Relating Siegel cusp forms to Siegel–Maaß forms

Abstract

In this paper we generalize a well-known isomorphism between the space of cusp forms of weight k for a Fuchsian subgroup of the first kind $\Gamma \subset \mathrm {SL}_{2}({\mathbb {R}})$ and the space of certain Maaß forms of weight k for $\Gamma$ to an isomorphism between the space of Siegel cusp forms of weight k for a subgroup $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$, which is commensurable with the Siegel modular group $\mathrm {Sp}_{n}({\mathbb {Z}})$, and a suitable space of Siegel–Maaß forms of weight k for $\Gamma$.

1 Introduction

Let ${\mathbb {H}}:=\{z=x+i y\in {\mathbb {C}}\,\vert \,y>0\}$ denote the upper half-plane and $\Gamma \subset \mathrm {SL}_{2}({\mathbb {R}})$ a Fuchsian subgroup of the first kind, which acts by fractional linear transformations on ${\mathbb {H}}$. Let ${\mathcal {S}}_{k}(\Gamma )$ denote the space of cusp forms of weight k for $\Gamma$ and let ${\mathcal {H}}_{k}(\Gamma )$ denote the space of real-analytic automorphic forms of weight k for $\Gamma$, on which the Maaß Laplacian

$$\begin{aligned} \Delta _{k}:=y^{2}\bigg (\frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{\partial y^{2}}\bigg )-iky\frac{\partial }{\partial x} \end{aligned}$$

of weight k acts. Then, it is well-known that there is an isomorphism

$$\begin{aligned} {\mathcal {S}}_{k}(\Gamma )\cong \ker \bigg (\Delta _{k}+\frac{k}{2}\bigg (1-\frac{k}{2}\bigg )\mathrm {id}\bigg ) \end{aligned}$$

(1.1)

of ${\mathbb {C}}$-vector spaces, induced by the assignment $f\mapsto y^{k/2}f$, where the right-hand side consists of the Maaß forms in ${\mathcal {H}}_{k}(\Gamma )$ with eigenvalue $k/2(1-k/2)$ of $\Delta _{k}$. This identification of two types of automorphic forms for $\Gamma$ has various useful applications. For example, in the article [2], the isomorphism (1.1) was crucial in relating the sup-norm bound problem for cusp forms of weight k for $\Gamma$ to bounds for the heat kernel for $\Delta _{k}$ on the quotient space $\Gamma \backslash {\mathbb {H}}$.

In this paper, we attempt a generalization of the isomorphism (1.1) to the Siegel modular setting, which, to our surprise, we could not find in the literature. Letting $\mathrm {Sym}_{n}({\mathbb {C}})$ be the set of complex symmetric $(n\times n)$-matrices, we let ${\mathbb {H}}_{n}:=\{Z=X+iY\in \mathrm {Sym}_{n}({\mathbb {C}})\,\vert \,Y>0\}$ denote the Siegel upper half-space of degree $n\ge 1$ and we let $\Gamma \subset \mathrm {Sp}_{n} ({\mathbb {R}})$ denote a subgroup acting by generalized fractional linear transformations on ${\mathbb {H}}_{n}$, which is commensurable with the Siegel modular group $\mathrm {Sp}_{n}({\mathbb {Z}})$. Then, let ${\mathcal {S}}_{k}^{n}(\Gamma )$ denote the space of Siegel cusp forms of weight k and degree n for $\Gamma$ and let ${\mathcal {H}}_{k}^{n}(\Gamma )$ be the space of real-analytic automorphic forms of weight k and degree n for $\Gamma$, on which the Siegel–Maaß Laplacian

$$\begin{aligned} \Delta _{k}:={{\,\mathrm{\mathrm {tr}}\,}}\bigg (Y\bigg (\bigg (Y\frac{\partial }{\partial X}\bigg )^{t}\frac{\partial }{\partial X}+\bigg (Y\frac{\partial }{\partial Y}\bigg )^{t}\frac{\partial }{\partial Y}\bigg )-ikY\frac{\partial }{\partial X}\bigg ) \end{aligned}$$

(1.2)

of weight k acts. As the main result of this paper, we show in Corollary 5.4 the isomorphism

$$\begin{aligned} {\mathcal {S}}_{k}^{n}(\Gamma )\cong \ker \bigg (\Delta _{k}+\frac{nk}{4}(n-k+1)\mathrm {id}\bigg ), \end{aligned}$$

(1.3)

of ${\mathbb {C}}$-vector spaces, induced by the assignment $f\mapsto \det (Y)^{k/2}f$, thereby generalizing the isomorphism (1.1) to the Siegel modular setting. The right-hand side of (1.3) now consists of the Siegel–Maaß forms in ${\mathcal {H}}_{k}^{n}(\Gamma )$ with eigenvalue $nk(n-k+1)/4)$ of $\Delta _{k}$.

In case $n=1$, the isomorphism (1.1) is obtained as a by-product of the proof of the symmetry of the Maaß Laplacian $\Delta _{k}$ (see [6]). The most straightforward proof of the symmetry of $\Delta _{k}$ is obtained by constructing a suitable $\mathrm {SL}_{2}({\mathbb {R}})$-invariant differential form using the raising or the lowering operators on ${\mathbb {H}}$, and then integrating it over the quotient space $\Gamma \backslash {\mathbb {H}}$ (see [1], p. 135). Generalizations of all these operators as well as their transformation behaviour under the action of the symplectic group $\mathrm {Sp}_{n}({\mathbb {R}})$ to the Siegel modular setting have been provided by Maaß in [3]. However, in spite of all these crucial ingredients being around for a long time, we could not find in the literature a precise proof of the symmetry of the Siegel–Maaß Laplacian $\Delta _{k}$ of weight k. We provide a complete proof of the symmetry of $\Delta _{k}$ in Theorem 5.1, where we construct the appropriate $\mathrm {Sp}_{n}({\mathbb {R}})$-invariant differential form on ${\mathbb {H}}_{n}$, which, while calculationally a bit cumbersome, is conceptually a rather straightforward piecing-together of Maaß’s calculations. Our main result in Corollary 5.4 is then a consequence of Theorem 5.1.

As indicated above, the generalization of the isomorphism (1.1) will perspectively allow us, among others, to relate the sup-norm bound problem for Siegel cusp forms of weight k and degree n for $\Gamma$ to bounds for the heat kernel for the Siegel–Maaß Laplacian $\Delta _ {k}$ on the quotient space $\Gamma \backslash {\mathbb {H}}_{n}$.

This paper is organized as follows: In Sect. 2, we provide a quick summary of the basics of the Siegel upper half-space and Siegel modular forms. In the subsequent two Sects. 3 and 4, we introduce and discuss the transformation behaviour of the Maaß operators $K_{\alpha }, \Lambda _ {\beta }$, and $\Omega _{\alpha ,\beta }$ in the Siegel modular setting that is central to our analysis. This material is already present in Chapter 15 of [3], but due to sub-optimal typesetting, at places, it is hard to decipher. The transformation behaviour of $\Omega _{\alpha ,\beta }$ can also be obtained via a somewhat different analysis on ${\mathbb {H}}_{n}\times {\mathbb {R}}/2\pi {\mathbb {Z}}$ given in Chapter 8 of Maaß’s later book [4]. As the two analyses turn out to be essentially equivalent, the transformation behaviours of $K_{\alpha }$ and $\Lambda _{\beta }$ can also be derived from the alternative method in [4]. However, for the transformation behaviours of $K_{\alpha }$ and $\Lambda _{\beta }$, we stick to Maaß’s original analysis in [3] as it is more direct and we redo these calculations for the reader’s convenience, but to keep the exposition short, we refer to [4] for the transformation behaviour of $\Omega _{\alpha ,\beta }$. Finally in Sect. 5, piecing together Maaß’s results, we construct the appropriate $\mathrm {Sp}_{n}({\mathbb {R}})$-invariant differential form on ${\mathbb {H}}_{n}$ to show the symmetry of the Siegel–Maaß Laplacian $\Delta _{k}$, and then use it to show the generalization (1.3) of the isomorphism (1.1).

