# Estimates for the Fourier coefficients of the Duke-Imamoglu-Ikeda lift

Tamotsu IKEDA  Graduate school of mathematics, Kyoto University,
Kitashirakawa, Kyoto, 606-8502, Japan<EMAIL_ADDRESS>and  Hidenori
KATSURADA  Department of Mathematics, Hokkaido, University, Kita 10, Nishi 8,
Kitaku, Sapporo, Hokkaido, 060-0810, Japan, and Muroran Institute of
Technology 27-1 Mizumoto, Muroran, 050-8585, Japan<EMAIL_ADDRESS>it.ac.jp

(Date: July 5, 2022)

###### Abstract.

Let $k$ and $n$ be positive even integers. For a Hecke eigenform $h$ in the
Kohnen plus subspace of weight $k-n/2+1/2$ for $\varGamma_{0}(4)$, let
$I_{n}(h)$ be the Duke-Imamoglu-Ikeda lift of $h$ to the space of cusp forms
of weight $k$ for $Sp_{n}({\mathbb{Z}})$. We then give an estimate of the
Fourier coefficients of $I_{n}(h)$. It is better than the usual Hecke bound
for the Fourier coefficients of a Siegel cusp form.

###### Key words and phrases:

Fourier coefficient, Duke-Imamoglu-Ikeda lift, Siegel series

###### 2020 Mathematics Subject Classification:

11F30, 11F46

The research was partially supported by JSPS KAKENHI Grant Number 17H02834,
16H03919, 22K03228 and 21K03152.

##  1\. Introduction

Let $\varGamma^{(n)}=Sp_{n}({\mathbb{Z}})$ be the Siegel modular group of
genus $n$, and $S_{k}(\varGamma^{(n)})$ the space of cusp forms of weight $k$
for $\varGamma^{(n)}$. Then we have the following Fourier expansion:

$F(Z)=\sum_{B}c_{F}(B)\exp(2\pi\sqrt{-1}\mathrm{tr}(BZ)),$

where $B$ runs over all positive definite half-integral matrices of size $n$.
It is an interesting problem to estimate $c_{F}(B)$. By the standard method we
obtain the following estimate, called the Hecke bound, of $c_{F}(B)$ for
$B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$:

$c_{F}(B)\ll_{F}(\det(B))^{k/2}.$

However it is weak in general. In the case $n=1$, we have Deligne’s estimate
(cf. [7]), which is best possible, and from now on we consider the case $n\geq
2$. Then there are several improvements of the Hecke bound (cf. [1], [2], [3],
[4], [6],[12],[14],[16]). Among others, Böcherer and Kohnen [1] proved that

($*$) $\displaystyle c_{F}(B)\ll_{\varepsilon,F}\det(B)^{{k\over 2}-{1\over
2n}-(1-{1\over n})\alpha_{n}+\varepsilon}\ (\varepsilon>0)$

if $k>n+1$. Here

$\alpha_{n}=\Big{(}4(n-1)+4\Big{[}{n-1\over 2}\Big{]}+{2\over
n+2}\Big{)}^{-1}.$

In this paper, we improve this bound for the Duke-Imamoglu-Ikeda lift
$I_{n}(h)$ of a cuspidal Hecke eigenform $h$ in
$S_{k-n/2+1/2}^{+}(\varGamma_{0}(4))$ to $S_{k}(\varGamma^{(n)})$. (As for a
precise definition of the Duke-Imamoglu-Ikeda lift, see Section 3.) That is,
we prove the following estimate:

(Theorem 3.1) We have

(**) $\displaystyle
c_{I_{n}(h)}(B)\ll_{\varepsilon,I_{n}(h)}{\mathfrak{d}}_{B}^{-n/4+5/12}(\det(2B))^{(k-1)/2+\varepsilon}\
(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$, where ${\mathfrak{d}}_{B}$
is the discriminant of ${\mathbb{Q}}(\sqrt{(-1)^{n/2}\det B})/{\mathbb{Q}}$.

From the above result, we have

(***) $\displaystyle
c_{I_{n}(h)}(B)|\ll_{\varepsilon,I_{n}(h)}(\det(2B))^{(k-1)/2+\varepsilon}\
(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. We note that our estimate
is slightly stronger than (*). We explain how to obtain the estimate (**). By
definition, $c_{I_{n}(h)}(B)$ is expressed in terms of the
$|{\mathfrak{d}}_{B}|$-th Fourier coefficient $c_{h}(|{\mathfrak{d}}_{B}|)$ of
$h$, and $\prod_{p}\widetilde{F}_{p}(B,\alpha_{p})$, where for a prime number
$p$, $\widetilde{F}_{p}(B,X)$ is the polynomial in $X$ and $X^{-1}$ defined in
[10], and $\alpha_{p}$ is a certain complex number such that $|\alpha_{p}|=1$
(cf. Section 3). In view of Corollary 5.9, Theorem 5.10, we can estimate
$\widetilde{F}_{p}(B,a)$ for any $a\in{\mathbb{C}}$ in purely combinatorial
method (cf. Theorem 5.11), and therefore we obtain the following estimate (cf.
Theorem 6.4 (1)):

$\displaystyle|c_{I_{n}(h)}(B)|\leq|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{f}}_{B}^{k-1}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})),$

where ${\mathfrak{f}}_{B}=\sqrt{\det(2B)/|{\mathfrak{d}}_{B}|}$, and
${\mathfrak{e}}_{i,B}^{(p)}$ is that defined after Remark 6.2. On the other
hand, by [5], we obtain a reasonable estimate of
$c_{h}(|{\mathfrak{d}}_{B}|)$. Combining these two estimates, we obtain the
estimate (**). We also obtain another estimate for $c_{I_{n}(h)}(B)$ (cf.
Theorem 3.2).

It is expected that we can obtain a similar estimate for the Fourier
coefficient of the lift constructed in [11].

This paper is organized as follows. In Section 2, we review the Siegel series.
In Section 3, we state our main result. In Section 4, we review the Gross-
Keating invariant. In Section 5, we give an estimate of
$\widetilde{F}_{p}(B,\alpha_{p})$, and in Section 6, we prove our main result.

We thank Valentin Blomer for many valuable discussions, by which this paper is
motivated. We also thank him for many useful comments, by which our main
result has been improved greatly.

