bar{b}}_2^{'}d_1^{'}-{\bar{b}}_1^{'}d_2^{'})=&|\det (g)|^{-4}[(a_2{\bar{x}}z+{\bar{b}}_2{\bar{x}}w+b_2{\bar{y}}z)(a_1z{\bar{z}}+{\bar{b}}_1{\bar{z}}w+b_1z{\bar{w}})\nonumber \\&-(a_1{\bar{x}}z+{\bar{b}}_1{\bar{x}}w+b_1{\bar{y}}z)(a_2z{\bar{z}}+{\bar{b}}_2{\bar{z}}w+b_2z{\bar{w}})]\nonumber \\ =&|\det (g)|^{-4}[a_2b_1({\bar{x}}zz{\bar{w}}-{\bar{y}}zz{\bar{z}})-a_1b_2({\bar{x}}zz{\bar{w}}-{\bar{y}}zz{\bar{z}})]\nonumber \\&+b_1{\bar{b}}_2({\bar{x}}wz{\bar{w}}-{\bar{y}}z{\bar{z}}w)-{\bar{b}}_1b_2({\bar{x}}wz{\bar{w}}-{\bar{y}}z{\bar{z}}w)\nonumber \\ =&\det (g)^{-2}\det ({\bar{g}})^{-1}[(a_2b_1-a_1b_2)z^2+(b_1{\bar{b}}_2-{\bar{b}}_1b_2)wz]. \end{aligned}$$

(45)

Remark 5.15

With our choice of $\beta$, the pair $({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _{\infty }\backslash \Omega _{\beta ,\infty ,+}$ satisfies the condition $\dagger$ as in Remark 3.6. It means that $b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}\times \bar{{\mathfrak {q}}}{\mathcal {O}}_{\bar{{\mathfrak {q}}}}$ (i.e. $q|b_i$) for split q|m with $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$, and $b_i\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ for inert q|m with $(q)={\mathfrak {q}}$. So $b_1{\bar{b}}_2-{\bar{b}}_1b_2$ appearing in (44) and (45) turns out to be divisible by $q^2$ for each prime q|m.

Recall the ordered decomposition ${\mathfrak {f}}{\mathfrak {n}}={\mathfrak {M}}_i{\mathfrak {L}}_{i}$ (or ${\mathfrak {f}}{\mathfrak {n}}{\mathfrak {q}}_2={\mathfrak {M}}_i{\mathfrak {L}}_{i}$ with ${\mathfrak {q}}_2$ above 2 when $d\equiv 2,3{{\,\textrm{mod}\,}}4$) and its corresponding representative $\kappa _i=\frac{\kappa _{i,2}}{\kappa _{i,1}}$ of equivalence class of cusps with $\kappa _{i,1}$ and $\kappa _{i,2}$ coprime.

Lemma 5.16

For $({\textbf{x}}_1,{\textbf{x}}_2)\in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$ and non-trivial ${\mathfrak {M}}_i$ dividing ${\mathfrak {f}}{\mathfrak {n}}$ when $d\equiv 1{{\,\textrm{mod}\,}}4$ (or ${\mathfrak {f}}{\mathfrak {n}}{\mathfrak {q}}_2$ with ${\mathfrak {q}}_2$ above 2 when $d\equiv 2,3{{\,\textrm{mod}\,}}4$), we have that $\varphi ^{{{\,\textrm{new}\,}}}_f$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$.

Proof

Write

$$\begin{aligned} W_{\kappa _i}^{-1}=\left( \begin{matrix}\kappa _{i,2}&{}-\frac{\lambda _2}{M_i}\\ \kappa _{i,1}&{}\lambda _1\end{matrix}\right) =:\begin{pmatrix}x&{}y\\ z&{}w\end{pmatrix}\quad \text { with }\det (W_{\kappa _i}^{-1})=\frac{1}{M_i} \end{aligned}$$

and for $({\textbf{x}}_1,{\textbf{x}}_2)=\left( \begin{pmatrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{pmatrix},\begin{pmatrix}a_2&{}b_2\\ {\bar{b}}_2&{}0\end{pmatrix}\right) \in \Gamma _\infty \backslash \Omega _{\beta ,\infty ,+}$, set

$$\begin{aligned} ({\textbf{x}}^{'}_1,{\textbf{x}}^{'}_2)=\left( \left( \begin{matrix}a^{'}_1&{}b^{'}_1\\ {\bar{b}}^{'}_1&{}d^{'}_1\end{matrix}\right) ,\left( \begin{matrix}a^{'}_2&{}b^{'}_2\\ {\bar{b}}^{'}_2&{}d^{'}_2\end{matrix}\right) \right) :=|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot \left( \begin{pmatrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{pmatrix},\begin{pmatrix}a_1&{}b_1\\ {\bar{b}}_1&{}0\end{pmatrix}\right) . \end{aligned}$$

By (43), we have for $j=1,2$

$$\begin{aligned} a^{'}_j=&|M_i|^2(a_jx{\bar{x}}+{\bar{b}}_j{\bar{x}}y+b_jx{\bar{y}})=|M_i|^2a_j\kappa _{i,2}{\bar{\kappa }}_{i,2}-{\overline{M}}_i{\bar{b}}_j{\bar{\kappa }}_{i,2}\lambda _2-M_ib_j\kappa _{i,2}{\bar{\lambda }}_2,\\ b^{'}_j=&|M_i|^2(a_jx{\bar{z}}+{\bar{b}}_jy{\bar{z}}+b_jx{\bar{w}})=|M_i|^2a_j\kappa _{i,2}{\bar{\kappa }}_{i,1}-{\overline{M}}_i{\bar{b}}_j\lambda _2{\bar{\kappa }}_{i,1}+|M_i|^2b_j\kappa _{i,2}{\bar{\lambda }}_1,\\ {\bar{b}}^{'}_j=&|M_i|^2(a_j{\bar{x}}z+b_j{\bar{y}}z+{\bar{b}}_j{\bar{x}}w)=|M_i|^2a_j{\bar{\kappa }}_{i,2}\kappa _{i,1}-M_ib_j{\bar{\lambda }}_2\kappa _{i,1}+|M_i|^2{\bar{b}}_j{\bar{\kappa }}_{i,2}\lambda _1,\\ d^{'}_j=&|M_i|^2(a_jz{\bar{z}}+{\bar{b}}_j{\bar{z}}w+b_jz{\bar{w}})=|M_i|^2(a_j\kappa _{i,1}{\bar{\kappa }}_{i,1}+{\bar{b}}_j{\bar{\kappa }}_{i,1}\lambda _1+b_j\kappa _{i,1}{\bar{\lambda }}_1). \end{aligned}$$

• (I) Let q a prime dividing $m|d_F|$ which is split, inert or ramified. We will only treat in details the case when q is split with $(q)={\mathfrak {q}}\bar{{\mathfrak {q}}}$ and other cases can be treated similarly. We want to show that if

  $$\begin{aligned} ({\mathfrak {M}}_i,{\mathfrak {m}})={\mathfrak {q}},({\mathfrak {M}}_i,(m))=\bar{{\mathfrak {q}}}\text { or }({\mathfrak {M}}_i,{\mathfrak {m}})=(q) \end{aligned}$$

  then $\omega (1,\gamma )\varphi _q^\chi$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ for $[\gamma ]\in {\overline{\Gamma }}_0(q)/{\overline{\Gamma }}(q)$.