2 Basic notations and definitions

For $n\in {\mathbb {N}}_{>0}$ and a commutative ring R, let $\mathrm {M}_{n}(R)$ denote the set of $(n\times n)$-matrices with entries in R and $\mathrm {Sym}_{n}(R)$ the set of symmetric matrices in $\mathrm {M}_{n}(R)$. The Siegel upper half-space ${\mathbb {H}}_{n}$ of degree n is then defined by

$$\begin{aligned} {\mathbb {H}}_{n}:=\{Z=X+iY\in \mathrm {M}_{n}({\mathbb {C}})\,\vert \,X,Y\in \mathrm {Sym}_{n}({\mathbb {R}}):\,Y>0\}. \end{aligned}$$

The symplectic group $\mathrm {Sp}_{n}({\mathbb {R}})$ of degree n is defined by

$$\begin{aligned} \mathrm {Sp}_{n}({\mathbb {R}}):=\{\gamma \in \mathrm {M}_{2n}({\mathbb {R}})\,\vert \,\gamma ^{t}J_{n}\gamma =J_{n}\}, \end{aligned}$$

where $J_{n}\in \mathrm {M}_{2n}({\mathbb {R}})$ is the skew-symmetric matrix

with $\mathbb {1}_{n}$ denoting the identity matrix of $\mathrm {M}_{n}({\mathbb {R}})$. The group $\mathrm {Sp}_{n}({\mathbb {R}})$ acts by the symplectic action

$$\begin{aligned} {\mathbb {H}}_{n}\ni Z\mapsto \gamma Z=(AZ+B)(CZ+D)^{-1}\qquad \big (\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})\big ) \end{aligned}$$

(2.1)

on ${\mathbb {H}}_{n}$. Under this action $\mathrm {Im}(Z)$ transforms as

$$\begin{aligned} \mathrm {Im}(\gamma Z)=(CZ+D)^{-t}\,\mathrm {Im}(Z)(C{\overline{Z}}+D)^{-1}, \end{aligned}$$

(2.2)

which, on taking determinants on both sides, gives

$$\begin{aligned} \det (\mathrm {Im}(\gamma Z))=\frac{\det (\mathrm {Im}(Z))}{\vert \det (CZ+D)\vert ^{2}}. \end{aligned}$$

Similarly, taking matrix-differentials on both sides of the symplectic action (2.1), it is easy to see that under this action, the matrix-differential form $\mathrm {d}Z$ transforms as

$$\begin{aligned} \mathrm {d}(\gamma Z)=(CZ+D)^{-t}\,\mathrm {d}Z\,(CZ+D)^{-1}. \end{aligned}$$

(2.3)

The arclength $\mathrm {d}s^{2}_{n}$ and the volume form $\mathrm {d}\mu _{n}$ on ${\mathbb {H}}_{n}$ in terms of $Z=(z_{j,k})_{1\le j\le k\le n}\in {\mathbb {H}}_{n}$ are given by

$$\begin{aligned} \mathrm {d}s^{2}_{n}(Z)&={{\,\mathrm{\mathrm {tr}}\,}}(Y^{-1}\,\mathrm {d}Z\,Y^{-1}\,\mathrm {d}{\overline{Z}})\quad \qquad (Z=X+iY), \\ \mathrm {d}\mu _{n}(Z)&=\frac{\bigwedge \limits _{1\le j\le k\le n}\mathrm {d}x_{j,k}\wedge \mathrm {d}y_{j,k}}{\det (Y)^{n+1}}\qquad (z_{j,k}=x_{j,k}+ iy_{j,k}). \end{aligned}$$

From Eqs. (2.2) and (2.3) it is obvious that the arclength $\mathrm {d}s^{2}_{n}$ and the volume form $\mathrm {d}\mu _{n}$ on ${\mathbb {H}}_{n}$ given by the above equations are invariant under the symplectic action. Corresponding to this metric, we have the Laplace–Beltrami operator $\Delta$ on ${\mathbb {H}}_{n}$, called the Siegel Laplacian, which is also invariant under the symplectic action.

Definition 2.1

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$, i.e., the intersection $\Gamma \cap \mathrm {Sp}_{n}({\mathbb {Z}})$ is a finite index subgroup of $\Gamma$ as well as of $\mathrm {Sp}_{n}({\mathbb {Z}})$. We let $\gamma _ {j}\in \mathrm {Sp}_{n}({\mathbb {Z}})$ ($j=1,\ldots ,h$) denote a set of representatives for the left cosets of $\Gamma \cap \mathrm {Sp}_{n}({\mathbb {Z}})$ in $\mathrm {Sp}_{n}({\mathbb {Z}})$. Then, a Siegel modular form of weight k and degree n for $\Gamma$ is a function $f:{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ satisfying the following conditions:

• (i) f is holomorphic;

• (ii) $f(\gamma Z)=\det (CZ+D)^{k}f(Z)$ for all $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$;

• (iii) given $Y_{0}\in \mathrm {Sym}_{n}({\mathbb {R}})$ with $Y_{0}>0$, the quantities $\det (C_{j}Z+D_{j})^{-k}f(\gamma _{j} Z)$ are bounded in the region $\{Z=X+iY\in {\mathbb {H}}_{n}\,\vert \,Y\ge Y_{0}\}$ for the set of representatives $\gamma _{j}=\big ({\begin{matrix}A_{j}&{}B_{j}\\ C_{j}&{}D_{j}\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {Z}})$ ($j=1,\ldots ,h$).

Moreover, a Siegel modular form f as above is called a Siegel cusp form of weight k and degree n for $\Gamma$ if condition (iii) above is strengthened to the condition

• (iii’) given $Y_{0}\in \mathrm {Sym}_{n}({\mathbb {R}})$ with $Y_{0}\gg 0$, the quantities $\det (C_{j}Z+D_{j})^{-k}f(\gamma _{j} Z)$ become arbitrarily small in the region $\{Z=X+iY\in {\mathbb {H}}_{n}\,\vert \,Y\ge Y_{0}\}$ for the set of representatives $\gamma _{j}=\big ({\begin{matrix}A_{j}&{}B_{j}\\ C_ {j}&{}D_{j}\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {Z}})$ ($j=1,\ldots ,h$).

Remark 2.2

The sets of Siegel modular forms and Siegel cusp forms of weight k and degree n for $\Gamma$ have the structure of ${\mathbb {C}}$-vector spaces, which we denote by ${\mathcal {M}}_{k}^{n}(\Gamma )$ and ${\mathcal {S}}_{k}^{n}(\Gamma )$, respectively, and which turn out to be finite dimensional. Moreover, the space ${\mathcal {S}}_{k}^{n}(\Gamma )$ is equipped with the so-called Petersson inner product given by

$$\begin{aligned} \langle f,g\rangle :=\int _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{k}f(Z){\overline{g}}(Z)\,\mathrm {d}\mu _{n}(Z)\qquad (f,g\in {\mathcal {S}}_{k}^{n} (\Gamma )), \end{aligned}$$

making ${\mathcal {S}}_{k}^{n}(\Gamma )$ into a hermitian inner product space.

3 Siegel–Maaß Laplacian of weight $(\alpha ,\beta )$

In this section, we will recall from [3] various differential operators acting on smooth complex valued functions defined on ${\mathbb {H}}_ {n}$. In particular, we will define the Siegel–Maaß Laplacian of weight $(\alpha ,\beta )$, where $\alpha ,\beta \in {\mathbb {R}}$. Letting $\alpha =k/2$ and $\beta =-k/2$ will then lead us to the Siegel–Maaß Laplacian $\Delta _{k}$ mentioned in formula (1.2) in the introduction. We point out that the Siegel Laplacian $\Delta$ mentioned in the previous section and the Siegel–Maaß Laplacian $\Delta _{k}$ are related by the formula

$$\begin{aligned} \Delta _{k}=\Delta -ik{{\,\mathrm{\mathrm {tr}}\,}}\bigg (Y\frac{\partial }{\partial X}\bigg ) \end{aligned}$$

with the symmetric $(n\times n)$-matrix $\partial /\partial X$ of partial derivatives being defined below.

Given $Z=X+iY\in {\mathbb {H}}_{n}$, we start by introducing the following symmetric $(n\times n)$-matrices of partial derivatives:

$$\begin{aligned} \mathrm {(i)}&\quad \bigg (\frac{\partial }{\partial X}\bigg )_{j,k}:=\frac{1+\delta _{j,k}}{2}\frac{\partial }{\partial x_{j,k}}, \\ \mathrm {(ii)}&\quad \bigg (\frac{\partial }{\partial Y}\bigg )_{j,k}:=\frac{1+\delta _{j,k}}{2}\frac{\partial }{\partial y_{j,k}}, \\ \mathrm {(iii)}&\quad \frac{\partial }{\partial Z}:=\frac{1}{2}\bigg (\frac{\partial }{\partial X}-i\frac{\partial }{\partial Y}\bigg ), \\ \mathrm {(iv)}&\quad \frac{\partial }{\partial {\overline{Z}}}:=\frac{1}{2}\bigg (\frac{\partial }{\partial X}+i\frac{\partial }{\partial Y}\bigg ), \end{aligned}$$

where $\delta _{j,k}$ is the Kornecker delta symbol.

Definition 3.1

Following Maaß [3], we define, using the above notations, for arbitrary real numbers $\alpha ,\beta \in {\mathbb {R}}$, the following $(n \times n)$-matrices of differential operators acting on smooth complex valued functions on ${\mathbb {H}}_{n}$:

$$\begin{aligned} \mathrm {(i)}&\quad K_{\alpha }:=(Z-{\overline{Z}})\frac{\partial }{\partial Z}+\alpha \mathbb {1}_{n}, \\ \mathrm {(ii)}&\quad \Lambda _{\beta }:=(Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}-\beta \mathbb {1}_{n}, \\ \mathrm {(iii)}&\quad \Omega _{\alpha ,\beta }:=\Lambda _{\beta -(n+1)/2}K_{\alpha }+\alpha (\beta -(n+1)/2)\mathbb {1}_{n}, \\ \mathrm {(iv)}&\quad {\widetilde{\Omega }}_{\alpha ,\beta }:=K_{\alpha -(n+1)/2}\Lambda _{\beta }+\beta (\alpha -(n+1)/2)\mathbb {1}_{n}. \end{aligned}$$

Next, we want to expand $\Omega _{\alpha ,\beta }$ and ${\widetilde{\Omega }}_{\alpha ,\beta }$ in terms of $Z,\overline{ Z},\partial /\partial Z$, and $\partial /\partial {{\overline{Z}}}$. For that we need the following lemma.