Notation Let $R$ be a commutative ring. We denote by $R^{\times}$ the group of
units in $R$. We denote by $M_{mn}(R)$ the set of $(m,n)$ matrices with
entries in $R$, and especially write $M_{n}(R)=M_{nn}(R)$. We often identify
an element $a$ of $R$ and the matrix $(a)$ of degree 1 whose component is $a$.
If $m$ or $n$ is 0, we understand an element of $M_{mn}(R)$ is the empty
matrix and denote it by $\emptyset$. Let $GL_{n}(R)$ be the group consisting
of all invertible elements of $M_{n}(R)$, and $\mathrm{Sym}_{n}(R)$ the set of
symmetric matrices of degree $n$ with entries in $R$. For a semigroup $S$ we
put $S^{\Box}=\\{s^{2}\ |\ s\in S\\}$. Let $R$ be an integral domain of
characteristic different from $2$, and $K$ its quotient field. We say that an
element $A$ of $\mathrm{Sym}_{n}(K)$ is non-degenerate if the determinant
$\det A$ of $A$ is non-zero. For a subset $S$ of $\mathrm{Sym}_{n}(K)$, we
denote by $S^{{\rm{nd}}}$ the subset of $S$ consisting of non-degenerate
matrices. We say that a symmetric matrix $A=(a_{ij})$ of degree $n$ with
entries in $K$ is half-integral over $R$ if $a_{ii}\ (i=1,...,n)$ and
$2a_{ij}\ (1\leq i\not=j\leq n)$ belong to $R$. We denote by
${\mathcal{H}}_{n}(R)$ the set of half-integral matrices of degree $n$ over
$R$. We note that ${\mathcal{H}}_{n}(R)=\mathrm{Sym}_{n}(R)$ if $R$ contains
the inverse of $2$. We denote by ${\mathbb{Z}}_{>0}$ and ${\mathbb{Z}}_{\geq
0}$ the set of positive integers and the set of non-negative integers,
respectively. For an $(m,n)$ matrix $X$ and an $(m,m)$ matrix $A$, we write
$A[X]={}^{t}XAX$, where ${}^{t}X$ denotes the transpose of $X$. Let $G$ be a
subgroup of $GL_{n}(K)$. Then we say that two elements $B$ and $B^{\prime}$ in
$\mathrm{Sym}_{n}(K)$ are $G$-equivalent if there is an element $g$ of $G$
such that $B^{\prime}=B[g]$. We denote by $1_{m}$ the unit matrix of degree
$m$ and by $O_{m,n}$ the zero matrix of type $(m,n)$. We sometimes abbreviate
$O_{m,n}$ as $O$ if there is no fear of confusion. For two square matrices $X$
and $Y$ we write $X\bot Y=\left(\begin{matrix}X&O\\\ O&Y\end{matrix}\right)$.
We often write $x\bot Y$ instead of $(x)\bot Y$ if $(x)$ is a matrix of degree
1. For an $m\times n$ matrix, $B=(b_{ij})$ and sequences ${\bf
i}=(i_{1},\ldots,i_{r}),(j_{1},\ldots,j_{r})$ of integers such that $1\leq
i_{1},\ldots,i_{r}\leq m,1\leq j_{1},\ldots,j_{r}\leq n$, we put

$B\begin{pmatrix}{\bf i}\\\ {\bf j}\end{pmatrix}=(b_{i_{k},j_{l}})_{1\leq
k,l\leq r}.$

##  2\. Siegel series

Let $F$ be a non-archimedean local field of characteristic $0$, and
${\mathfrak{o}}={\mathfrak{o}}_{F}$ its ring of integers. The maximal ideal
and the residue field of ${\mathfrak{o}}$ is denoted by ${\mathfrak{p}}$ and
${\mathfrak{k}}$, respectively. We fix a prime element $\varpi$ of
${\mathfrak{o}}$ once and for all. The cardinality of ${\mathfrak{k}}$ is
denoted by $q$. Let $\mathrm{ord}=\mathrm{ord}_{{\mathfrak{p}}}$ denote
additive valuation on $F$ normalized so that $\mathrm{ord}(\varpi)=1$. If
$a=0$, We write $\mathrm{ord}(0)=\infty$ and we make the convention that
$\mathrm{ord}(0)>\mathrm{ord}(b)$ for any $b\in F^{\times}$. We also denote by
$|*|_{{\mathfrak{p}}}$ denote the valuation on $F$ normalized so that
$|\varpi|_{{\mathfrak{p}}}=q^{-1}$. We put
$e_{0}=\mathrm{ord}_{{\mathfrak{p}}}(2)$.

For a non-degenerate element $B\in{\mathcal{H}}_{n}({\mathfrak{o}})$, we put
$D_{B}=(-4)^{[n/2]}\det B$. If $n$ is even, we denote the discriminant ideal
of $F(\sqrt{D_{B}})/F$ by ${\mathfrak{D}}_{B}$. We also put

$\xi_{B}=\begin{cases}1&\text{ if $D_{B}\in F^{\times 2}$,}\\\ -1&\text{ if
$F(\sqrt{D_{B}})/F$ is unramified quadratic,}\\\ 0&\text{ if
$F(\sqrt{D_{B}})/F$ is ramified quadratic.}\end{cases}$

Put

${\mathfrak{e}}_{B}=\begin{cases}\mathrm{ord}(D_{B})-\mathrm{ord}({\mathfrak{D}}_{B})&\text{
if $n$ is even}\\\ \mathrm{ord}(D_{B})&\text{ if $n$ is odd.}\end{cases}$

We make the convention that $\xi_{B}=1,{\mathfrak{e}}_{B}=0$ if $B$ is the
empty matrix. Once for all, we fix an additive character $\psi$ of $F$ of
order zero, that is, a character such that

${\mathfrak{o}}=\\{a\in F\ |\ \psi(ax)=1\ \text{ for any}\
x\in{\mathfrak{o}}\\}.$

For a half-integral matrix $B$ of degree $n$ over ${\mathfrak{o}}$ define the
local Siegel series $b_{{\mathfrak{p}}}(B,s)$ by

$b_{{{\mathfrak{p}}}}(B,s)=\sum_{R}\psi(\mathrm{tr}(BR))\mu(R)^{-s},$

where $R$ runs over a complete set of representatives of
$\mathrm{Sym}_{n}(F)/\mathrm{Sym}_{n}({\mathfrak{o}})$ and
$\mu(R)=[R{\mathfrak{o}}^{n}+{\mathfrak{o}}^{n}:{\mathfrak{o}}^{n}]$. The
series $b_{{\mathfrak{p}}}(B,s)$ converges absolutely if the real part of $s$
is large enough, and it has a meromorphic continuation to the whole $s$-plane.
Now for a non-degenerate half-integral matrix $B$ of degree $n$ over
${\mathfrak{o}}$ define a polynomial $\gamma_{q}(B,X)$ in $X$ by

$\gamma_{q}(B,X)=\begin{cases}(1-X)\prod_{i=1}^{n/2}(1-q^{2i}X^{2})(1-q^{n/2}\xi_{B}X)^{-1}&\text{
if $n$ is even}\\\ (1-X)\prod_{i=1}^{(n-1)/2}(1-q^{2i}X^{2})&\text{ if $n$ is
odd.}\end{cases}$

Then it is shown by [19] that there exists a polynomial
$F_{{\mathfrak{p}}}(B,X)$ in $X$ such that

$F_{{\mathfrak{p}}}(B,q^{-s})={b_{{\mathfrak{p}}}(B,s)\over\gamma_{q}(B,q^{-s})}.$

We define a symbol $X^{1/2}$ so that $(X^{1/2})^{2}=X$. We define
$\widetilde{F}_{{\mathfrak{p}}}(B,X)$ as

$\widetilde{F}_{{\mathfrak{p}}}(B,X)=X^{-{\mathfrak{e}}_{B}/2}F(B,q^{-(n+1)/2}X).$

We note that
$\widetilde{F}_{{\mathfrak{p}}}(B,X)\in{\mathbb{Q}}(q^{1/2})[X,X^{-1}]$ if $n$
is even, and
$\widetilde{F}_{{\mathfrak{p}}}(B,X)\in{\mathbb{Q}}[X^{1/2},X^{-1/2}]$ if $n$
is odd.