  • (I.1) Let $({\mathfrak {M}}_i,{\mathfrak {m}})={\mathfrak {q}}$ and then we have

    $$\begin{aligned} {\mathfrak {q}}|{\mathfrak {M}}_i,\ \bar{{\mathfrak {q}}}|\overline{{\mathfrak {M}}}_i,\ \bar{{\mathfrak {q}}}|(\kappa _{i,1}),\ {\mathfrak {q}}|({\bar{\kappa }}_{i,1}),\ \bar{{\mathfrak {q}}}\not \mid (\kappa _{i,2}),\ {\mathfrak {q}}\not \mid ({\bar{\kappa }}_{i,2}) \end{aligned}$$

    By Remark 5.15, there is no need to discuss the integrality of $b_j$ but we care for that of $a_j$. Suppose that $a_j\in {\mathbb {Z}}_q$. It is easy to observe that $a^{'}_j,b^{'}_j,{\bar{b}}^{'}_j,d^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$. Set

    $$\begin{aligned} \gamma =(\gamma _1,\gamma _2)=\left( \begin{pmatrix}u_1&{}v_1\\ 0&{}u_1^{-1}\end{pmatrix},\begin{pmatrix}u_2&{}v_2\\ 0&{}u_2^{-1}\end{pmatrix}\right) \end{aligned}$$

    with $[u_1],[u_2]\in ({\mathcal {O}}/(q))^\times$ and $[v_1],[v_2]\in {\mathcal {O}}/(q)$. We write

    $$\begin{aligned} ({\textbf{x}}^{''}_1,{\textbf{x}}^{''}_2)=\left( \begin{pmatrix}a^{''}_1&{}b^{''}_1\\ c^{''}_1&{}d^{''}_1\end{pmatrix},\begin{pmatrix}a^{''}_2&{}b^{''}_2\\ c^{''}_2&{}d^{''}_2\end{pmatrix}\right) :=(\gamma ^{-1}_1{\textbf{x}}^{'}_1{}^t(\gamma _2^{-1})^*,\gamma ^{-1}_1{\textbf{x}}^{'}_2{}^t(\gamma _2^{-1})^*) \end{aligned}$$

    and compute

    $$\begin{aligned} \begin{pmatrix}a^{''}_j&{}b^{''}_j\\ c^{''}_j&{}d^{''}_j\end{pmatrix}=&\begin{pmatrix}u_1^{-1}&{}-v_1\\ 0&{}u_1\end{pmatrix}\begin{pmatrix}a^{'}_j&{}b^{'}_j\\ {\bar{b}}^{'}_j&{}d^{'}_j\end{pmatrix}\begin{pmatrix}u_2&{}0\\ v_2&{}u_2^{-1}\end{pmatrix}\\ =&\begin{pmatrix}u_1^{-1}u_2a^{'}_j-v_1u_2{\bar{b}}^{'}_j+u_1^{-1}v_2b^{'}_j-v_1v_2d^{'}_j&{}u_1^{-1}u_2^{-1}b^{'}_j-v_1u_2^{-1}d^{'}_j\\ u_1u_2{\bar{b}}^{'}_j+u_1v_2d^{'}_j&{}u_1u_2^{-1}d^{'}_j\end{pmatrix}. \end{aligned}$$

    Then, as $a^{'}_j,b^{'}_j,{\bar{b}}^{'}_j,d^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, we can observe that $a^{''}_j,b^{''}_j,c^{''}_j,d^{''}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ as well which implies that

    $$\begin{aligned} \frac{a^{''}_2b^{''}_1-a^{''}_1b^{''}_2}{m}+\frac{c^{''}_2d_1^{'}-c^{''}_1d_2'}{m}\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}. \end{aligned}$$

    It immediately follows that $\omega (1,\gamma )\varphi _q^{\chi ,{{\,\textrm{new}\,}}}$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$. Suppose that $a_j\notin {\mathbb {Z}}_q$ and set $l_q=\text {min}\{\text {ord}_{q}(a_j)\}\le -1$. Assume that $\omega (1,\gamma )\varphi _q^{\chi ,{{\,\textrm{new}\,}}}$ is non-vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ which requires that $a^{''}_j,d^{''}_j\in {\mathbb {Z}}_q$ and $b^{''}_j,c^{''}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$.

    • We first consider $v_2\in ({\mathcal {O}}/(q))^\times$. Observing

      $$\begin{aligned} c^{''}_j=u_1u_2{\bar{b}}^{'}_j+u_1v_2d^{'}_j\quad \text {and}\quad d^{''}_j=u_1u_2^{-1}d^{'}_j, \end{aligned}$$

      we know that for $c^{''}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ and $d^{''}_j\in {\mathbb {Z}}_q$ we need $d^{'}_j\in {\mathbb {Z}}_q$ and at least ${\bar{b}}^{'}_j\in {\mathcal {O}}_{\mathfrak {q}}$. As

      $$\begin{aligned} {\bar{b}}^{'}_j=|M_i|^2a_j{\bar{\kappa }}_{i,2}\kappa _{i,1}-M_ib_j{\bar{\lambda }}_2\kappa _{i,1}+|M_i|^2{\bar{b}}_j{\bar{\kappa }}_{i,2}\lambda _1 \end{aligned}$$

      with ${\bar{\kappa }}_{i,2},\kappa _{i,1}\in {\mathcal {O}}_{\mathfrak {q}}^\times$, we then need $M_i\in {\mathfrak {q}}^{-l_q}$ which makes

      $$\begin{aligned} d^{'}_j=|M_i|^2(a_j\kappa _{i,1}{\bar{\kappa }}_{i,1}+{\bar{b}}_j{\bar{\kappa }}_{i,1}\lambda _1+b_j\kappa _{i,1}{\bar{\lambda }}_1)\quad (q|\kappa _{i,1}{\bar{\kappa }}_{i,1}) \end{aligned}$$

      lie in $q{\mathbb {Z}}_q$. Looking back to $c^{''}_j=u_1u_2{\bar{b}}^{'}_j+u_1v_2d^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, that $d^{'}_j\in q{\mathbb {Z}}_q$ makes ${\bar{b}}^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$. It follows that we need $M_i\in {\mathfrak {q}}^{-l_q+1}{\mathcal {O}}_{\mathfrak {q}}$, a contradiction to that ${\mathfrak {M}}_i$ is square-free.