Lemma 3.2

Let $C,D:{\mathbb {H}}_{n}\longrightarrow \mathrm {M}_{n}({\mathbb {C}})$ be smooth matrix valued functions depending on Z and ${\overline{Z}}$. Then, the following matrix operator identities hold:

(i) Assuming that $\partial C/\partial Z=0$ and $\partial D/\partial Z=0$, we have

$$\begin{aligned} \frac{\partial }{\partial Z}(CZ+D)^{t}=\bigg ((CZ+D)\frac{\partial }{\partial Z}\bigg )^{t}+\frac{1}{2}(n+1)C^{t}. \end{aligned}$$

(ii) Assuming that $\partial C/\partial {\overline{Z}}=0$ and $\partial D/\partial {\overline{Z}}=0$, we have

$$\begin{aligned} \frac{\partial }{\partial {\overline{Z}}}(C{\overline{Z}}+D)^{t}=\bigg ((C{\overline{Z}}+D)\frac{\partial }{\partial {\overline{Z}}}\bigg )^{t}+\frac{1}{2}(n+1)C^{t}. \end{aligned}$$

Proof

Since part (ii) follows from part (i) by conjugation, we prove only (i). Let $\Phi$ be a matrix depending on Z and ${\overline{Z}}$ such that the product $(CZ+D)^{t}\Phi$ makes sense. Then, writing the (j, k)-th entry of the matrix $\partial /\partial Z(CZ+D)^{t}\Phi$ as the sum

$$\begin{aligned} \bigg (\frac{\partial }{\partial Z}(CZ+D)^{t}\Phi \bigg )_{j,k}=\sum \limits _{l,m=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{j,l}\big ((CZ+D)^{t}_{l,m}\Phi _{m,k}\big ) \end{aligned}$$

and noting that $\partial Z/\partial z_{j,l}=(1-\delta _{j,l})E_{j,l}+E_{l,j}$, where $E_{j,k}\in \mathrm {M}_{n}({\mathbb {C}})$ is the matrix with its (j, k)-th entry being 1 and the remaining entries being 0, elementary calculations lead us to the operator identity

$$\begin{aligned} \frac{\partial }{\partial Z}(CZ+D)^{t}=\bigg ((CZ+D)\frac{\partial }{\partial Z}\bigg )^{t}+\frac{1}{2}(n+1)C^{t}, \end{aligned}$$

which is what we needed to prove. $\square$

Corollary 3.3

For $Z\in {\mathbb {H}}_{n}$, the following operator identities hold:

$$\begin{aligned} \mathrm {(i)}\quad \frac{\partial }{\partial Z}(Z-{\overline{Z}})=\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}\bigg )^{t}+\frac{1}{2}(n+1)\mathbb {1}_{n}, \\ \mathrm {(ii)}\quad \frac{\partial }{\partial {\overline{Z}}}(Z-{\overline{Z}})=\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}\bigg )^{t}-\frac{1}{2}(n+1)\mathbb {1}_{n}. \end{aligned}$$

Proof

As $\partial {\overline{Z}}/\partial Z=0$, we can choose $C=\mathbb {1}_{n}$ and $D=-{\overline{Z}}$ in Lemma 3.2 (i), from which the first claimed formula follows. The second formula follows analogously. $\square$

Using the above corollary, one can expand $\Omega _{\alpha ,\beta }$ and ${\widetilde{\Omega }}_{\alpha ,\beta }$ as

$$\begin{aligned} \Omega _{\alpha ,\beta }&=(Z-{\overline{Z}})\bigg ((Z-\overline{ Z})\frac{\partial }{\partial {\overline{Z}}}\bigg )^{t}\frac{\partial }{\partial Z}+\alpha (Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}- \beta (Z-{\overline{Z}})\frac{\partial }{\partial Z}, \\ {\widetilde{\Omega }}_{\alpha ,\beta }&=(Z-{\overline{Z}})\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}\bigg )^{t}\frac{\partial }{\partial {\overline{Z}}}+\alpha (Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}-\beta (Z-{\overline{Z}})\frac{\partial }{\partial Z}. \end{aligned}$$

Then, $\Omega _{\alpha ,\beta }$ and ${\widetilde{\Omega }}_{\alpha ,\beta }$ are related by the identity

$$\begin{aligned} {\widetilde{\Omega }}_{\alpha ,\beta }=(Z-{\overline{Z}})\big ((Z-{\overline{Z}})^{-1}\Omega _{\alpha ,\beta }\big )^{t}. \end{aligned}$$

Definition 3.4

The operator $\Delta _{\alpha ,\beta }:=-{{\,\mathrm{\mathrm {tr}}\,}}(\Omega _{\alpha ,\beta })=-{{\,\mathrm{\mathrm {tr}}\,}}({\widetilde{\Omega }}_{\alpha ,\beta })$ is called the Siegel–Maaß Laplacian of weight $(\alpha ,\beta )$.

4 Transformation behaviour of Maaß operators

Recall that the symplectic action of $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$ on the point $Z\in {\mathbb {H}}_{n}$ is given by

$$\begin{aligned} \gamma Z=(AZ+B)(CZ+D)^{-1}=(CZ+D)^{-t}(AZ+B)^{t}; \end{aligned}$$

to avoid cumbersome notation, we will use in this paper sometimes the shorthand $Z^{\gamma }:=\gamma Z$. In this section, we will then study the transformation behaviour of the Maaß operators introduced in Definition 3.1 by expressing the operators $K^{\gamma }_ {\alpha }$, $\Lambda ^{\gamma }_{\beta }$, $\Omega ^{\gamma }_{\alpha ,\beta }$ obtained by replacing $Z,{\overline{Z}}$ by $Z^{\gamma },{\overline{Z}}^{\gamma }$ in ${K}_{\alpha }$, ${\Lambda }_{\beta }$, ${\Omega }_{\alpha ,\beta }$, respectively, as they operate on smooth complex valued functions defined on ${\mathbb {H}}_{n}$.

We begin by investigating how the matrix differential operators $\partial /\partial Z$ and $\partial /\partial {\overline{Z}}$ transform under the symplectic action of $\gamma$ on Z. From equation (2.3), we know that the differential $\mathrm {d}Z$ transforms like

$$\begin{aligned} \mathrm {d}Z^{\gamma }=(CZ+D)^{-t}\,\mathrm {d}Z\,(CZ+D)^{-1}. \end{aligned}$$

Therefore, as the differential of a smooth function $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ depending only on Z is given by $\mathrm {d}\varphi ={{\,\mathrm{\mathrm {tr}}\,}}(\partial \varphi /\partial Z\,\mathrm {d}Z)$, we have

$$\begin{aligned} \frac{\partial \varphi }{\partial Z}=(CZ+D)^{-1}\frac{\partial \varphi }{\partial {Z^{\gamma }}}(C Z+D)^{-t}, \end{aligned}$$

i.e., the operator $\partial /\partial Z$ transforms as

$$\begin{aligned} \frac{\partial }{\partial {Z^{\gamma }}}=(C Z+D)\bigg ((CZ+D)\frac{\partial }{\partial Z}\bigg )^{t}. \end{aligned}$$

(4.1)

By conjugation, the operator $\partial /\partial {\overline{Z}}$ transforms as

$$\begin{aligned} \frac{\partial }{\partial {{\overline{Z}}^{\gamma }}}=(C{\overline{Z}}+D)\bigg ((C{\overline{Z}}+D)\frac{\partial }{\partial {\overline{Z}}}\bigg )^{t}. \end{aligned}$$

(4.2)

Next we need to know how to differentiate $\det (Z-{\overline{Z}})$ and $\det (CZ+D)$ with respect to Z, which we carry out in the next two lemmas.

Lemma 4.1

The matrix identity

$$\begin{aligned} \frac{\partial \det (Z-{\overline{Z}})}{\partial Z}=\det (Z-{\overline{Z}})(Z-{\overline{Z}})^{-1} \end{aligned}$$

holds.