##  3\. The Duke-Imamoglu-Ikeda lift and main result

Put $J_{n}=\begin{pmatrix}O_{n}&-1_{n}\\\ 1_{n}&O_{n}\end{pmatrix}$.
Furthermore, put

$\varGamma^{(n)}=Sp_{n}({{\mathbb{Z}}})=\\{M\in GL_{2n}({{\mathbb{Z}}})\ |\
J_{n}[M]=J_{n}\\}.$

Let ${\mathbb{H}}_{n}$ be Siegel’s upper half-space of degree $n$. We define
$j(\gamma,Z)=\det(CZ+D)$ for $\gamma=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}$
and $Z\in{\mathbb{H}}_{n}$. We note that
$\varGamma^{(1)}=SL_{2}({\mathbb{Z}})$. Let $l$ be an integer or half-integer.
For a congruence subgroup $\varGamma$ of $\varGamma^{(n)}$, we denote by
$M_{l}(\varGamma)$ the space of Siegel modular forms of weight $l$ with
respect to $\varGamma$, and by $S_{l}(\varGamma)$ its subspace consisting of
cusp forms. Let $T$ be an element of ${{\mathcal{H}}_{n}}({\mathbb{Z}})_{>0}$
with $n$ even. Let ${\mathfrak{d}}_{T}$ be the discriminant of
${\mathbb{Q}}(\sqrt{(-1)^{n/2}\det(T)})/{\mathbb{Q}}$. Then we have
$(-1)^{n/2}\det(2T)/{\mathfrak{d}}_{T}={\mathfrak{f}}_{T}^{2}$ with
${\mathfrak{f}}_{T}\in{\mathbb{Z}}_{>0}$. Now let $k$ be a positive even
integer, and $\varGamma_{0}(4)=\Bigl{\\{}\bigl{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\bigr{)}\in SL_{2}({\mathbb{Z}})\ |\ c\equiv 0\text{
mod}4.\bigr{\\}}$. Let

$h(z)=\sum_{m\in{{\mathbb{Z}}}_{>0}\atop(-1)^{n/2}m\equiv 0,1\ {\rm mod}\
4}c_{h}(m){\bf e}(mz)$

be a Hecke eigenform in the Kohnen plus space
$S_{k-n/2+1/2}^{+}(\varGamma_{0}(4))$ and
$f(z)=\sum_{m=1}^{\infty}c_{f}(m){\bf e}(mz)$ be the primitive form in
$S_{2k-n}(SL_{2}(\mathbb{Z}))$ corresponding to $h$ under the Shimura
correspondence (cf. Kohnen [13]). We define a Fourier series $I_{n}(h)(Z)$ in
$Z\in{\mathbb{H}}_{n}$ by

$I_{n}(h)(Z)=\sum_{T\in{{\mathcal{H}}_{n}}({\mathbb{Z}})_{>0}}c_{I_{n}(h)}(T){\bf
e}({\rm tr}(TZ))$

where
$c_{I_{n}(h)}(T)=c_{h}(|{\textfrak{d}}_{T}|){\textfrak{f}}_{T}^{k-(n+1)/2}\prod_{p}\widetilde{F}_{p}(T,\alpha_{f}(p)).$
Then the first named author [8] showed that $I_{n}(h)(Z)$ is a Hecke eigenform
in $S_{k}(\varGamma^{(n)})$ whose standard $L$-function coincides with
$\zeta(s)\prod_{i=1}^{n}L(s+k-i,f)$ (see also [11]). We call $I_{n}(h)$ the
Duke-Imamoglu-Ikeda lift (D-I-I lift for short) of $h$.

######  Theorem 3.1.

Let the notation be as above. Then we have

$\displaystyle
c_{I_{n}(h)}(B)\ll_{\varepsilon,I_{n}(h)}{\mathfrak{d}}_{B}^{-n/4+5/12}(\det(2B))^{(k-1)/2+\varepsilon}\
(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. In particular, we have

$\displaystyle
c_{I_{n}(h)}(B)\ll_{\varepsilon,I_{n}(h)}(\det(2B))^{(k-1)/2+\varepsilon}\
(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$.

We give another estimate of $c_{I_{n}(h)}(B)$ in terms of minors of $B$. We
denote by ${\mathcal{I}}_{r}$ the set of sequences $(i_{1},\ldots,i_{r})$ of
integers such that $1\leq i_{1}<\cdots<i_{r}\leq n$. Let $R$ be an integral
domain of characteristic different from $2$. For an element
$B=(b_{ij})\in{\mathcal{H}}_{n}(R)$, and ${\bf i}=(i_{1},\ldots,i_{r}),{\bf
j}=(j_{1},\ldots,j_{r})\in{\mathcal{I}}_{r}$ we define $b_{{\bf i},{\bf
j}}^{(r)}=b_{{\bf i},{\bf j}}^{(r)}(B)$ as

$b_{{\bf i},{\bf j}}^{(r)}=2^{2[r/2]+1-\delta_{{\bf i},{\bf j}}}\det
B\begin{pmatrix}{\bf i}\\\ {\bf j}\end{pmatrix},$

where

$\delta_{{\bf i},{\bf j}}=\begin{cases}1&\text{ if }{\bf i}={\bf j}\\\
0&\text{ otherwise}.\end{cases}.$

For $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$ put

$G_{r}(B)=\mathrm{GCD}_{({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}}b_{{\bf i},{\bf j}}^{(r)}.$

######  Theorem 3.2.

Let the notation be as above. Then we have

$\displaystyle c_{I_{n}(h)}(B)$
$\displaystyle\ll_{\varepsilon,I_{n}(h)}{\mathfrak{d}}_{B}^{1/6}(\det(2B))^{k/2-(n+1)/4+\varepsilon}\prod_{i=1}^{n-1}G_{i}(B)^{1/2}\quad(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. In particular, we have

$\displaystyle c_{I_{n}(h)}(B)$
$\displaystyle\ll_{\varepsilon,I_{n}(h)}(\det(2B))^{k/2-n/4-1/12+\varepsilon}\prod_{i=1}^{n-1}G_{i}(B)^{1/2}\quad(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$.

##  4\. The Gross-Keating invariant

We first recall the definition of the Gross-Keating invariant of a quadratic
form over ${\mathfrak{o}}$ following [9].