    • Let $v_2=0$. Then we have

      $$\begin{aligned} \begin{pmatrix}a^{''}_j&{}b^{''}_j\\ c^{''}_j&{}d^{''}_j\end{pmatrix}=\begin{pmatrix}*&{}*\\ u_1u_2{\bar{b}}^{'}_j&{}u_1u_2^{-1}d^{'}_j\end{pmatrix}. \end{aligned}$$

      For $c^{''}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, we need

      $$\begin{aligned} {\bar{b}}^{'}_j=|M_i|^2a_j{\bar{\kappa }}_{i,2}\kappa _{i,1}-M_ib_j{\bar{\lambda }}_2\kappa _{i,1}+|M_i|^2{\bar{b}}_j{\bar{\kappa }}_{i,2}\lambda _1\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}\quad ({\bar{\kappa }}_{i,2},\kappa _{i,1}\in {\mathcal {O}}_{\mathfrak {q}}^\times ) \end{aligned}$$

      which requires $M_i\in {\mathfrak {q}}^{-l_q+1}{\mathcal {O}}_{\mathfrak {q}}$ contradicting to that ${\mathfrak {M}}_i$ is square-free.

    Therefore, when $({\mathfrak {M}}_i,{\mathfrak {m}})={\mathfrak {q}}$, we can deduce that $\omega (1,\gamma )\varphi _q^\chi$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$.

  • (I.2) When $({\mathfrak {M}}_i,{\mathfrak {m}})=\bar{{\mathfrak {q}}}$, we can prove it in the same way.

  • (I.3) Let $({\mathfrak {M}}_i,{\mathfrak {m}})=q$ and then we have $q|M_i$ and $\kappa _{i,1},{\bar{\kappa }}_{i,1}\in {\mathcal {O}}_{\mathfrak {q}}^\times$. It is clear for $a_j\in {\mathbb {Z}}_q$. Suppose that $a_j\notin {\mathbb {Z}}_q$ and set $l_q=\text {min}\{\text {ord}_{q}(a_j)\}\le -1$. First assume $v_1$ is a unit. Let $l_q=-1$. Then we have that all $a^{'}_j,d^{'}_j, b^{'}_j,{\bar{b}}^{'}_j$ are divisible by q and so are $a^{''}_j,d^{''}_j, b^{''}_j,c^{''}_j$. Let $l_q=-2$. It is clear that $d^{'}_j\in {\mathbb {Z}}_q$. Then, for $b^{''}_j,{\bar{b}}^{''}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$, we need $b^{'}_j,{\bar{b}}^{'}_j\in {\mathcal {O}}_{\mathfrak {q}}$. Expand

    $$\begin{aligned} a^{''}_2b^{''}_1-a^{''}_1b^{''}_2=&u_1^{-2}(a^{'}_2b^{'}_1-a^{'}_1b^{'}_2)+u_1^{-1}v_1(a^{'}_2d^{'}_1-a^{'}_1d^{'}_2)\\&+u_1^{-1}v_1(b^{'}_1{\bar{b}}^{'}_2-{\bar{b}}^{'}_1b^{'}_2)+v_1^2({\bar{b}}^{'}_2d^{'}_1-{\bar{b}}^{'}_1d^{'}_2) \end{aligned}$$

    and

    $$\begin{aligned} c^{''}_2d^{''}_1-c^{''}_1d^{''}_2=u_1^{2}({\bar{b}}^{'}_2d^{'}_1-{\bar{b}}^{'}_1d^{'}_2). \end{aligned}$$

    It is not hard to observe that $a^{'}_2d^{'}_1-a^{'}_1d^{'}_2\in q^2{\mathbb {Z}}_q$ and $b^{'}_1{\bar{b}}^{'}_2-{\bar{b}}^{'}_1b^{'}_2\in {\mathfrak {q}}^2{\mathcal {O}}_{\mathfrak {q}}$. By (44) and (45), we have

    $$\begin{aligned} a^{'}_2d^{'}_1-a^{'}_1d^{'}_2=M_i|M_i|^2((a_2b_1-a_1b_2)\kappa ^2_{i,2}+(b^{'}_1{\bar{b}}^{'}_2-{\bar{b}}^{'}_1b^{'}_2)\kappa ^2_{i,2}(-\lambda /M_i)) \end{aligned}$$

    and

    $$\begin{aligned} {\bar{b}}^{'}_2d^{'}_1-{\bar{b}}^{'}_1d^{'}_2=M_i|M_i|^2((a_2b_1-a_1b_2)\kappa ^2_{i,1}+(b^{'}_1{\bar{b}}^{'}_2-{\bar{b}}^{'}_1b^{'}_2)\kappa _{i,1}\lambda _1), \end{aligned}$$

    both of which lie in ${\mathfrak {q}}^2{\mathcal {O}}_{\mathfrak {q}}$. It follows again that

    $$\begin{aligned} \frac{a^{''}_2b^{''}_1-a^{''}_1b^{''}_2}{m}+\frac{c^{''}_2d_1^{'}-c^{''}_1d_2^{'}}{m}\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}. \end{aligned}$$

    For $l_q<-2$ there is no chance for $d^{''}_j\in {\mathbb {Z}}_q$ as $M_{i,1}$ is square-free. It is clear for $v_1=0$. Therefore $\omega (1,\gamma )\varphi _q^{\chi }$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ in this case.