Proof

By writing $\mathrm {Im}(Z)=Y=(y_{j,k})_{1\le j\le k\le n}\in \mathrm {Sym}_{n}({\mathbb {R}})$ as before, we obtain from equation (140) in subsection 2.8.2 of [5] the equality

$$\begin{aligned} \frac{\partial \det (Y)}{\partial y_{j,k}}=\det (Y)(2-\delta _{j,k})(Y^{-1})_{j,k}. \end{aligned}$$

Now since $((1+\delta _{j,k})/2)(2-\delta _{j,k})=1$, we have $\partial \det (Y)/\partial Y=\det (Y)Y^{-1}$, which is equivalent to the identity stated in the lemma. $\square$

Lemma 4.2

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$. Then, the matrix identity

$$\begin{aligned} \frac{\partial \det (CZ+D)}{\partial Z}=\det (CZ+D)(CZ+D)^{-1}C=\det (CZ+D)C^{t}(C Z+D)^{-t} \end{aligned}$$

holds.

Proof

By writing $Z=(z_{j,k})_{1\le j\le k\le n}\in \mathrm {Sym}_{n}({\mathbb {C}})$, we compute, using the chain rule of differentiation

$$\begin{aligned} \frac{\partial \det (CZ+D)}{\partial z_{j,k}}={{\,\mathrm{\mathrm {tr}}\,}}\bigg (\bigg (\frac{\partial \det (CZ+D)}{\partial (CZ+D)}\bigg )^{t}\frac{\partial (CZ+D)}{\partial z_{j,k}}\bigg ). \end{aligned}$$

(4.3)

By equation (49) in subsection 2.1.2 of [5], we note

$$\begin{aligned} \frac{\partial \det (CZ+D)}{\partial (CZ+D)}=\det (CZ+D)(CZ+D)^{-t}. \end{aligned}$$

(4.4)

Then, plugging (4.4) into (4.3), we arrive at

$$\begin{aligned} \frac{\partial \det (CZ+D)}{\partial z_{j,k}}&=\det (CZ+D){{\,\mathrm{\mathrm {tr}}\,}}\bigg ((CZ+D)^{-1}\frac{\partial (CZ+D)}{\partial z_{j,k}}\bigg ) \\&=\det (CZ+D)\sum \limits _{l,m=1}^{n}(CZ+D)^{-1}_{l,m}\frac{\partial (CZ+D)_{m,l}}{\partial z_{j,k}}. \end{aligned}$$

Now, entrywise partial differentiation with respect to the entries $z_{j,k}$ of Z followed by an elementary calculation with taking care of the ensuing Kronecker delta symbols leads us to the matrix identity

$$\begin{aligned} \frac{\partial \det (CZ+D)}{\partial Z}=\det (CZ+D)(CZ+D)^{-1}C=\det (CZ+D)C^{t}(CZ+D)^{-t}, \end{aligned}$$

which is what we needed to prove. $\square$

Lemmas 4.1 and 4.2 prepare the groundwork for calculating the transformation behaviour of the Maaß operators, which we undertake one by one in the subsequent three propositions.

Proposition 4.3

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$ and $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ be a smooth function. Then, the operator $K^{\gamma }_{\alpha }$ obtained by replacing $Z\in {\mathbb {H}}_{n}$ in ${K}_{\alpha }$ by $Z^ {\gamma }=\gamma Z$ is related to the operator ${K}_{\alpha }$ by the identity

$$\begin{aligned}&K^{\gamma }_{\alpha }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(C{\overline{Z}}+D)^{-t}K_{\alpha }\varphi (Z)(CZ+D)^{t}. \end{aligned}$$

Proof

From the definition of $K^{\gamma }_{\alpha }$, we have

$$\begin{aligned}&K^{\gamma }_{\alpha }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\bigg ((Z^{\gamma }-{\overline{Z}}^{\gamma })\frac{\partial }{\partial Z^{\gamma }}+\alpha \mathbb {1}_{n}\bigg )\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+ D)^{\beta }\varphi (Z). \end{aligned}$$

Then, expanding $\partial /\partial Z^{\gamma }$ by means of equation (4.1) gives

$$\begin{aligned} \nonumber&K^{\gamma }_{\alpha }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big )=\alpha \det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\mathbb {1}_{n} \\&\qquad +(C{\overline{Z}}+D)^{-t}(Z-{\overline{Z}})\frac{\partial }{\partial Z}\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big )(CZ+D)^{t}. \end{aligned}$$

(4.5)

Now, focusing on the second line of the above equality and using Lemma 4.2, we get

$$\begin{aligned}&\frac{\partial }{\partial Z}\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\bigg (\alpha \varphi (Z)C^{t}(CZ+D)^{-t}+\frac{\partial \varphi }{\partial Z}\bigg ). \end{aligned}$$

Multiplying the above equation from the left by $(Z-{\overline{Z}})$ gives

$$\begin{aligned}&(Z-{\overline{Z}})\frac{\partial }{\partial Z}\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\bigg (\alpha \varphi (Z)(ZC^{t}-{\overline{Z}}C^{t})(CZ+D)^{-t}+(Z-{\overline{Z}})\frac{\partial \varphi }{\partial Z}\bigg ). \end{aligned}$$

Now writing $(ZC^{t}-{\overline{Z}}C^{t})=(CZ+D)^{t}-(C{\overline{Z}}+D)^{t}$ and using the definition of $K_{\alpha }$ on the right-hand side of the above equation, we have

$$\begin{aligned}&(Z-{\overline{Z}})\frac{\partial }{\partial Z}\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (C Z+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta } \big (K_{\alpha }\varphi ( Z)-\alpha \varphi ( Z)(C{\overline{Z}}+D)^t(C Z+D)^{-t}\big ). \end{aligned}$$

Therefore, multiplying on the left by $(C{\overline{Z}}+D)^{-t}$ and on the right by $(CZ+D)^{t}$, we obtain

$$\begin{aligned}&(C{\overline{Z}}+D)^{-t}(Z-{\overline{Z}})\frac{\partial }{\partial Z}\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big )(CZ+D)^{t} \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\big ((C{\overline{Z}}+D)^{-t}K_{\alpha }\varphi (Z)(CZ+D)^{t}-\alpha \varphi (Z)\mathbb {1}_{n}\big ). \end{aligned}$$

Combining the last equality with Eq. (4.5), leads to the identity

$$\begin{aligned}&K^{\gamma }_{\alpha }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(C{\overline{Z}}+D)^{-t}{K}_{\alpha }\varphi (Z)(CZ+D)^{t}, \end{aligned}$$

which is what we had set out to prove. $\square$

Proposition 4.4

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$ and $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ be a smooth function. Then, the operator $\Lambda ^{\gamma }_{\beta }$ obtained by replacing $Z\in {\mathbb {H}}_{n}$ in ${\Lambda }_{\beta }$ by $Z^{\gamma }=\gamma Z$ is related to the operator ${\Lambda }_{\beta }$ by the identity

$$\begin{aligned}&\Lambda ^{\gamma }_{\beta }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(CZ+D)^{-t}{\Lambda }_{\beta }\varphi (Z)(C{\overline{Z}}+D)^{t}. \end{aligned}$$

Proof

Since ${\overline{K}}_{\beta }=-\Lambda _{\beta }$, the required identity follows from Proposition 4.3 by complex conjugation. $\square$

Proposition 4.5

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$ and $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ be a smooth function. Then, the operator $\Omega ^{\gamma }_{\alpha ,\beta }$ obtained by replacing $Z\in {\mathbb {H}}_{n}$ in ${\Omega }_{\alpha , \beta }$ by $Z^{\gamma }=\gamma Z$ is related to the operator ${\Omega }_{\alpha ,\beta }$ by the identity

$$\begin{aligned} \nonumber&\Omega ^{\gamma }_{\alpha ,\beta }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big ) \\&\qquad =\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(CZ+D)^{-t}{\Omega }_{\alpha ,\beta }\varphi (Z)(CZ+D)^{t}. \end{aligned}$$

(4.6)

Proof

The proof of this proposition can be obtained by suitably combining Propositions 4.3 and 4.4 according to the definition of the operator $\Omega _{\alpha ,\beta }$. Alternatively, the claimed transformation formula can also be found on p. 120 of [4]. $\square$

Remark 4.6

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$ and $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ be a smooth function. Taking traces on both sides of Eq. (4.6) leads to the following transformation behaviour of the Siegel–Maaß Laplacian $\Delta _{\alpha ,\beta }$

$$\begin{aligned} \Delta ^{\gamma }_{\alpha ,\beta }\big (\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)\big )=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\Delta _ {\alpha ,\beta }\varphi (Z). \end{aligned}$$

Now, if the smooth function $\varphi$ satisfies the functional equation

$$\begin{aligned} \varphi (Z^{\gamma })=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z), \end{aligned}$$

the transformation behaviour of $\Delta _{\alpha ,\beta }$ leads to the identity

$$\begin{aligned} \Delta ^{\gamma }_{\alpha ,\beta }\varphi (Z^{\gamma })=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\Delta _{\alpha ,\beta }\varphi (Z). \end{aligned}$$