For two matrices $B,B^{\prime}\in{\mathcal{H}}_{n}({\mathfrak{o}})$, we
sometimes write $B\sim B^{\prime}$ if $B$ and $B^{\prime}$ are
$GL_{n}({\mathfrak{o}})$-equivalent. The $GL_{n}({\mathfrak{o}})$-equivalence
class of $B$ is denoted by $\\{B\\}$. Let
$B=(b_{ij})\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$. Let $S(B)$ be the
set of all non-decreasing sequences $(a_{1},\ldots,a_{n})\in{\mathbb{Z}}_{\geq
0}^{n}$ such that

$\displaystyle\mathrm{ord}(b_{i})$ $\displaystyle\geq a_{i},$
$\displaystyle\mathrm{ord}(2b_{ij})$
$\displaystyle\geq(a_{i}+a_{j})/2\qquad(1\leq i,j\leq n).$

Set

$S(\\{B\\})=\bigcup_{B^{\prime}\in\\{B\\}}S(B^{\prime})=\bigcup_{U\in{\mathrm{GL}}_{n}({\mathfrak{o}})}S(B[U]).$

The Gross-Keating invariant (or the GK-invariant for short)
${\underline{a}}=(a_{1},a_{2},\ldots,a_{n})$ of $B$ is the greatest element of
$S(\\{B\\})$ with respect to the lexicographic order $\succ$ on
${\mathbb{Z}}_{\geq 0}^{n}$. Here, the lexicographic order $\succ$ is, as
usual, defined as follows. For
$(y_{1},y_{2},\ldots,y_{n}),(z_{1},z_{2},\ldots,z_{n})\in{\mathbb{Z}}_{\geq
0}^{n}$, let $j$ be the largest integer such that $y_{i}=z_{i}$ for $i<j$.
Then $(y_{1},y_{2},\ldots,y_{n})\succ(z_{1},z_{2},\ldots,z_{n})$ if
$y_{j}>z_{j}$. The Gross-Keating invariant is denoted by $\mathrm{GK}(B)$. A
sequence of length $0$ is denoted by $\emptyset$. When $B$ is a matrix of
degree $0$, we understand $\mathrm{GK}(B)=\emptyset$. By definition, the
Gross-Keating invariant $\mathrm{GK}(B)$ is determined only by the
$GL_{n}({\mathfrak{o}})$-equivalence class of $B$. We say that
$B\in{\mathcal{H}}_{n}({\mathfrak{o}})$ is an optimal form if
$\mathrm{GK}(B)\in S(B)$. Let $B\in{\mathcal{H}}_{n}({\mathfrak{o}})$. Then
$B$ is $GL_{n}({\mathfrak{o}})$-equivalent to an optimal form $B^{\prime}$.
Then we say that $B$ has an optimal decomposition $B^{\prime}$. We say that
$B\in{\mathcal{H}}_{n}({\mathfrak{o}})$ is a diagonal Jordan form if $B$ is
expressed as

$B=\varpi^{a_{1}}u_{1}\bot\cdots\bot\varpi^{a_{n}}u_{n}$

with $a_{1}\leq\cdots\leq a_{n}$ and
$u_{1},\cdots,u_{n}\in{\mathfrak{o}}^{\times}$. Then, in the non-dyadic case,
the diagonal Jordan form $B$ above is optimal, and
$\mathrm{GK}(B)=(a_{1},\ldots,a_{n})$. Therefore, the diagonal Jordan
decomposition is an optimal decomposition. However, in the dyadic case, not
all half-integral symmetric matrices have a diagonal Jordan decomposition, and
the Jordan decomposition is not necessarily an optimal decomposition. Let
$B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$, and let
$\mathrm{GK}(B)=(a_{1},\ldots,a_{n})$. For $1\leq i\leq n$ put

${\mathfrak{e}}_{i}={\mathfrak{e}}_{i}(B)=\begin{cases}a_{1}+\cdots+a_{i}&\text{
if }i\text{ is odd}\\\ 2[(a_{1}+\cdots+a_{i})/2]&\text{ if }i\text{ is
even}\end{cases}$

The following result is due to [9, Theorem 0.1], and plays an important role
in proving our main result:

######  Theorem 4.1.

Let $B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\mathrm{nd}}$. Then

$\displaystyle{\mathfrak{e}}_{n}(B)={\mathfrak{e}}_{B}.$

For our later purpose, we give an estimate of ${\mathfrak{e}}_{i}$. We put

${\bf f}_{r}(B)=\min_{X\in M_{nr}({\mathfrak{o}})\atop\det
B[X]\not=0}{\mathfrak{e}}_{B[X]},$

and

${\bf d}_{r}(B)=\min_{X\in
M_{nr}({\mathfrak{o}})}\mathrm{ord}_{{\mathfrak{p}}}(\det B[X]).$

Clearly we have ${\bf f}_{r}(B)\leq{\bf d}_{r}(B)$.

######  Theorem 4.2.

Let $B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$, and $r\leq n-1$ be a
positive integer. Then,

${\mathfrak{e}}_{r}\leq{\bf f}_{r}(B),$

and in particular

${\mathfrak{e}}_{r}\leq{\bf d}_{r}(B).$

###### Proof.

The assertion follows from [9, Lemma 3.8]. ∎

We express ${\bf d}_{r}(B)$ in terms of minors of $B$. For a sequence ${\bf
i}=(i_{1},\ldots,i_{r})$ of integers let $\mathrm{supp}({\bf i})$ denote the
set $\\{i_{1},\ldots,i_{r}\\}$. For $B\in{\mathcal{H}}_{n}({\mathfrak{o}})$

${\bf g}_{r}(B)=\min_{({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}}\mathrm{ord}_{{\mathfrak{p}}}(b_{{\bf
i},{\bf j}}^{(r)}).$

For $X=(x_{ij})\in M_{nr}(R)$, and ${\bf
i}=(i_{1},\ldots,i_{r})\in{\mathcal{I}}_{r}$ let $X({\bf i})=\det
X\begin{pmatrix}1,\ldots,r\\\ i_{1},\ldots,i_{r}\end{pmatrix}$.

######  Lemma 4.3.

Let $B\in{\mathcal{H}}_{n}(R)$. Then, for any $X,Y\in M_{nr}(R)$, we have

$\displaystyle 2^{2[r/2]+1-\delta_{X,Y}}\det({}^{t}XBY)=\sum_{({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}}2^{\delta_{{\bf i},{\bf
j}}-\delta_{X,Y}}b_{{\bf i},{\bf j}}^{(r)}X({\bf i})Y({\bf j}),$

where $\delta_{X,Y}=\begin{cases}1&\text{ if }X=Y\\\ 0&\text{otherwise}\\\
\end{cases}$. In particular, we have

$\displaystyle 2^{2[r/2]}\det B[X]=\sum_{({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}\atop{\bf j}\succ{\bf i}}b_{{\bf
i},{\bf j}}^{(r)}X({\bf i})X({\bf j})$

###### Proof.

The first assertion can be proved using [17, II, Theorem 9 (ii)] twice, and
the second assertion can be proved remarking that $\det B\begin{pmatrix}{\bf
j}\\\ {\bf i}\end{pmatrix}=\det B\begin{pmatrix}{\bf i}\\\ {\bf
j}\end{pmatrix}$. ∎

Let $B$ and $B^{\prime}$ be elements of
${\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$ and suppose that $B^{\prime}$ is
$GL_{n}({\mathfrak{o}})$-equivalent to $B$. Then clearly we have ${\bf
d}_{r}(B^{\prime})={\bf d}_{r}(B)$. Moreover, by the above lemma, we have
${\bf g}_{r}(B)={\bf g}_{r}(B^{\prime})$.