• (II) Next we consider $\varphi _q^{\mathfrak {n}}$ for split $q|N({\mathfrak {n}})$ and omit details for inert q. As discussed in Subsect. 4.3.4, we can have $({\mathfrak {n}},(q))={\mathfrak {q}}$, $({\mathfrak {n}},(q))=\bar{{\mathfrak {q}}}$ or $({\mathfrak {n}},(q))=(q)$. Again we want to show that if $({\mathfrak {n}},(q))|{\mathfrak {M}}_i$ then $\omega (1,\gamma )\varphi _q^{\mathfrak {n}}$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$ for $[\gamma ]\in {\overline{\Gamma }}_0({\mathfrak {q}})/{\overline{\Gamma }}({\mathfrak {q}})$, ${\overline{\Gamma }}_0(\bar{{\mathfrak {q}}})/{\overline{\Gamma }}(\bar{{\mathfrak {q}}})$ or ${\overline{\Gamma }}_0(q)/{\overline{\Gamma }}(q)$ respectively. Let $a_j\in {\mathbb {Z}}_q$. Assume that $({\mathfrak {n}},(q))={\mathfrak {q}}$ and ${\mathfrak {q}}|{\mathfrak {M}}_i$. Then it is clear that ${\bar{b}}^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ and $d^{'}_j\in q{\mathbb {Z}}_q$. Expanding

  $$\begin{aligned} b^{''}_1c^{''}_2+b^{''}_2c^{''}_1=&(u_1^{-1}u_2^{-1}b^{'}_1-v_1u_2^{-1}d^{'}_1)(u_1u_2{\bar{b}}^{'}_2+u_1v_2d^{'}_2)\\&+(u_1^{-1}u_2^{-1}b^{'}_2-v_1u_2^{-1}d^{'}_2)(u_1u_2{\bar{b}}^{'}_1+u_1v_2d^{'}_1), \end{aligned}$$

  we see it is in ${\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$. So $\omega (1,\gamma )\varphi _q^{\mathfrak {n}}$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$. Also it is clear for $\bar{{\mathfrak {q}}}|{\mathfrak {M}}_i$ or $(q)|{\mathfrak {M}}_i$. Let $a_j\notin {\mathbb {Z}}_q$ and set $l_q=\text {min}\{\text {ord}_{q}(a_j)\}\le -1$. Assume that $({\mathfrak {n}},(q))={\mathfrak {q}}$ and ${\mathfrak {q}}|{\mathfrak {M}}_i$. Then we have $q\not \mid \kappa _{i,1}$. Look at

  $$\begin{aligned} d^{'}_j=|M_i|^2(a_j\kappa _{i,1}{\bar{\kappa }}_{i,1}+{\bar{b}}_j{\bar{\kappa }}_{i,1}\lambda _1+b_j\kappa _{i,1}{\bar{\lambda }}_1). \end{aligned}$$

  Then there is no chance for $d^{''}_j=d^{'}_j$ to be in $q{\mathbb {Z}}_q$ as $M_i$ is square-free. So $\omega (1,\gamma )\varphi _q^{\mathfrak {n}}$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$. This also occurs in the case that $({\mathfrak {n}},(q))=\bar{{\mathfrak {q}}}$ and $\bar{{\mathfrak {q}}}|{\mathfrak {M}}_i$. Now assume that $({\mathfrak {n}},(q))=(q)$ and $(q)|{\mathfrak {M}}_i$. If $l_q=-1$, then we have $b^{'}_j,{\bar{b}}^{'}_j\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$ which implies that $b^{''}_1c^{''}_2+b^{''}_2c^{''}_1\in {\mathfrak {q}}{\mathcal {O}}_{\mathfrak {q}}$. If $l_q\le -2$, we can observe that there is no room for $d^{''}_j=u_1u_2^{-1}d^{'}_j\in q{\mathbb {Z}}$ for square-free $M_i$. Let $q^{'}$ be another prime dividing $N({\mathfrak {n}})$ with $q^{'}={\mathfrak {q}}^{'}\bar{{\mathfrak {q}}}^{'}$. Similarly, if ${\mathfrak {q}}^{'}|{\mathfrak {M}}_{i}$, $\bar{{\mathfrak {q}}}^{'}|{\mathfrak {M}}_{i}$ or $(q)|{\mathfrak {M}}_{i}$, we can show that $\omega (1,\gamma )\varphi ^{\mathfrak {n}}_{q^{'}}$ is vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$.

• (III) To finish our proof we consider $\varphi _2$ if 2 is ramified with $(2)={\mathfrak {q}}_2^2$. Set $\gamma =\begin{pmatrix}u&{}v\\ 0&{}u^{-1}\end{pmatrix}$ with $[u]\in ({\mathcal {O}}/{\mathfrak {q}}_2)^\times$ and $[v]\in {\mathcal {O}}/{\mathfrak {q}}_2$. We write

  $$\begin{aligned} ({\textbf{x}}^{''}_1,{\textbf{x}}^{''}_2)=\left( \begin{pmatrix}{\bar{b}}^{''}_1&{}a^{''}_1\sqrt{d}\\ c^{''}_1\sqrt{d}&{}{\bar{b}}^{''}_1\end{pmatrix},\begin{pmatrix}{\bar{b}}^{''}_2&{}a^{''}_2\sqrt{d}\\ c^{''}_2\sqrt{d}&{}{\bar{b}}^{''}_2\end{pmatrix}\right) :=(\gamma {\textbf{x}}^{'}_1{\bar{\gamma }}^*,\gamma {\textbf{x}}^{'}_2{\bar{\gamma }}^*). \end{aligned}$$

  Suppose that ${\mathfrak {q}}_2|{\mathfrak {M}}_i$. If $l_q=\text {min}\{\text {ord}_{q}(a_j)\}\ge -1$, then we have $b^{'}_j\in \frac{1}{{\mathfrak {q}}_2}{\mathcal {O}}_{{\mathfrak {q}}_2}$ and then $b^{''}_j\in \frac{1}{{\mathfrak {q}}_2}{\mathcal {O}}_{{\mathfrak {q}}_2}$ as well. So $b^{''}_1{\bar{b}}^{''}_2\in \frac{1}{2}{\mathcal {O}}_{{\mathfrak {q}}_2}$ and then $b^{''}_1{\bar{b}}^{''}_2+{\bar{b}}^{''}_1b^{''}_2=2\text {Re}(b^{''}_1{\bar{b}}^{''}_2)\in {\mathbb {Z}}_2$ which makes $\varphi _2$ vanish on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$. If $l_q=\text {min}\{\text {ord}_{q}(a_j)\}\le -2$, then there is no chance for $d^{'}_j\in {\mathbb {Z}}_2$ as ${\mathfrak {M}}_i$ is square-free, and so for $c^{''}_j$. So again we have $\omega (1,\gamma )\varphi _2$ vanishing on $|\det (W_{\kappa _i})|\cdot W_{\kappa _i}^{-1}\cdot ({\textbf{x}}_1,{\textbf{x}}_2)$.