Definition 4.7

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$, i.e., the intersection $\Gamma \cap \mathrm {Sp}_{n}({\mathbb {Z}})$ is a finite index subgroup of $\Gamma$ as well as of $\mathrm {Sp}_{n}({\mathbb {Z}})$. We let $\gamma _{j}\in \mathrm {Sp}_{n} ({\mathbb {Z}})$ ($j=1,\ldots ,h$) denote a set of representatives for the left cosets of $\Gamma \cap \mathrm {Sp}_{n}({\mathbb {Z}})$ in $\mathrm {Sp}_{n}({\mathbb {Z}})$. We then let ${\mathcal {V}}_{\alpha ,\beta }^{n}(\Gamma )$ denote the space of all functions $\varphi :{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ satisfying the following conditions:

• (i) $\varphi$ is real-analytic;

• (ii) $\varphi (\gamma Z)=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z)$ for all $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$;

• (iii) given $Y_{0}\in \mathrm {Sym}_{n}({\mathbb {R}})$ with $Y_{0}>0$, there exist $M\in {\mathbb {R}}_{>0}$ and $N\in {\mathbb {N}}$ such that the inequalities

  $$\begin{aligned} \vert \det (C_{j}Z+D_{j})^{-\alpha }\det (C_{j}{\overline{Z}}+D_{j})^{-\beta }\varphi (\gamma _{j} Z)\vert \le M{{\,\mathrm{\mathrm {tr}}\,}}(Y)^{N} \end{aligned}$$

  hold in the region $\{Z=X+iY\in {\mathbb {H}}_{n}\,\vert \,Y\ge Y_{0}\}$ for the set of representatives $\gamma _{j}=\big ({\begin{matrix}A_{j}&{}B_{j}\\ C_{j}&{} D_{j}\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {Z}})$ ($j=1,\ldots ,h$).

Remark 4.8

For $\varphi \in {\mathcal {V}}_{\alpha ,\beta }^{n}(\Gamma )$, we set

$$\begin{aligned} \Vert {\varphi }\Vert ^{2}:=\int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }\vert \varphi (Z)\vert ^{2}\,\mathrm {d}\mu _{n}(Z), \end{aligned}$$

whenever it is defined. In this way we obtain the Hilbert space

$$\begin{aligned} {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma ):=\big \{\varphi \in {\mathcal {V}}_{\alpha ,\beta }^{n}(\Gamma )\,\big \vert \,\Vert \varphi \Vert <\infty \big \} \end{aligned}$$

equipped with the inner product

$$\begin{aligned} \langle \varphi ,\psi \rangle =\int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }\varphi (Z){\overline{\psi }}(Z)\,\mathrm {d}\mu _{n}(Z)\qquad (\varphi , \psi \in {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )). \end{aligned}$$

We note that in order to enable $\Vert \varphi \Vert <\infty$, the exponent $N\in {\mathbb {N}}$ in part (iii) of Definition 4.7 has to be 0. Moreover, we note that Remark 4.6 shows that the Siegel–Maaß Laplacian $\Delta _{\alpha ,\beta }$ acts on the Hilbert space ${\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$.

Definition 4.9

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$. The elements of the Hilbert space ${\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ are called automorphic forms of weight $(\alpha ,\beta )$ and degree n for $\Gamma$. Moreover, if $\varphi \in {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ is an eigenform of $\Delta _{\alpha ,\beta }$, it is called a Siegel–Maaß form of weight $(\alpha , \beta )$ and degree n for $\Gamma$.

Corollary 4.10

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$ and $\varphi \in {\mathcal {H}}_{\alpha , \beta }^{n}(\Gamma )$. Then, we have for all $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$

$$\begin{aligned} \mathrm {(i)}&\quad K^{\gamma }_{\alpha }\varphi (Z^{\gamma })=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(C{\overline{Z}}+D)^{-t}K_{\alpha }\varphi (Z)(CZ+D) ^{t}, \\ \mathrm {(ii)}&\quad \Lambda ^{\gamma }_{\beta }\varphi (Z^{\gamma })=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(CZ+D)^{-t}{\Lambda }_{\beta }\varphi (Z)(C {\overline{Z}}+D)^{t}, \\ \mathrm {(iii)}&\quad \Omega ^{\gamma }_{\alpha ,\beta }\varphi (Z^{\gamma })=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }(CZ+D)^{-t}{\Omega }_{\alpha ,\beta } \varphi ( Z)(CZ+D)^{t}. \end{aligned}$$

Proof

The proof is an immediate consequence of Propositions 4.3–4.5 and the definition of the Hilbert space ${\mathcal {H}}_{\alpha , \beta }^{n}(\Gamma )$. $\square$

5 Symmetry of the Siegel–Maaß Laplacian of weight $(\alpha ,\beta )$

Let $\mathrm {d}Z:=(\mathrm {d}z_{j,k})_{1\le j,k\le n}$ denote the $(n\times n)$-matrix consisting of differential forms of degree 1 and let $[\mathrm {d}Z]:= \bigwedge _{1\le j\le k\le n}\mathrm {d}z_{j,k}$ denote the differential form of degree $n(n+1)/2$ at $Z\in {\mathbb {H}}_{n}$. We introduce an $(n\times n)$-matrix $\{\mathrm {d}Z\}$ consisting of differential forms of degree $(n(n+1)/2-1)$, namely

$$\begin{aligned} \{\mathrm {d}Z\}_{j,k}:=\frac{1+\delta _{j,k}}{2}\,\varpi _{j,k}, \end{aligned}$$

where $\varpi _{j,k}$ is defined by

$$\begin{aligned} \varpi _{j,k}:=\varepsilon _{j,k}\bigwedge \limits _{\begin{array}{c} 1\le l\le m \le n\\ (l,m)\ne (j,k) \end{array}}\mathrm {d}z_{l,m}\qquad (1\le j\le k\le n) \end{aligned}$$

in case $j\le k$ and $\varpi _{j,k}=\varpi _{k,j}$ in case $j>k$ with the sign $\varepsilon _{j,k}=\pm 1$ determined by ${\mathrm {d}}z_{j,k}\wedge \varpi _{j,k}=[{\mathrm {d}}Z]$. It is easy to see that

$$\begin{aligned} \mathrm {d}Z\wedge \{\mathrm {d}Z\}=\frac{1}{2}(n+1)[\mathrm {d}Z]\mathbb {1}_{n}. \end{aligned}$$

Let now $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {R}})$. Since we have $\mathrm {d}Z^{\gamma }=(CZ+D)^{-t} \,\mathrm {d}Z\,(CZ+D)^{-1}$ and $[\mathrm {d}Z^{\gamma }]=\det (CZ+D)^{-(n+1)}[\mathrm {d}Z]$, we derive from the relation

$$\begin{aligned} \mathrm {d}Z^{\gamma }\wedge \{\mathrm {d}Z^{\gamma }\}=\frac{1}{2}(n+1)[\mathrm {d}Z^{\gamma }]\mathbb {1}_{n} \end{aligned}$$

that the matrix $\{\mathrm {d}Z\}$ has the transformation behaviour

$$\begin{aligned} \{\mathrm {d}Z^{\gamma }\}=\det (C Z+D)^{-(n+1)}(CZ+D)\{\mathrm {d}Z\}(CZ+D)^{t}. \end{aligned}$$

Next we shall use these differential forms to show that the Siegel–Maaß Laplacian $\Delta _{\alpha ,\beta }$ acts as a symmetric operator on the Hilbert space ${\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$.

Theorem 5.1

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$ and let $\varphi ,\psi \in {\mathcal {H}}_{\alpha , \beta }^{n}(\Gamma )$ be compactly supported. Then, we have the formula

$$\begin{aligned} \langle -\Delta _{\alpha ,\beta }\varphi ,\psi \rangle =\int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\Lambda }}_ {\beta }{\overline{\psi }}(Z)\big )\,\mathrm {d}{\mu }_{n}(Z)+n\beta (\alpha -(n+1)/2)\langle \varphi ,\psi \rangle . \end{aligned}$$

In particular, this formula establishes the relation

$$\begin{aligned} \langle \Delta _{\alpha ,\beta }\varphi ,\psi \rangle =\langle \varphi ,\Delta _{\alpha ,\beta }\psi \rangle , \end{aligned}$$

which shows that the Siegel–Maaß Laplacian $\Delta _{\alpha ,\beta }$ acts as a symmetric operator on the Hilbert space ${\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$.