######  Theorem 4.4.

Let $B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$. Then any positive
integer $r\leq n$, we have

${\bf d}_{r}(B)={\bf g}_{r}(B).$

###### Proof.

The assertion clearly holds if $r=n$, and therefore we assume $r\leq n-1$. By
Lemma 4.3, we have ${\bf g}_{r}(B)\leq{\bf d}_{r}(B)$. For each ${\bf
i}=(i_{1},\ldots,i_{r}),{\bf j}=(j_{1},\ldots,j_{r})\in{\mathcal{I}}_{r}$ put
$s({\bf i},{\bf j})=\\#(\mathrm{supp}({\bf i})\cup\mathrm{supp}({\bf j}))$,
and

$s(B)=\min_{({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}\atop\mathrm{ord}_{\mathfrak{p}}(b_{{\bf
i},{\bf j}}^{(r)})={\bf g}_{r}(B)}s({\bf i},{\bf j}).$

Then we have $r\leq s(B)\leq\min(n,2r)$.

First let $F$ be a non-dyadic field. In view of the remark before this
theorem, we may assume that $B$ is a diagonal matrix. Let $({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}$ such that
$\mathrm{ord}_{\mathfrak{p}}(b_{{\bf i},{\bf j}}^{(r)})={\bf g}_{r}(B)$.
Clearly we have $s(B)=r$, and hence ${\bf i}={\bf j}$. Permutating the rows
and columns of $B$ appropriately, we may assume ${\bf i}={\bf
j}=(1,\ldots,r)$. Then by Lemma 4.3, we have

$B\Big{[}\begin{pmatrix}1_{r}\\\ O_{n-r,r}\end{pmatrix}\Big{]}=b_{{\bf i},{\bf
i}}^{(r)}.$

Hence we have ${\bf d}_{r}(B)\leq{\bf g}_{r}(B)$. This proves the assertion.

Next let $F$ be a dyadic field, and put $e_{0}=\mathrm{ord}_{\mathfrak{p}}(2)$
as in Section 2. Then, by [9, Section 2] and [15, Section 93], we may assume
that

($\bullet$) $\displaystyle
B=\varpi^{a_{1}}K_{1}\bot\varpi^{a_{n_{1}}}K_{n_{1}}\bot\varpi^{a_{n_{1}}+1}u_{n_{1}+1}\bot\varpi^{a_{n_{2}}}u_{n_{2}},$

where $a_{i}\in{\mathbb{Z}}_{\geq 0}\
(i=1,\ldots,n_{2}),u_{i}\in{\mathfrak{o}}^{\times}\ (i=n_{1}+1,\ldots,n_{2})$
and

$K_{i}=\begin{pmatrix}\alpha_{i}&\varpi^{f_{i}}/2\\\
\varpi^{f_{i}}/2&\beta_{i}\end{pmatrix}$

with $\alpha_{i}\in{\mathfrak{o}}^{\times},\beta_{i}\in{\mathfrak{o}},0\leq
f_{i}\leq e_{0}-1$ ($i=1,\ldots,n_{1})$. We claim that $s(B)\leq r+1$. Suppose
that $s(B)\geq r+2$. Then, clearly we have $n_{1}\geq 2$. Let $({\bf i},{\bf
j})\in{\mathcal{I}}_{r}\times{\mathcal{I}}_{r}$ such that $s({\bf i},{\bf
j})=s(B)$ and $\mathrm{ord}_{\mathfrak{p}}(b_{{\bf i},{\bf j}}^{(r)})={\bf
g}_{r}(B)$. Let $i_{k}$ be the least integer such that
$i_{k}\in\mathrm{supp}({\bf i})\setminus(\mathrm{supp}({\bf
i})\cap\mathrm{supp}({\bf j}))$. Then, we have $i_{k}\leq 2n_{1}$. By
($\bullet$), we have $j_{k}=i_{k}+1$ or $j_{k}=i_{k}-1$. Without loss of
generality, we may assume $j_{k}=i_{k}+1$. Then, $i_{k}=2i-1$ and $j_{k}=2i$
with some $1\leq i\leq n_{1}$. Let $i_{l}$ be the least integer such that
$i_{l}>i_{k}$ and $i_{l}\in\mathrm{supp}({\bf i})\setminus(\mathrm{supp}({\bf
i})\cap\mathrm{supp}({\bf j}))$. Again by ($\bullet$), we have
$(i_{l},j_{l})=(2j-1,2j)$ or $(i_{l},j_{l})=(2j,2j-1)$ with some $i<j\leq
n_{1}$. In the former case,

$\displaystyle\det B\begin{pmatrix}{\bf i}\\\ {\bf j}\end{pmatrix}=$
$\displaystyle\det B\begin{pmatrix}2i-1,2j-1\\\ 2i,2j\end{pmatrix}\det
B\begin{pmatrix}{\bf i}^{\prime\prime}\\\ {\bf j}^{\prime\prime}\end{pmatrix}$
$\displaystyle=4^{-1}\varpi^{a_{i}+a_{j}+f_{i}+f_{j}}\det B\begin{pmatrix}{\bf
i}^{\prime\prime}\\\ {\bf j}^{\prime\prime}\end{pmatrix}$

where $({\bf i}^{\prime\prime},{\bf
j}^{\prime\prime})\in{\mathcal{I}}_{r-2}\times{\mathcal{I}}_{r-2}$ such that
$\mathrm{supp}({\bf i}^{\prime\prime})=\mathrm{supp}({\bf
i})\setminus\\{2i-1,2j-1\\}$ and $\mathrm{supp}({\bf
j}^{\prime\prime})=\mathrm{supp}({\bf j})\setminus\\{2i,2j\\}$. Without loss
of generality, we may assume $a_{i}+f_{i}\leq a_{j}+f_{j}$. Let $({\bf
i}^{\prime},{\bf j}^{\prime})$ be an element of
${\mathcal{I}}_{r}\times{\mathcal{I}}_{r}$ such that $\mathrm{supp}({\bf
i}^{\prime})=\mathrm{supp}({\bf i}^{\prime\prime})\cup\\{2i,2i-1\\}$. and
$\mathrm{supp}({\bf j}^{\prime})=\mathrm{supp}({\bf
i}^{\prime\prime})\cup\\{2i,2i-1\\}$. Then, $s({\bf i}^{\prime},{\bf
j}^{\prime})=s({\bf i},{\bf j})-2$ and

$\displaystyle\det B\begin{pmatrix}{\bf i}^{\prime}\\\ {\bf
j}^{\prime}\end{pmatrix}=\det B\begin{pmatrix}2i-1,2i\\\
2i-1,2i\end{pmatrix}\det B\begin{pmatrix}{\bf i}^{\prime\prime}\\\ {\bf
j}^{\prime\prime}\end{pmatrix}=\det(\varpi^{a_{i}}K_{i})\det
B\begin{pmatrix}{\bf i}^{\prime\prime}\\\ {\bf
j}^{\prime\prime}\end{pmatrix},$

and hence

$\mathrm{ord}_{\mathfrak{p}}(b_{{\bf i}^{\prime},{\bf
j}^{\prime}}^{(r)})\leq\mathrm{ord}_{\mathfrak{p}}(b_{{\bf i},{\bf
j}}^{(r)}).$