$\square$

If follows that $I_{\kappa _{i}}$ is vanishing for $\kappa _i\ne \infty$. So we have proven our main theorem:

Theorem 5.17

Suppose that $F={\mathbb {Q}}(\sqrt{d})$ is an imaginary quadratic field of class number 1 with the discriminant $d_F$ and denote its ring of integers by ${\mathcal {O}}$. Let m be a square-free product of inert or split primes, and put ${\mathfrak {m}}=m{\mathcal {O}}$ and ${\mathfrak {f}}=\sqrt{d}{\mathfrak {m}}$. Choose a quadratic Hecke character $\chi _{\mathfrak {m}}$ of conductor ${\mathfrak {f}}$. Given a square-free ideal ${\mathfrak {n}}$ coprime to $(m|d_F|)$, let ${\mathcal {F}}$ be a weight 2 Bianchi cusp form of level $\Gamma _0({\mathfrak {n}})$. Choose the Schwartz function as in Remark 5.1 and $\beta$ as in Remark 5.2. Then the Fourier coefficient of the theta lift at $\beta$ as in (25) is

$$\begin{aligned} I_\infty =\frac{\mu _\beta \lambda _{{\mathfrak {m}},{\mathfrak {n}}}L({\mathcal {F}},\chi _{\mathfrak {m}},1)}{2A(1,1,\chi _{\mathfrak {m}},1)}. \end{aligned}$$

So, if $L({\mathcal {F}},\chi _{\mathfrak {m}},1)\ne 0$, we can deduce the non-vanishing of our theta lifting as above.

6 Non-vanishing of theta lifting

Recall from [6] that a new form in $S_2(\Gamma _0({\mathfrak {n}}))$ is an eigenform for all the Hecke operators $T_{\mathfrak {p}}$ for ${\mathfrak {p}}$ not dividing ${\mathfrak {n}}$, which is not induced from in $S_2(\Gamma _0({\mathfrak {m}}))$ for any level ${\mathfrak {m}}$ properly dividing ${\mathfrak {n}}$. There is an involution J induced by the action on ${\mathbb {H}}_3$ of the matrix $\left( {\begin{matrix}\epsilon &{}0\\ 0&{}1\end{matrix}}\right)$, where $\epsilon$ generates the unit group of ${\mathcal {O}}$. The effect of J on Fourier coefficients is $c(\alpha )\rightarrow c(\epsilon \alpha )$; the involution commutes with the Hecke operators, and splits $S_2(\Gamma _0({\mathfrak {n}}))$ into two eigenspaces,

$$\begin{aligned} S_2(\Gamma _0({\mathfrak {n}}))=S_2^+(\Gamma _0({\mathfrak {n}}))\oplus S_2^-(\Gamma _0({\mathfrak {n}})). \end{aligned}$$

Newforms in $S_2^+(\Gamma _0({\mathfrak {n}}))$ were called plusforms, and their Fourier coefficients satisfy the additional condition $c(\epsilon \alpha )=c(\alpha )$ for all $\alpha \in {\mathcal {O}}$. Denote by $S_2^{{{\,\textrm{new}\,}}}(\Gamma _0({\mathfrak {n}}))$ the space of newforms in $S_2(\Gamma _0({\mathfrak {n}}))$ and by $S_2^{{{\,\textrm{new}\,}},+}(\Gamma _0({\mathfrak {n}}))$ the space of plusforms in $S_2^{{{\,\textrm{new}\,}}}(\Gamma _0({\mathfrak {n}}))$. More discussions on newforms and plusforms of weight 2 Bianchi modular forms can be found in [6].

We can choose

$$\begin{aligned} \beta =\left( -\frac{dz}{r},\frac{dr}{r},\frac{d{\bar{z}}}{r}\right) \quad \text { for }(z,r)\in {\mathbb {H}}_3 \end{aligned}$$

as a basis for the left-invariant differential forms on ${\mathbb {H}}_3$. Let ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}}}(\Gamma _0({\mathfrak {n}}))$ and recall its Mellin transform from [6, Sect. 2.5]

$$\begin{aligned} \Lambda ({\mathcal {F}},s)=\frac{(4\pi )^2}{|d_F|}\cdot \int _0^\infty t^{2s-2}{\mathcal {F}}\cdot \beta \end{aligned}$$

(46)

for ${\mathcal {F}}=({\mathcal {F}}_0,{\mathcal {F}}_1,{\mathcal {F}}_2)$.

Proposition 6.1

[6, Proposition 2.1] Let ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}(\Gamma _0({\mathfrak {n}}))$. Then

• (1) For $\text {Re}(s)>3/2$ we have

  $$\begin{aligned} \Lambda ({\mathcal {F}},s)=(2\pi )^{2-2s}|d_F|^{s-1}\Gamma (s)^2L({\mathcal {F}},s). \end{aligned}$$

  (47)

• (2) Assume that ${\mathcal {F}}$ is an eigenform for the Fricke involution $\omega _{{\mathfrak {n}}}=\begin{pmatrix}0&{}-1\\ {\mathfrak {n}}&{}0\end{pmatrix}$, i.e., ${\mathcal {F}}|\omega _{{\mathfrak {n}}}=\varepsilon _{\mathfrak {n}}{\mathcal {F}}$ with $\varepsilon _{\mathfrak {n}}=\pm 1$. Then $\Lambda ({\mathcal {F}},s)$ satisfies the functional equation

  $$\begin{aligned} \Lambda ({\mathcal {F}},s)=-\varepsilon _{\mathfrak {n}} N({\mathfrak {n}})^{1-s}\Lambda ({\mathcal {F}},2-s). \end{aligned}$$

  (48)

Put $\alpha (a)=\left( \begin{matrix}1&{}a\\ 0&{}1\end{matrix}\right)$ and $\omega _N=\omega (N)=\begin{pmatrix}0&{}-1\\ N&{}0\end{pmatrix}$. Let $\psi$ be a character of $({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times$ with conductor ${\mathfrak {m}}_\psi$. Similar to the twisted Hilbert modular forms [22, Sect. 5], the twist of ${\mathcal {F}}$ by $\psi$ can be defined as, for $m\in {\mathfrak {m}}_\psi$,

$$\begin{aligned} {\mathcal {F}}_\psi =G(\psi ^{-1},1/m)^{-1}\sum _{u\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times }\psi ^{-1}(u){\mathcal {F}}|_2\alpha (u/m) \end{aligned}$$

where $G(\psi ^{-1},1/m)^{-1}$ is the Gauss sum of $\psi ^{-1}$.