Proof

We start by considering the differential form

$$\begin{aligned} \omega (Z):=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{\overline{\psi }}(Z){{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\{\mathrm {d}Z\}\big )\wedge [\mathrm {d}\overline{Z}]. \end{aligned}$$

Let $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$. Then, the transformation formulas

$$\begin{aligned} \mathrm {(a})&\quad \det (Z^{\gamma }-{\overline{Z}}^{\gamma })^{\alpha +\beta -(n+1)} \\&\qquad =\det (CZ+D)^{-(\alpha +\beta -(n+1))}\det (C{\overline{Z}}+D)^{-(\alpha +\beta -(n+1))}\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}, \\ \mathrm {(b})&\quad {\overline{\psi }}(Z^{\gamma })=\det (CZ+D)^{\beta }\det (C{\overline{Z}}+D)^{\alpha }{\overline{\psi }}(Z), \\ \mathrm {(c})&\quad {{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda ^{\gamma }_{\beta }\varphi (Z^{\gamma })(Z^{\gamma }-{\overline{Z}}^{\gamma })\{\mathrm {d}Z^{\gamma }\}\big ) \\&\qquad =\det (CZ+D)^{\alpha -(n+1)}\det (C{\overline{Z}}+D)^{\beta }{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\{\mathrm {d}Z\}\big ), \\ \mathrm {(d})&\quad [\mathrm {d}{\overline{Z}}^{\gamma }]=\det (C{\overline{Z}}+D)^{-(n+1)}[\mathrm {d}{\overline{Z}}] \end{aligned}$$

show that $\omega (Z^{\gamma })=\omega (Z)$ for all $\gamma \in \Gamma$, i.e., $\omega (Z)$ is a $\Gamma$-invariant differential form on ${\mathbb {H}}_{n}$, and hence can be considered as a differential form on the quotient space $\Gamma \backslash {\mathbb {H}}_{n}$. Since the automorphic forms $\varphi ,\psi$ are real-analytic, the differential form $\omega$ is a smooth differential form. Therefore, by Stokes’ theorem, we have

$$\begin{aligned} \int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\mathrm {d}\omega (Z)=\int \limits _{\partial \Gamma \backslash {\mathbb {H}}_{n}}\omega (Z). \end{aligned}$$

As $\varphi ,\psi$ are compactly supported, the integral on the right-hand side of the above equation vanishes, which gives

$$\begin{aligned} \int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\mathrm {d}\omega (Z)=0. \end{aligned}$$

(5.1)

As we shall see, by explicitly computing $\mathrm {d}\omega (Z)$, the vanishing of the above integral will lead to the formula claimed in the theorem.

For the computation of $\mathrm {d}\omega (Z)$, we set $\rho :=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{\overline{\psi }}(Z)$, $P:=\Lambda _{\beta }\varphi (Z)$, and $Q:=(Z-{\overline{Z}})$. Then, we obtain

$$\begin{aligned} \omega (Z)=\rho {{\,\mathrm{\mathrm {tr}}\,}}(P\,Q\,\{\mathrm {d}Z\})\wedge [\mathrm {d}{\overline{Z}}]=\sum \limits _{j,k,l=1}^{n}\rho \,p_{j,k}\,q_{k,l}\,\{\mathrm {d}Z\}_{l,j}\wedge [\mathrm {d}\overline{Z}]. \end{aligned}$$

Taking exterior derivatives on both sides leads to

$$\begin{aligned} \nonumber \mathrm {d}\omega (Z)&=\sum \limits _{j,k,l=1}^{n}\frac{\partial }{\partial z_{l,j}}(\rho \,p_{j,k}\,q_{k,l})\,\mathrm {d}z_{l,j}\wedge \frac{1+\delta _{l,j}}{2}\varpi _{l,j}\wedge [\mathrm {d}\overline{Z}] \\ \nonumber&=\sum \limits _{j,k,l=1}^{n}\frac{1+\delta _{l,j}}{2}\frac{\partial }{\partial z_{l,j}}(\rho \,p_{j,k}\,q_{k,l})\,[\mathrm {d}Z]\wedge [\mathrm {d}{\overline{Z}}] \\&=\sum \limits _{j,k,l=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{l,j}(\rho \,p_{j,k}\,q_{k,l})\,[\mathrm {d}Z]\wedge [\mathrm {d}{\overline{Z}}]. \end{aligned}$$

(5.2)

Now a term by term differentiation in the last expression on the right-hand side of the above equation allows us to write it as the sum of the three traces

$$\begin{aligned} \sum \limits _{j,k,l=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{l,j}(\rho \,p_{j,k}\,q_{k,l})={{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial \rho }{\partial Z}P\,Q\bigg )+\rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial }{\partial Z}P\,Q\bigg )+\rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (P^t\frac{\partial }{\partial Z}Q\bigg ), \end{aligned}$$

(5.3)

which we calculate one by one next.

(i) We begin by considering

$$\begin{aligned} \frac{\partial \rho }{\partial Z}=\frac{\partial }{\partial Z}\big (\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{\overline{\psi }}(Z)\big ), \end{aligned}$$

which, by Lemma 4.1, calculates to

$$\begin{aligned} \frac{\partial \rho }{\partial Z}&=(\alpha +\beta -(n+1)) \det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}(Z-{\overline{Z}})^{-1}\,{\overline{\psi }}(Z)\\&\quad +\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\frac{\partial {\overline{\psi }}(Z)}{\partial Z}. \end{aligned}$$

Now multiplying both sides of the above equation on the right by $P\,Q=\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})$ and taking the trace gives

$$\begin{aligned}&{{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial \rho }{\partial Z}P\,Q\bigg )\\&\quad =\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\bigg ((\alpha +\beta -(n+1)){{\,\mathrm{\mathrm {tr}}\,}}\big ((Z-{\overline{Z}})^{-1}{\overline{\psi }}(Z)\Lambda _{\beta } \varphi (Z)(Z-{\overline{Z}})\big ) \\&\qquad +{{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial {\overline{\psi }}(Z)}{\partial Z}\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\bigg )\bigg ), \end{aligned}$$

which, upon rearranging the terms inside the traces on the right-hand side by cyclically permuting them, becomes

$$\begin{aligned} {{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial \rho }{\partial Z}P\,Q\bigg )=&\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\bigg ((\alpha +\beta -(n+1)){{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\psi }}(Z)\big ) \nonumber \\&+{{\,\mathrm{\mathrm {tr}}\,}}\bigg (\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\frac{\partial {\overline{\psi }}(Z)}{\partial Z}\bigg )\bigg ). \end{aligned}$$

(5.4)

(ii) Next, we consider the second trace

$$\begin{aligned} \rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial }{\partial Z}P\,Q\bigg )=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{\overline{\psi }}(Z){{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial }{\partial Z}\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\bigg ) \end{aligned}$$

in Eq. (5.3), which, again through rearrangement of the terms inside the trace by a cyclical permutation, takes the form

$$\begin{aligned} \rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (\frac{\partial }{\partial Z}P\,Q\bigg )=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{{\,\mathrm{\mathrm {tr}}\,}}\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}\Lambda _{\beta }\varphi (Z){\overline{\psi }}(Z)\bigg ). \end{aligned}$$

(5.5)

(iii) Finally, we consider the third trace

$$\begin{aligned} \rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (P^{t}\frac{\partial }{\partial Z}Q\bigg )=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}{\overline{\psi }}(Z){{\,\mathrm{\mathrm {tr}}\,}}\bigg (\big (\Lambda _{\beta }\varphi (Z)\big )^{t}\bigg (\frac{\partial }{\partial Z}(Z-\overline{Z})\bigg )\mathbb {1}_{n}\bigg ) \end{aligned}$$

in Eq. (5.3). By the first operator identity in Corollary 3.3, we have the matrix identity

$$\begin{aligned} \bigg (\frac{\partial }{\partial Z}(Z-{\overline{Z}})\bigg )\mathbb {1}_{n}=\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}\bigg )^{t}\mathbb {1}_{n}+\frac{1}{2}(n+1)\mathbb {1}_{n}=\frac{1}{2}(n+1) \mathbb {1}_{n}, \end{aligned}$$

which gives, upon rearrangement of the scalar quantities, the identity

$$\begin{aligned} \rho {{\,\mathrm{\mathrm {tr}}\,}}\bigg (P^{t}\frac{\partial }{\partial Z}Q\bigg )=\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\frac{1}{2}(n+1){{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\psi }}(Z)\big ). \end{aligned}$$

(5.6)

Now, adding up Eqs. (5.4)–(5.6), it follows from Eq. (5.3) that

$$\begin{aligned}&\sum \limits _{j,k,l=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{l,j}(\rho \,p_{j,k}\,q_{k,l})\\&\quad =\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\bigg ((\alpha +\beta -(n+1)/2){{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta } \varphi (Z){\overline{\psi }}(Z)\big ) \\&\qquad +{{\,\mathrm{\mathrm {tr}}\,}}\bigg (\Lambda _{\beta }\varphi (Z)(Z-{\overline{Z}})\frac{\partial {\overline{\psi }}(Z)}{\partial Z}\bigg )+{{\,\mathrm{\mathrm {tr}}\,}}\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}\Lambda _{\beta }\varphi (Z) {\overline{\psi }}(Z)\bigg )\bigg ). \end{aligned}$$

Rearranging terms on the right-hand side of the last expression, leads to

$$\begin{aligned}&\sum \limits _{j,k,l=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{l,j}(\rho \,p_{j,k}\,q_{k,l})\\&\quad =\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\bigg ({{\,\mathrm{\mathrm {tr}}\,}}\bigg (\Lambda _{\beta }\varphi (Z)\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}+\beta \mathbb {1}_{n}\bigg ){\overline{\psi }}(Z)\bigg ) \\&\quad \quad +{{\,\mathrm{\mathrm {tr}}\,}}\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial Z}+(\alpha -(n+1)/2)\mathbb {1}_{n}\bigg )\Lambda _{\beta }\varphi (Z){\overline{\psi }}(Z)\bigg ). \end{aligned}$$