In the latter case, we also obtain a similar inequality. This is a
contradiction, and we prove the claim. Suppose that $s(B)=r$. Then, in the
same way as in the non-dyadic case, we prove ${\bf d}_{r}(B)\leq{\bf
g}_{r}(B)$. Next suppose that $s(B)=r+1$. Then, we may assume ${\bf
i}=(1,\ldots,r)$ and ${\bf j}=(1,\ldots,r-1,r+1)$. If
$\mathrm{ord}_{{\mathfrak{p}}}(b_{{\bf j},{\bf j}}^{(r)})={\bf g}_{r}(B)$,
then the assertion can be proved in the same manner as above, and we may
assume that $\mathrm{ord}_{{\mathfrak{p}}}(b_{{\bf j},{\bf j}}^{(r)})>{\bf
g}_{r}(B)$. Put $X=\begin{pmatrix}1_{r-1}&0\\\ 0&1\\\ 0&1\\\
O_{n-r-1,r-1}&0\end{pmatrix}$. Then, again by Lemma 4.3, we have

$2^{2[r/2]}\det B[X]=b_{{\bf i},{\bf i}}^{(r)}+b_{{\bf i},{\bf
j}}^{(r)}+b_{{\bf j},{\bf j}}^{(r)},$

and hence

$\mathrm{ord}_{\mathfrak{p}}(2^{2[r/2]}\det
B[X])=\mathrm{ord}_{\mathfrak{p}}(b_{{\bf i},{\bf j}}^{(r)}).$

Hence we have ${\bf d}_{r}(B)\leq{\bf g}_{r}(B)$ also in this case.

This completes the assertion.

∎

######  Remark 4.5.

For $B=(b_{ij})\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\rm nd}$, we have

${\mathfrak{e}}_{1}=\min_{1\leq i,j\leq
n}\mathrm{ord}_{\mathfrak{p}}(b_{i,j}^{(1)}).$

This has been proved in the case $F={\mathbb{Q}}_{p}$ (cf. [20]), and can be
proved in the same manner in the general case.

##  5\. Estimate of $\widetilde{F}_{\mathfrak{p}}(B,\alpha)$.

In this section we estimate $\widetilde{F}_{\mathfrak{p}}(B,X)$ for
$B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\mathrm{nd}}$ with $n$ even and
$\alpha\in{\mathbb{C}}^{\times}$. This is one of key ingredients in the proof
of our main result. We recall the definition of a naive $\mathrm{EGK}$ datum
(cf. [9]). Let ${\mathcal{Z}}_{3}=\\{0,1,-1\\}$.

######  Definition 5.1.

An element $(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ of
${\mathbb{Z}}_{\geq 0}^{n}\times{\mathcal{Z}}_{3}^{n}$ is said to be a naive
$\mathrm{EGK}$ datum of length $n$ if the following conditions hold:

  * (N1)

$a_{1}\leq\cdots\leq a_{n}$.

  * (N2)

Assume that $i$ is even. Then $\varepsilon_{i}\not=0$ if and only if
$a_{1}+\cdots+a_{i}$ is even.

  * (N3)

Assume that $i$ is odd. Then $\varepsilon_{i}\not=0$.

  * (N4)

$\varepsilon_{1}=1$.

  * (N5)

Let $i\geq 3$ be an odd integer and assume that $a_{1}+\cdots+a_{i-1}$ is
even. Then
$\varepsilon_{i}=\varepsilon_{i-1}^{a_{i}+a_{i-1}}\varepsilon_{i-2}$.

We denote by $\mathcal{NEGK}_{n}$ the set of all naive $\mathrm{EGK}$ data of
length $n$.

######  Definition 5.2.

For integers $e,\widetilde{e}$, a real number $\xi$, define rational functions
$C(e,\widetilde{e},\xi;Y,X)$ and $D(e,\widetilde{e},\xi;Y,X)$ in $Y^{1/2}$ and
$X^{1/2}$ by

$C(e,\widetilde{e},\xi;Y,X)={Y^{\widetilde{e}/2}X^{-(e-\widetilde{e})/2-1}(1-\xi
Y^{-1}X)\over X^{-1}-X}$

and

$D(e,\widetilde{e},\xi;Y,X)={Y^{\widetilde{e}/2}X^{-(e-\widetilde{e})/2}\over
1-\xi X}.$

For a positive integer $i$ put

$C_{i}(e,\widetilde{e},\xi;Y,X)=\begin{cases}C(e,\widetilde{e},\xi;Y,X)&\text{
if $i$ is even }\\\ D(e,\widetilde{e},\xi;Y,X)&\text{ if $i$ is
odd.}\end{cases}.$

######  Definition 5.3.

For a sequence $\underline{a}=(a_{1},\ldots,a_{n})$ of integers and an integer
$1\leq i\leq n$, we define
${\mathfrak{e}}_{i}={\mathfrak{e}}_{i}(\underline{a})$ as

${\mathfrak{e}}_{i}=\begin{cases}a_{1}+\cdots+a_{i}&\text{ if $i$ is odd}\\\
2[(a_{1}+\cdots+a_{i})/2]&\text{ if $i$ is even.}\end{cases}$

We also put ${\mathfrak{e}}_{0}=0$.

For a naive $\mathrm{EGK}$ datum
$(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ and an integer
$1\leq m\leq n$, put
$H_{m}=(a_{1},\ldots,a_{m};\varepsilon_{1},\ldots,\varepsilon_{m})$. Then
$H_{m}$ is also a naive $\mathrm{EGK}$ datum of length $m$.

######  Definition 5.4.

For a naive $\mathrm{EGK}$ datum
$H=(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ we define a
rational function ${\mathcal{F}}(H;Y,X)$ in $X^{1/2}$ and $Y^{1/2}$ as
follows: First we define

${\mathcal{F}}(H;Y,X)=X^{-a_{1}/2}+X^{-a_{1}/2+1}+\cdots+X^{a_{1}/2-1}+X^{a_{1}/2}$

if $n=1$. Let $n>1$. Then
$H^{\prime}=(a_{1},\ldots,a_{n-1};\varepsilon_{1},\ldots,\varepsilon_{n-1})$
is a naive $\mathrm{EGK}$ datum of length $n-1$. Assume that
${\mathcal{F}}(H^{\prime};Y,X)$ is defined for $H^{\prime}$. Then, we define
${\mathcal{F}}(H;Y,X)$ as

$\displaystyle{\mathcal{F}}(H;Y,X)=C_{n}({\mathfrak{e}}_{n},{\mathfrak{e}}_{n-1},\xi;Y,X){\mathcal{F}}(H^{\prime};Y,YX)$
$\displaystyle+\zeta
C_{n}({\mathfrak{e}}_{n},{\mathfrak{e}}_{n-1},\xi;Y,X^{-1}){\mathcal{F}}(H^{\prime};Y,YX^{-1}),$

where $\xi=\varepsilon_{n}$ or $\varepsilon_{n-1}$ according as $n$ is even or
odd, and $\zeta=1$ or $\varepsilon_{n}$ according as $n$ is even or odd.