Lemma 6.2

Let ${\mathcal {F}}\in S_2(\Gamma _0({\mathfrak {n}}))$, $\psi$ a character of $({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times$, and ${\mathfrak {M}}$ the least common multiple of ${\mathfrak {n}}$, ${\mathfrak {m}}_\psi ^2$, and ${\mathfrak {m}}_\psi$. Then ${\mathcal {F}}_\psi \in S_2(\Gamma _0({\mathfrak {M}}),\psi ^2)$.

Proof

We will apply Miyake’s treatment in [16, Lemma 4.3.10] to our case without any new techniques.

Let $\gamma =\left( \begin{matrix}a&{}b\\ cM&{}d\end{matrix}\right) \in \Gamma _0({\mathfrak {M}})$ where $M\in {\mathfrak {M}}$ and put

$$\begin{aligned} \gamma '=\alpha (u/m)\gamma \alpha (d^2u/m)^{-1}, \end{aligned}$$

then $\gamma '\in \Gamma _0({\mathfrak {M}})\subset \Gamma _0({\mathfrak {n}})$. Writing $\gamma '=\left( \begin{matrix}a^{'}&{}b^{'}\\ c^{'}&{}d^{'}\end{matrix}\right)$, we have

$$\begin{aligned} {\mathcal {F}}|_2\alpha (u/m)\gamma ={\mathcal {F}}|_2\gamma '\alpha (d^2u/m)={\mathcal {F}}|_2\alpha (d^2u/m). \end{aligned}$$

Therefore

$$\begin{aligned} {\mathcal {F}}_\psi |_2\gamma&=G(\psi ^{-1},1/m)^{-1}\sum _{u\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^*}\psi ^{-1}(u){\mathcal {F}}|_2\alpha (u/m)\gamma \\&=G(\psi ^{-1},1/m)^{-1}\sum _{u\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^*}\psi ^{-1}(u){\mathcal {F}}|_2\alpha (d^2u/m)\\&=\psi (d^2)G(\psi ^{-1},1/m)^{-1}\sum _{u\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^*}\psi ^{-1}(d^2u){\mathcal {F}}|_2\alpha (d^2u/m)\\&=\psi ^2(d){\mathcal {F}}_\psi \end{aligned}$$

which implies that $f_\psi \in S_2(\Gamma _0({\mathfrak {M}}),\psi ^2)$. $\square$

Lemma 6.3

Let ${\mathcal {F}}\in S_2(\Gamma _0({\mathfrak {n}}))$ and $\psi$ a character of $({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times$. If $({\mathfrak {n}},{\mathfrak {m}}_\psi )=1$, then

$$\begin{aligned} {\mathcal {F}}_\psi |_2\omega ({\mathfrak {n}}{\mathfrak {m}}_\psi ^2)=C_\psi {\mathcal {G}}_{\psi ^{-1}} \end{aligned}$$

where ${\mathcal {G}}={\mathcal {F}}|_2\omega _{\mathfrak {n}}$ and

$$\begin{aligned} C_\psi =C_{\psi ,{\mathfrak {n}}}=\psi ({\mathfrak {n}})G(\psi )/G(\psi ^{-1}). \end{aligned}$$

Proof

We will apply Miyake’s treatment in [16, Lemma 4.3.11] to our case without any new techniques.

For $u\in {\mathcal {O}}$ prime to $m\in {\mathfrak {m}}_\psi$, take $n,v\in {\mathcal {O}}$ and $N\in {\mathfrak {n}}$ so that $nm-Nuv=1$. Then

$$\begin{aligned} \alpha (u/m)\omega (Nm^2)=m\cdot \omega (N)\left( \begin{matrix}m&{}-v\\ -uN&{}n\end{matrix}\right) \alpha (v/m). \end{aligned}$$

(49)

Since ${\mathcal {G}}={\mathcal {F}}|_2\omega _{\mathfrak {n}}$ belongs to $S_2(\Gamma _0({\mathfrak {n}}))$, (49) implies

$$\begin{aligned} {\mathcal {F}}|_2\alpha (u/m)\omega (Nm^2)={\mathcal {G}}|_2\alpha (v/m), \end{aligned}$$

so that

$$\begin{aligned} G(\psi ^{-1}){\mathcal {F}}_\psi |_2\omega (Nm^2)=&\sum _{u\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times }\psi ^{-1}(u){\mathcal {F}}|_2\alpha (u/m)\omega (Nm^2)\\ =&\sum _{v\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times }\psi (-Nv){\mathcal {G}}|_2\alpha (v/m)\\ =&\psi (-N)\sum _{v\in ({\mathcal {O}}/{\mathfrak {m}}_\psi )^\times }\psi (v){\mathcal {G}}|_2\alpha (v/m). \end{aligned}$$

Then the assertion follows immediately. $\square$

Combining Lemma 6.3 and Proposition 6.1, for the central value at $s=1$ we obtain:

Proposition 6.4

For ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}(\Gamma _0({\mathfrak {n}}))$ and $\psi$ a quadratic Hecke character, we have

$$\begin{aligned} L({\mathcal {F}}_\psi ,1)=-\varepsilon _{\mathfrak {n}}\psi ({\mathfrak {n}})L({\mathcal {F}}_\psi ,1). \end{aligned}$$

Let ${\mathfrak {n}}$, $\chi _{\mathfrak {m}}$ and ${\mathfrak {m}}$ be as in Theorem 5.17. For ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}(\Gamma _0({\mathfrak {n}}))$, it follows that for the non-vanishing of $L({\mathcal {F}},\chi _{\mathfrak {m}},1)=L({\mathcal {F}}_{\chi _{\mathfrak {m}}},1)$, we need at least $\varepsilon _{\mathfrak {n}}\chi _{\mathfrak {m}}({\mathfrak {n}})=-1$.

Lemma 6.5

Given a Bianchi modular form ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}(\Gamma _0({\mathfrak {n}}))$, there always exists a quadratic Hecke character $\chi _{\mathfrak {m}}$ of conductor ${\mathfrak {m}}$ such that $\varepsilon _{\mathfrak {n}}\chi _{\mathfrak {m}}({\mathfrak {n}})=-1$.