Identifying the operator $(Z-{\overline{Z}})\partial /\partial Z+\beta \mathbb {1}_{n}$ on the right-hand side of the above equation as $-{\overline{\Lambda }}_{\beta }$ and the operator $(Z-{\overline{Z}})\partial /\partial Z+(\alpha -(n+1)/2)\mathbb {1}_{n}$ as $K_{\alpha -(n+1)/2}$, we can rewrite the right-hand side of the above equation as

$$\begin{aligned} \det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\Big (-{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z)\,{\overline{\Lambda }}_{\beta }{\overline{\psi }}(Z)\big )+{{\,\mathrm{\mathrm {tr}}\,}}\big (K_{\alpha -(n+1)/2}\,\Lambda _ {\beta }\varphi (Z){\overline{\psi }}(Z)\big )\Big ), \end{aligned}$$

which, by definition of ${\widetilde{\Omega }}_{\alpha ,\beta }$, is equal to

$$\begin{aligned}&\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\\&\quad \Big ({{\,\mathrm{\mathrm {tr}}\,}}\big ({\widetilde{\Omega }}_{\alpha ,\beta }-\beta (\alpha -(n+1)/2)\mathbb {1}_{n}\big )\varphi (Z){\overline{\psi }}(Z)-{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\Lambda }}_{\beta }{\overline{\psi }}(Z)\big )\Big ). \end{aligned}$$

In total, we get

$$\begin{aligned}&\sum \limits _{j,k,l=1}^{n}\bigg (\frac{\partial }{\partial Z}\bigg )_{l,j}(\rho \,p_{j,k}\,q_{k,l})\\&\quad =\det (Z-{\overline{Z}})^{\alpha +\beta -(n+1)}\Big (-\Delta _{\alpha ,\beta }\,\varphi (Z){\overline{\psi }} (Z)-{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\Lambda }}_{\beta }{\overline{\psi }}(Z)\big ) \\&\qquad -n\beta (\alpha -(n+1)/2)\,\varphi (Z){\overline{\psi }}(Z)\Big ). \end{aligned}$$

Thus, substituting $\sum \limits _{j,k,l=1}^{n}(\partial /\partial Z)_{l,j}(\rho \,p_{j,k}\,q_{k,l})$ back into equation (5.2), we arrive at

$$\begin{aligned} \mathrm {d}\omega (Z)&=\det (Z-{\overline{Z}})^{\alpha +\beta }\Big (-\Delta _{\alpha ,\beta }\,\varphi (Z){\overline{\psi }}(Z)-{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z)\overline{\Lambda }_{\beta }{\overline{\psi }}(Z)\big ) \\&\quad -n\beta (\alpha -(n+1)/2)\,\varphi (Z){\overline{\psi }}(Z)\Big )\frac{[\mathrm {d}Z]\wedge [\mathrm {d}{\overline{Z}}]}{\det (Z-{\overline{Z}})^{n+1}}. \end{aligned}$$

Now, noting that the volume form

$$\begin{aligned} \det (Z-{\overline{Z}})^{\alpha +\beta }\frac{[\mathrm {d}Z]\wedge [\mathrm {d}{\overline{Z}}]}{\det (Z-{\overline{Z}})^{n+1}} \end{aligned}$$

is just a constant multiple of $\det (Y)^{\alpha +\beta }\mathrm {d}\mu _{n}(Z)$, it follows readily from the vanishing result (5.1) that

$$\begin{aligned}&\langle -\Delta _{\alpha ,\beta }\varphi ,\psi \rangle \\&\quad =\int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }{{\,\mathrm{\mathrm {tr}}\,}}\big (\Lambda _{\beta }\varphi (Z){\overline{\Lambda }}_ {\beta }{\overline{\psi }}(Z)\big )\,\mathrm {d}{\mu }_{n}(Z)+n\beta (\alpha -(n+1)/2)\langle \varphi ,\psi \rangle , \end{aligned}$$

which is the claimed formula.

Using the latter formula, we compute

which proves the claimed symmetry of the Siegel–Maaß Laplacian $\Delta _{\alpha ,\beta }$. $\square$

Corollary 5.2

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$ and let $\varphi \in {\mathcal {H}}_{\alpha , \beta }^{n}(\Gamma )$ be a Siegel–Maaß form of weight $(\alpha ,\beta )$ and degree n for $\Gamma$. Then, if $\varphi$ is an eigenform of $\Delta _{\alpha , \beta }$ with eigenvalue $\lambda$, we have $\lambda \in {\mathbb {R}}$ and $\lambda \ge n\beta (\alpha -(n+1)/2)$.

Furthermore, $\varphi$ has eigenvalue $\lambda =\beta (\alpha -(n+1)/2)$ if and only if $\varphi (Z)=\det (Y)^{-\beta }f(Z)$, where $f:{\mathbb {H}}_{n}\longrightarrow {\mathbb {C}}$ is a holomorphic function satisfying

$$\begin{aligned} f(\gamma Z)=\det (CZ+D)^{\alpha -\beta } f(Z) \end{aligned}$$

for all $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$. Moreover, if $\beta <0$, then f is a Siegel cusp form of weight $\alpha - \beta$ and degree n for $\Gamma$.

Proof

Since $\varphi \in {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ is an eigenform of $\Delta _{\alpha ,\beta }$ with eigenvalue $\lambda$, i.e., we have $(\Delta _{\alpha ,\beta }+ \lambda \,\mathrm {id})\varphi =0$, we compute using Theorem 5.1

$$\begin{aligned} \lambda \langle \varphi ,\varphi \rangle&=\langle -\Delta _{\alpha ,\beta }\varphi ,\varphi \rangle \\&=\int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }{{\,\mathrm{\mathrm {tr}}\,}}\big (\vert \Lambda _{\beta }\varphi (Z)\vert ^{2}\big )\,\mathrm {d}\mu _{n}(Z)+n\beta (\alpha -(n+1)/2) \langle \varphi ,\varphi \rangle . \end{aligned}$$

This immediately implies that $\lambda \in {\mathbb {R}}$. Furthermore, since ${{\,\mathrm{\mathrm {tr}}\,}}(\vert \Lambda _{\beta }\varphi (Z)\vert ^{2})\ge 0$, we conclude that

$$\begin{aligned} \lambda \ge n\beta (\alpha -(n+1)/2). \end{aligned}$$

To prove the second part of the corollary, we observe that the above equation shows that the equality $\lambda =n\beta (\alpha -(n+1)/2)$ is equivalent to

$$\begin{aligned} \int \limits _{\Gamma \backslash {\mathbb {H}}_{n}}\det (Y)^{\alpha +\beta }{{\,\mathrm{\mathrm {tr}}\,}}\big (\vert \Lambda _{\beta }\varphi (Z)\vert ^{2}\big )\,\mathrm {d}\mu _{n}(Z)=0. \end{aligned}$$

Since ${{\,\mathrm{\mathrm {tr}}\,}}(\vert \Lambda _{\beta }\varphi (Z)\vert ^{2})\ge 0$, the above integral vanishes if and only if ${{\,\mathrm{\mathrm {tr}}\,}}(\vert \Lambda _{\beta }\varphi (Z)\vert ^{2})=0$. Now, as the matrix

$$\begin{aligned} \Lambda _{\beta }\varphi (Z)=(Z-{\overline{Z}})\frac{\partial \varphi }{\partial {\overline{Z}}}-\beta \varphi (Z)\mathbb {1}_{n} \end{aligned}$$

is similar to the complex symmetric matrix

$$\begin{aligned} S(Z):=2i\,Y^{1/2}\frac{\partial \varphi }{\partial {\overline{Z}}}Y^{1/2}-\beta \varphi (Z)\mathbb {1}_{n}, \end{aligned}$$

as we have the relation $\Lambda _{\beta }\varphi (Z)=Y^{1/2}S(Z)Y^{-1/2}$, the matrix $\vert \Lambda _{\beta }\varphi (Z)\vert ^{2}$ becomes similar to the positive semidefinite hermitian matrix $S(Z){\overline{S}}(Z)$, which is diagonalizable with non-negative real eigenvalues. Therefore, the condition ${{\,\mathrm{\mathrm {tr}}\,}}(S(Z){\overline{S}}(Z))= {{\,\mathrm{\mathrm {tr}}\,}}(\vert \Lambda _{\beta }\varphi (Z)\vert ^{2})=0$ is equivalent to the vanishing of all the eigenvalues of $S(Z){\overline{S}}(Z)$, which is equivalent to the vanishing of S(Z) and hence of $\Lambda _{\beta }\varphi (Z)$. All in all, this proves that the equality $\lambda =n\beta (\alpha -(n+1)/2)$ is equivalent to the vanishing condition $\Lambda _{\beta }\varphi =0$.