The following result is due to [10, Proposition 4.1].

######  Proposition 5.5.

Let $H$ be a naive $\mathrm{EGK}$ datum of length $n$. Then we have

$\displaystyle{\mathcal{F}}(H;Y,X^{-1})=\zeta{\mathcal{F}}(H;Y,X),$

where $\zeta=\varepsilon_{n}$ or $1$ according as $n$ is odd or even.

For a naive $\mathrm{EGK}$ datum $H$, let
${\mathcal{G}}(H;Y,X)=X^{{\mathfrak{e}}_{n}/2}{\mathcal{F}}(H;Y,X)$. It
follows from the proof of [10, Proposition 4.2], ${\mathcal{G}}(H;Y,X)$ is a
polynomial in $X$ of degree ${\mathfrak{e}}_{n}$ with coefficients in
${\mathbb{Q}}[Y,Y^{-1}]$, write

$\displaystyle{\mathcal{G}}(H;Y,X)=\sum_{i=0}^{{\mathfrak{e}}_{n}}a_{i}(H,Y)X^{i}.$

We give an induction formula for ${\mathcal{G}}(H;Y,X)$.

######  Theorem 5.6.

Let $H=(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ be a naive
$\mathrm{EGK}$ datum of length $n$, and put $a_{i}(Y)=a_{i}(H,Y)$ and
$b_{i}(Y)=a_{i}(H_{n-1};Y)$.

  * (1)

Let $n$ be an even integer such that $n\geq 2$, Then, for any
$l=0,\ldots,{\mathfrak{e}}_{n}$, $a_{i}(Y)$ is expressed as

$\displaystyle a_{i}(Y)=$
$\displaystyle\sum_{\max((l-{\mathfrak{e}}_{n-1})/2,0)\leq j\leq
l/2}b_{l-2j}(Y)Y^{l-2j}$
$\displaystyle-\varepsilon_{n}\sum_{\max((l-1-{\mathfrak{e}}_{n-1})/2,0)\leq
j\leq(l-1)/2}b_{l-1-2j}(Y)Y^{l-2-2j}$ $\displaystyle-\sum_{0\leq
j\leq({\mathfrak{e}}_{n-1}+l-{\mathfrak{e}}_{n}-2)/2}b_{{\mathfrak{e}}_{n}-l-2j+2}(Y)Y^{{\mathfrak{e}}_{n}-l+2j+2}$
$\displaystyle+\varepsilon_{n}\sum_{0\leq
j\leq({\mathfrak{e}}_{n-1}+l-{\mathfrak{e}}_{n}-1)/2}b_{{\mathfrak{e}}_{n}-l-2j+1}(Y)Y^{{\mathfrak{e}}_{n}-l+2j}.$

  * (2)

Let $n$ be odd such that $n\geq 3$.

    * (2.1)

Assume that $\varepsilon_{n-1}\not=0$. Then, for any
$l=0,\ldots,{\mathfrak{e}}_{n}$, $a_{i}(Y)$ is expressed as

$\displaystyle a_{i}(Y)$
$\displaystyle=\sum_{\max(l-{\mathfrak{e}}_{n-1},0)\leq j\leq
l}b_{l-j}(Y)Y^{l-j}$ $\displaystyle-\varepsilon_{n}\sum_{0\leq
j\leq{\mathfrak{e}}_{n-1}+l-{\mathfrak{e}}_{n}-1}b_{{\mathfrak{e}}_{n}-l-2j+2}(Y)Y^{{\mathfrak{e}}_{n}-l+j+1}\varepsilon_{n-1}^{j}.$

    * (2.2)

Assume that $\varepsilon_{n-1}=0$. Then, for any
$l=0,\ldots,{\mathfrak{e}}_{n}$, $a_{i}(Y)$ is expressed as

$a_{i}(Y)=b_{l}Y^{l}+\varepsilon_{n}b_{{\mathfrak{e}}_{n}-l}Y^{{\mathfrak{e}}_{n}-l}.$

Throughout (1),(2),(3), we make the convention that the sum $\sum_{0\leq j\leq
a}(*)$ is zero if $a<0$. We also understand $b_{j}=0$ if $j<0$ or
$j>{\mathfrak{e}}_{n-1}$.

###### Proof.

Let $n$ be even. Then

(A) $\displaystyle{\mathcal{G}}(H;Y,X)$
$\displaystyle={(1-Y^{-1}\varepsilon_{n}X){\mathcal{G}}(H_{n-1}:Y,YX)\over
1-X^{2}}$
$\displaystyle-{X^{e_{n}+2}(1-Y^{-1}\varepsilon_{n}X^{-1}){\mathcal{G}}(H_{n-1};Y,YX^{-1})\over
1-X^{2}}$

Let ${\mathcal{H}}(Y,X)$ denote the right-hand side of (A). Then, as a formal
power series in $X$, ${\mathcal{H}}(Y,X)$ can be written as

$\displaystyle{\mathcal{H}}(Y,X)$
$\displaystyle=(1-Y^{-1}\varepsilon_{n}X){\mathcal{G}}(H_{n-1}:Y,YX)\sum_{j=0}^{\infty}X^{2j}$
$\displaystyle-X^{e_{n}+2}(1-Y^{-1}\varepsilon_{n}X^{-1}){\mathcal{G}}(H_{n-1};Y,YX^{-1})\sum_{j=0}^{\infty}X^{2j}$
$\displaystyle=(1-Y^{-1}\varepsilon_{n}X)\sum_{i=0}^{{\mathfrak{e}}_{n-1}}b_{i}(Y)(YX)^{i}\sum_{j=0}^{\infty}X^{2j}$
$\displaystyle-X^{e_{n}+2}(1-Y^{-1}\varepsilon_{n}X^{-1})\sum_{i=0}^{{\mathfrak{e}}_{n-1}}b_{i}(Y)(YX^{-1})^{i}\sum_{j=0}^{\infty}X^{2j}$
$\displaystyle=\sum_{l=0}^{\infty}(\sum_{0\leq i\leq{\mathfrak{e}}_{n-1},j\geq
0\atop
i+2j=l}b_{i}(Y)Y^{i})X^{l}-Y^{-1}\varepsilon_{n}\sum_{l=0}^{\infty}(\sum_{0\leq
i\leq{\mathfrak{e}}_{n-1},j\geq 0\atop i+2j=l-1}b_{i}(Y)Y^{i})X^{l}$
$\displaystyle-\sum_{l=0}^{\infty}(\sum_{0\leq i\leq{\mathfrak{e}}_{n-1},j\geq
0\atop{\mathfrak{e}}_{n}-i+2j=l}b_{i}(Y)Y^{i})X^{l}+Y^{-1}\varepsilon_{n}\sum_{l=0}^{\infty}(\sum_{0\leq
i\leq{\mathfrak{e}}_{n-1},j\geq
0\atop{\mathfrak{e}}_{n}-i+2j=l+1}b_{i}(Y)Y^{i})X^{l}.$

Since ${\mathcal{G}}(H;Y,X)$ is a polynomial in $X$ of degree
${\mathfrak{e}}_{n}$, the $l$-th coefficient of ${\mathcal{H}}(Y,X)$ as a
power series in $X$ is $a_{i}(Y)$ or $0$ according as
$l\leq{\mathfrak{e}}_{n}$ or $l>{\mathfrak{e}}_{n}$, and by a simple
computation we prove the assertion.