Proof

Assume that

$$\begin{aligned} \varepsilon _{\mathfrak {n}}\chi _{\mathfrak {m}}({\mathfrak {n}})=\prod _{\text {prime }{\mathfrak {q}}_i|{\mathfrak {n}}}\varepsilon _{{\mathfrak {q}}_i}\chi _{\mathfrak {m}}({\mathfrak {q}}_i)=-1. \end{aligned}$$

We denote, for each prime ${\mathfrak {q}}_i$ dividing ${\mathfrak {n}}$,

$$\begin{aligned} \lambda _{{\mathfrak {q}}_i}:=\chi _{\mathfrak {m}}({\mathfrak {q}}_i)\varepsilon _{{\mathfrak {q}}_i}\in \{\pm 1\}. \end{aligned}$$

(50)

Recall the Chinese Remainder Theorem in the following. Let $N=\prod _i n_i$ with the $n_i$ being pairwise coprime. Given any integer $a_i$ there exists an integer x such that $x\equiv a_i$ (mod $n_i$) for every i. To solve the system of congruences consider $N_i=N/n_i$ and then there exists integers $M_i$ such that $N_iM_i\equiv 1$ (mod $n_i$). A solution of the system of congruences is $x=\sum _ia_iN_iM_i$. The way for computing the solution can also be applied into principal ideal domains.

Recall the quadratic residue symbol from [18, Chapter V]. The quadratic residue symbol for ${\mathcal {O}}$ is defined by, for a prime ideal ${\mathfrak {p}}\subset {\mathcal {O}}$,

$$\begin{aligned} \left( \frac{\alpha }{{\mathfrak {p}}}\right) =\alpha ^{\frac{N{\mathfrak {p}}-1}{2}}\text { mod }{\mathfrak {p}}. \end{aligned}$$

It has properties completely analogous to those of classical Legendre symbol

$$\begin{aligned} \left( \frac{\alpha }{{\mathfrak {p}}}\right) =\left\{ \begin{array}{rl}0,&{}\alpha \in {\mathfrak {p}},\\ 1,&{}\alpha \notin {\mathfrak {p}}\text { and }\exists \eta \in {\mathcal {O}}:\alpha \equiv \eta ^2\text { mod }{\mathfrak {p}},\\ -1,&{}\alpha \notin {\mathfrak {p}}\text { and there is no such }\eta .\end{array}\right. \end{aligned}$$

The quadratic residue symbol can be extended to take non-prime ideals or non-zero elements as its denominator, in the same way that the Jacobi symbol extends the Legendre symbol. For $0\ne \beta \in {\mathcal {O}}$ then we define $\left( \frac{\alpha }{\beta }\right) :=\left( \frac{\alpha }{(\beta )}\right)$ where $(\beta )$ is the principal ideal generated by $\beta$. Analogous to the Jacobi symbol, this symbol is multiplicative in the top and bottom parameters.

We are interested in the quadratic reciprocity law in the case of the imaginary quadratic field $F={\mathbb {Q}}(\sqrt{d})$ with class number one (see [12, Chapter VIII]. For any $\alpha \in {\mathcal {O}}$ with odd norm we define elements $t_\alpha ,t^{'}_\alpha \in {\mathbb {Z}}/2{\mathbb {Z}}$ by

$$\begin{aligned} \alpha \equiv \sqrt{d}^{t_\alpha }(1+2\sqrt{d})^{t'_\alpha }\xi ^2 \text { mod } 4 \quad \text { for }\xi \in {\mathcal {O}}. \end{aligned}$$

Then the quadratic reciprocity law for coprime elements of odd norm is given by

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) \left( \frac{\beta }{\alpha }\right) =(-1)^T \end{aligned}$$

where

$$\begin{aligned} T\equiv \left\{ \begin{array}{ll}t_\alpha t^{'}_\beta +t^{'}_\alpha t_\beta +t_\alpha t_\beta \text { mod }2,&{}\text { if }d\equiv 1,2\text { mod }4\\ t_\alpha t^{'}_\beta +t^{'}_\alpha t_\beta \text { mod }2,&{}\text { if }d\equiv 3\text { mod }4.\end{array}\right. \end{aligned}$$

In particular, if $\alpha \equiv 1$ mod 4, we can observe that $t_\alpha =t^{'}_\alpha =0$ which implies that $T\equiv 0$ mod 2. It follows that

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) \left( \frac{\beta }{\alpha }\right) =1\quad \text { for }\alpha \equiv 1\text { mod }4. \end{aligned}$$

(51)

We want to find a quadratic character defined by the quadratic residue symbol, $\chi _{{\mathfrak {m}}}=\left( \frac{\cdot }{m\sqrt{d}}\right)$, such that $\chi _{{\mathfrak {m}}}({\mathfrak {q}}_i)=\varepsilon _{{\mathfrak {q}}_i}\lambda _{{\mathfrak {q}}_i}$ for $\lambda _{{\mathfrak {q}}_i}$ given in (50). By our assumption m is the product of inert or split primes. We can impose that $m\equiv 1$ mod 4 to get $\left( \frac{\cdot }{m}\right) =\left( \frac{m}{\cdot }\right)$ by the above quadratic reciprocity law (51). To achieve $\chi _{{\mathfrak {m}}}({\mathfrak {q}}_i)=\left( \frac{m}{{\mathfrak {q}}_i}\right) \left( \frac{{\mathfrak {q}}_i}{\sqrt{d}}\right) =\varepsilon _{{\mathfrak {q}}_i}\lambda _{{\mathfrak {q}}_i}$, we need $\left( \frac{m}{{\mathfrak {q}}_i}\right) =\left( \frac{{\mathfrak {q}}_i}{\sqrt{d}}\right) \varepsilon _{{\mathfrak {q}}_i}\lambda _{{\mathfrak {q}}_i}$ which can be done via imposing congruence conditions $(*)$ on m modulo ${\mathfrak {q}}_i$. Therefore, by the Chinese remainder theorem, there exists a m satisfying

$$\begin{aligned} \left\{ \begin{array}{l} m\equiv 1 \text { mod }4\\ \text {congruence conditions }(*)\text { on } m \text { mod }{\mathfrak {q}}_i \text { for each prime } {\mathfrak {q}}_i|{\mathfrak {n}}.\end{array}\right. \end{aligned}$$

(52)