Continuing, we now set $f(Z):=\det (Y)^{\beta }\varphi (Z)$, and compute

$$\begin{aligned} \frac{\partial f}{\partial {\overline{Z}}}=\beta \det (Y)^{\beta -1}\frac{\partial \det (Y)}{\partial {\overline{Z}}}\varphi (Z)+\det (Y)^{\beta }\frac{\partial \varphi }{\partial {\overline{Z}}}. \end{aligned}$$

Since we have

$$\begin{aligned} \frac{\partial \det (Y)}{\partial {\overline{Z}}}=\frac{1}{2}\bigg (\frac{\partial }{\partial X}+i\frac{\partial }{\partial Y}\bigg )\det (Y)=\frac{i}{2}\frac{\partial \det (Y)}{\partial Y}=\frac{i}{2}\det (Y)Y^{-1}, \end{aligned}$$

the above equality becomes

$$\begin{aligned} \frac{\partial f}{\partial {\overline{Z}}}&=\frac{i\beta }{2}\det (Y)^{\beta }Y^{-1}\varphi (Z)+\det (Y)^{\beta }\frac{\partial \varphi }{\partial {\overline{Z}}} \\&=-\frac{i}{2}\det (Y)^{\beta }Y^{-1}\bigg (-\beta \varphi (Z)\mathbb {1}_{n}+2iY\frac{\partial \varphi }{\partial {\overline{Z}}}\bigg ) \\&=-\frac{i}{2}\det (Y)^{\beta }Y^{-1}\bigg ((Z-{\overline{Z}})\frac{\partial \varphi }{\partial {\overline{Z}}}-\beta \varphi (Z)\mathbb {1}_{n}\bigg ) \\&=-\frac{i}{2}\det (Y)^{\beta }Y^{-1}\Lambda _{\beta }\varphi (Z). \end{aligned}$$

In total, this shows that $\partial f/\partial {\overline{Z}}=0$, i.e., the function f is holomorphic, if and only if $\Lambda _{\beta }\varphi (Z)=0$, which, by the previous argument, is equivalent to $\varphi \in {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ being a Siegel–Maaß form with eigenvalue $\lambda =\beta (\alpha -(n+1)/2)$.

Furthermore, as the function $\varphi \in {\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ has the transformation behaviour

$$\begin{aligned} \varphi (\gamma Z)=\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z) \end{aligned}$$

for all $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$, the function $f(Z)=\det (Y)^{\beta }\varphi (Z)=\det ({{\,\mathrm{Im}\,}}(Z))^{\beta }\varphi (Z)$ has the transformation behaviour

$$\begin{aligned} f(\gamma Z)&=\det ({{\,\mathrm{Im}\,}}(\gamma Z))^{\beta }\varphi (\gamma Z) \\&=\bigg (\frac{\det ({{\,\mathrm{Im}\,}}(Z))}{\vert \det (CZ+D)\vert ^{2}}\bigg )^{\beta }\det (CZ+D)^{\alpha }\det (C{\overline{Z}}+D)^{\beta }\varphi (Z) \\&=\det (CZ+D)^{\alpha -\beta }\det ({{\,\mathrm{Im}\,}}(Z))^{\beta }\varphi (Z) \\&=\det (CZ+D)^{\alpha -\beta }f(Z), \end{aligned}$$

as claimed.

Finally, letting $\gamma _{j}=\big ({\begin{matrix}A_{j}&{}B_{j}\\ C_{j}&{}D_{j}\end{matrix}}\big )\in \mathrm {Sp}_{n}({\mathbb {Z}})$ ($j=1,\ldots ,h$) be a set of representatives for the left cosets of $\Gamma \cap \mathrm {Sp}_{n}({\mathbb {Z}})$ in $\mathrm {Sp}_{n}({\mathbb {Z}})$, Remark 4.8 shows that given $Y_{0}\in \mathrm {Sym}_{n}({\mathbb {R}})$ with $Y_{0}>0$, the quantities

$$\begin{aligned} \vert \det (C_{j}Z+D_{j})^{-\alpha }\det (C_{j}{\overline{Z}}+D_{j})^{-\beta }\varphi (\gamma _{j} Z)\vert \end{aligned}$$

have to be bounded in the region $\{Z=X+iY\in {\mathbb {H}}_{n}\,\vert \,Y\ge Y_{0}\}$. Therefore, if $\beta <0$, this implies that given $Y_{0}\in \mathrm {Sym}_{n} ({\mathbb {R}})$ with $Y_{0}\gg 0$, the quantities

$$\begin{aligned}&\vert \det (C_{j}Z+D_{j})^{-(\alpha -\beta )}f(\gamma _{j} Z)\vert \\&\quad =\vert \det (C_{j}Z+D_{j})^{-\alpha }\det (C_{j}{\overline{Z}}+D_{j})^{-\beta }\det ({{\,\mathrm{Im}\,}}(\gamma _{j} Z))^{\beta } \varphi (\gamma _{j} Z)\vert \end{aligned}$$

will become arbitrarily small in the region $\{Z=X+iY\in {\mathbb {H}}_{n}\,\vert \,Y\ge Y_{0}\}$. In other words, f is indeed a Siegel cusp form of weight $\alpha - \beta$ and degree n for $\Gamma$.

With all this, the proof of the corollary is complete. $\square$

Remark 5.3

For $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$ and $\alpha =k/2$, $\beta =-k/2$ with $k\in {\mathbb {N}}_{>0}$, we denote the Hilbert space ${\mathcal {H}}_{\alpha ,\beta }^{n}(\Gamma )$ simply by ${\mathcal {H}}_{k}^{n}(\Gamma )$. Similarly, we write for the operator $\Omega _{\alpha ,\beta }$ simply $\Omega _{k}$, which becomes

$$\begin{aligned} \Omega _{k}&=(Z-{\overline{Z}})\bigg ((Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}\bigg )^{t}\frac{\partial }{\partial Z}+\frac{k}{2}(Z-{\overline{Z}})\frac{\partial }{\partial {\overline{Z}}}+\frac{k}{2}(Z-{\overline{Z}}) \frac{\partial }{\partial Z} \\&=-Y\bigg (\bigg (Y\frac{\partial }{\partial X}\bigg )^{t}\frac{\partial }{\partial X}+\bigg (Y\frac{\partial }{\partial Y}\bigg )^{t}\frac{\partial }{\partial Y}\bigg )+ikY\frac{\partial }{\partial X}. \end{aligned}$$

Finally, we write for the operator $\Delta _{\alpha ,\beta }$ simply $\Delta _{k}$ and call it the Siegel–Maaß Laplacian of weight k; it is given as

$$\begin{aligned} \Delta _{k}={{\,\mathrm{\mathrm {tr}}\,}}\bigg (Y\bigg (\bigg (Y\frac{\partial }{\partial X}\bigg )^{t}\frac{\partial }{\partial X}+\bigg (Y\frac{\partial }{\partial Y}\bigg )^{t}\frac{\partial }{\partial Y}\bigg )-ikY\frac{\partial }{\partial X}\bigg ). \end{aligned}$$

We note that the transformation behaviour of a Siegel–Maaß form $\varphi$ of weight k and degree n for $\Gamma$ takes the form

$$\begin{aligned} \varphi (\gamma Z)=\bigg (\frac{\det (CZ+D)}{\det (C{\overline{Z}}+D)}\bigg )^{k/2}\varphi (Z), \end{aligned}$$

where $\gamma =\big ({\begin{matrix}A&{}B\\ C&{}D\end{matrix}}\big )\in \Gamma$.

In the last corollary, we summarize the main results about Siegel–Maaß forms of weight k and degree n for $\Gamma$.

Corollary 5.4

Let $\Gamma \subset \mathrm {Sp}_{n}({\mathbb {R}})$ be a subgroup commensurable with $\mathrm {Sp}_{n}({\mathbb {Z}})$ and let $\varphi \in {\mathcal {H}}_{k}^{n} (\Gamma )$ be a Siegel–Maaß form of weight k and degree n for $\Gamma$. Then, if $\varphi$ is an eigenform of $\Delta _{k}$ with eigenvalue $\lambda$, we have $\lambda \in {\mathbb {R}}$ and

$$\begin{aligned} \lambda \ge \frac{nk}{4}(n-k+1), \end{aligned}$$

with equality attained if and only if the function $\varphi$ is of the form $\varphi (Z)=\det (Y)^{k/2}f(Z)$ for some Siegel cusp form $f\in {\mathcal {S}}_{k}^{n}(\Gamma )$ of weight k and degree n for $\Gamma$. In other words, there is an isomorphism

$$\begin{aligned} {\mathcal {S}}_{k}^{n}(\Gamma )\cong \ker \bigg (\Delta _{k}+\frac{nk}{4}(n-k+1)\mathrm {id}\bigg ) \end{aligned}$$

of ${\mathbb {C}}$-vector spaces, induced by the assignment $f\mapsto \det (Y)^{k/2}f$.

Proof

The proof is an immediate consequence of Corollray 5.2 by setting $\alpha =k/2$ and $\beta =-k/2$. $\square$