Now we have proven this lemma. $\square$

Write $S:=\{\text {place }v:v\mid 2|d|{\mathfrak {n}}\}$. Let $\xi$ be a quadratic idelic Hecke character of conductor $M_\xi {\mathcal {O}}$ such that $M_\xi \equiv 1{{\,\textrm{mod}\,}}4$, $M_\xi \equiv m{{\,\textrm{mod}\,}}{\mathfrak {q}}_i$ for each ${\mathfrak {q}}_i|{\mathfrak {n}}$ and at v dividing $\sqrt{d}{\mathcal {O}}$ the local component $\xi _v$ is ramified with square-free conductor. Note that its conductor is coprime to $2{\mathfrak {n}}$ and divisible by $\sqrt{d}{\mathcal {O}}$, and so is its induced character $\chi _\xi$ of $({\mathcal {O}}/M_\xi {\mathcal {O}})^\times$. Also we can observe that $M_\xi$ satisfies the conditions in (52). So, by the preceding lemma there exists a $\chi _\xi$ attached to $\xi$ such that $\varepsilon _{{\mathfrak {n}}}\chi _\xi ({\mathfrak {n}})=-1$. Let $\Psi (S;\xi )$ denote the set of quadratic characters $\chi _\xi$ such that ${\widetilde{\chi }}_{\xi ,v}=\xi _v$ for all $v\in S$. Recall from [8, Theorem B(1)]

Proposition 6.6

Suppose $\pi$ is a cuspidal automorphic representation of ${{\,\textrm{GL}\,}}_2({\mathbb {A}})$ which is self-contragredient. Suppose that for some quadratic character $\chi \in \Psi (S;\xi )$ one has root number $\epsilon (\pi \otimes \chi )=1$. Then there exist infinitely many quadratic characters $\chi ^{'}\in \Psi (S;\xi )$ such that $L(\pi \otimes \chi ^{'},1)\ne 0$.

In Sect. 1.6 we have discussed the automorphic representation $\pi$ on the space of weight 2 Bianchi modular forms. Also, we have shown that there exists a $\chi _\xi \in \Psi (S;\xi )$ such that $\varepsilon _{{\mathfrak {n}}}\chi _\xi ({\mathfrak {n}})=-1$, i.e., $\epsilon (\pi \otimes \chi _\xi )=1$. So we can apply the above proposition to deduce that, for ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}$, there are infinitely many quadratic characters $\chi \in \Psi (S;\xi )$ such that $L({\mathcal {F}},\chi ,1)$ is non-vanishing.

We will explain that these infinitely many quadratic characters always include a quadratic character with square-free conductor. This is necessary since the quadratic character $\chi _{\mathfrak {m}}$ as in Theorem 5.17 has the square-free conductor ${\mathfrak {m}}$. Suppose that $\Psi (S;\xi )\ni \chi _{\mathfrak {M}}:({\mathcal {O}}/{\mathfrak {M}})^\times \rightarrow {\mathbb {C}}^\times$ is a quadratic Hecke character. Set ${\mathfrak {M}}=\prod _{\text {prime }{\mathfrak {p}}_i|{\mathfrak {M}}}{\mathfrak {p}}_i^{r_i}$ with $r_i\ge 1$. By the Chinese Remainder Theorem, we have $({\mathcal {O}}/{\mathfrak {M}})^\times \simeq \prod _{{\mathfrak {p}}_i|{\mathfrak {M}}}({\mathcal {O}}/{\mathfrak {p}}_i^{r_i})^\times$ and then can write $\chi _{\mathfrak {M}}=\prod \chi _{{\mathfrak {M}},{\mathfrak {p}}_i}$ with $\chi _{{\mathfrak {M}},{\mathfrak {p}}_i}$ defined on $({\mathcal {O}}/{\mathfrak {p}}_i^{r_i})^\times$. It is known that $({\mathcal {O}}/{\mathfrak {p}}^{r})^\times$ has cyclic order of either $p^r(p-1)$ for ${\mathfrak {p}}$ above split prime p or $p^{2(r-1)}(p^2-1)$ for ${\mathfrak {p}}$ above inert prime p. So $\chi _{{\mathfrak {M}},{\mathfrak {p}}_i}$ is induced from a character defined on $({\mathcal {O}}/{\mathfrak {p}})^\times$ which implies that $\chi _{\mathfrak {M}}$ is induced from a primitive character $\chi _{{\mathfrak {m}}_0}$ of square-free conductor ${\mathfrak {m}}_0$.

We will show that the non-vanishing of $L({\mathcal {F}},\chi _{\mathfrak {M}},1)$ is equivalent to that of $L({\mathcal {F}},\chi _{{\mathfrak {m}}_0},1)$. Write ${\mathfrak {M}}={\mathfrak {m}}_0{\mathfrak {n}}_0^2$. It is a fact that

$$\begin{aligned} L({\mathcal {F}},\chi _{\mathfrak {M}},s)=L({\mathcal {F}},\chi _{{\mathfrak {m}}_0},s)\prod _{v|{\mathfrak {n}}_0}(1-a_{\mathcal {F}}({\mathfrak {p}}_v)\chi _{{\mathfrak {m}}_0}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-s}+N({\mathfrak {p}}_v)^{1-2s}) \end{aligned}$$

where $a_{\mathcal {F}}$ denotes the Fourier coefficient of ${\mathcal {F}}$. It suffices to show the non-vanishing of

$$\begin{aligned} 1-a_{\mathcal {F}}({\mathfrak {p}}_v)\chi _{{\mathfrak {m}}_0}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-s}+N({\mathfrak {p}}_v)^{1-2s}\quad \text {at }s=1 \end{aligned}$$

which can be rewritten as the Hecke polynomial

$$\begin{aligned} (1-\alpha _{\mathcal {F}}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-s})(1-\beta _{\mathcal {F}}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-s}). \end{aligned}$$

As $|\alpha _{\mathcal {F}}({\mathfrak {p}}_v)|<N({\mathfrak {p}}_v)$ and $|\beta _{\mathcal {F}}({\mathfrak {p}}_v)|<N({\mathfrak {p}}_v)$, we can deduce the non-vanishing of

$$\begin{aligned} (1-\alpha _{\mathcal {F}}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-1})(1-\beta _{\mathcal {F}}({\mathfrak {p}}_v)N({\mathfrak {p}}_v)^{-1}). \end{aligned}$$

As ${\mathfrak {m}}_0$ is square-free, divisible by $\sqrt{d}$ and coprime to ${\mathfrak {n}}$ such that $L({\mathcal {F}},\chi _{{\mathfrak {m}}_0},1)$ is non-vanishing, following Theorem 5.17 we can deduce that

Theorem 6.7

Given a Bianchi modular form ${\mathcal {F}}\in S_2^{{{\,\textrm{new}\,}},+}$ with ${\mathfrak {n}}$ coprime to $d_F{\mathcal {O}}$, there always exists a quadratic Hecke character such that the theta lifting as in Theorem 5.17 is non-vanishing.