Let $n$ be odd and. Then we have

(B) $\displaystyle{\mathcal{G}}(H;Y,X)={{\mathcal{G}}(H_{n-1};Y,YX)\over
1-\varepsilon_{n-1}X}-\varepsilon_{n}{X^{{\mathfrak{e}}_{n}+1}{\mathcal{G}}(H_{n-1};Y,YX^{-1})\over
1-\varepsilon_{n-1}X}.$

Assume that $\varepsilon_{n-1}\not=0$. Then, as a formal power series in $X$,
the right-hand side of (B) can be written as

${\mathcal{G}}(H_{n-1};Y,YX)\sum_{j=0}^{\infty}(\varepsilon_{n-1}X)^{j}-\varepsilon_{n}X^{{\mathfrak{e}}_{n}+1}{\mathcal{G}}(H_{n-1};Y,YX^{-1})\sum_{j=0}^{\infty}(\varepsilon_{n-1}X)^{j}.$

Then the assertion can be proved in the same manner as above.

Let $n$ be odd, and $\varepsilon_{n-1}=0$. Then

${\mathcal{G}}(H;Y,X)={\mathcal{G}}(H_{n-1};Y,YX)+\varepsilon_{n}X^{{\mathfrak{e}}_{n}}{\mathcal{G}}(H_{n-1};Y,YX^{-1}).$

That is,

${\mathcal{G}}(H;Y,X)=\sum_{i=0}^{e_{n-1}}b_{i}(Y)(YX)^{i}+\varepsilon_{n}X^{{\mathfrak{e}}_{n}}\sum_{i=0}^{e_{n-1}}b_{i}(Y)(YX^{-1})^{i}.$

Thus the assertion directly follows. ∎

######  Theorem 5.7.

Let $H=(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ be a naive
$\mathrm{EGK}$ datum of length $n$. Let $q$ be a positive integer. Then

(C)
$\displaystyle|a_{i}(q^{1/2})|\leq\prod_{l=1}^{n-1}({\mathfrak{e}}_{l}+1)\prod_{l=1}^{n-1}q^{e_{l}/2}$

for any $i=0,\ldots,e_{n}$.

###### Proof.

We prove the assertion by induction on $n$. The assertion clearly holds if
$n=1$. Let $n\geq 2$ and assume that the assertion holds for any naive
$\mathrm{EGK}$ datum of length $n^{\prime}<n$. By the functional equation of
${\mathcal{F}}(H;X,Y)$, it suffices to prove the assertion for $0\leq i\leq
e_{n}/2$. Put $b_{i}=b_{i}(q^{1/2})$. Let $n$ be even. Then, by Theorem 5.6,
(1), for $0\leq i\leq{\mathfrak{e}}_{n}/2$, $a_{i}(q^{1/2})$ is given by

$\displaystyle a_{i}(q^{1/2})$
$\displaystyle=\sum_{\max((i-{\mathfrak{e}}_{n-1})/2,0)\leq j\leq
i/2}b_{i-2j}q^{(i-2j)/2}$
$\displaystyle-\varepsilon_{n}\sum_{\max(i-1-{\mathfrak{e}}_{n-1})/2,0)\leq
j\leq(i-1)/2}b_{i-2j-1}q^{(i-2j-2)/2}$ $\displaystyle-\sum_{0\leq
j\leq(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-2)/2}b_{{\mathfrak{e}}_{n}+2-2j-i}q^{({\mathfrak{e}}_{n}+2-2j-i)/2}$
$\displaystyle+\varepsilon_{n}\sum_{0\leq
j\leq(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-1)/2}b_{{\mathfrak{e}}_{n}+1-2j-i}q^{({\mathfrak{e}}_{n}-2j-i)/2}.$

Here we make the convention the sum $\sum_{0\leq j\leq a}(*)$ is zero if
$a<0$. We also understand $b_{j}=0$ if $j<0$ or $j>{\mathfrak{e}}_{n-1}$. For
each $i=0,\ldots,{\mathfrak{e}}_{n}/2$, put

$\displaystyle{\mathbb{B}}_{i,1}$ $\displaystyle=\\#(\\{j\in{\mathbb{Z}}\ |\
\max((i-{\mathfrak{e}}_{n-1})/2,0)\leq j\leq i/2\\}),$
$\displaystyle{\mathbb{B}}_{i,2}$ $\displaystyle=\\#(\\{j\in{\mathbb{Z}}\ |\
\max((i-1-{\mathfrak{e}}_{n-1})/2,0)\leq j\leq(i-1)/2\\}),$
$\displaystyle{\mathbb{B}}_{i,3}$ $\displaystyle=\\#(\\{j\in{\mathbb{Z}}\ |\
0\leq j\leq(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-2)/2\\}),$
$\displaystyle{\mathbb{B}}_{i,4}$ $\displaystyle=\\#(\\{j\in{\mathbb{Z}}\ |\
0\leq j\leq(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-1)/2\\},$

and
${\mathbb{B}}_{i}={\mathbb{B}}_{i,1}+{\mathbb{B}}_{i,2}+{\mathbb{B}}_{i,3}+{\mathbb{B}}_{i,4}$.
By the induction assumption, we have

$|b_{i}|\leq\prod_{l=1}^{n-2}({\mathfrak{e}}_{l}+1)\prod_{l=1}^{n-2}q^{e_{l}/2}.$

Hence we have

$\displaystyle|a_{i}(q^{1/2})|$
$\displaystyle\leq{\mathbb{B}}_{i}q^{{\mathfrak{e}}_{n-1}/2}\prod_{l=1}^{n-2}({\mathfrak{e}}_{l}+1)\prod_{l=1}^{n-2}q^{e_{l}/2}$

We claim that we have

(D) $\displaystyle{\mathbb{B}}_{i}\leq{\mathfrak{e}}_{n-1}+1.$

To prove this we note that

(E) $\displaystyle\\#(\\{j\in{\mathbb{Z}}\ |\ \alpha\leq
j\leq\beta)=\begin{cases}(\beta-\alpha+2)/2&\text{ if }\alpha,\beta\text{ are
even}\\\ (\beta-\alpha)/2&\text{ if }\alpha,\beta\text{ are odd}\\\
(\beta-\alpha+1)/2&\text{ otherwise}\end{cases}$

for any integers $0\leq\alpha\leq\beta$. Assume $i\geq{\mathfrak{e}}_{n-1}+1$.
Then, $\max((i-{\mathfrak{e}}_{n-1})/2,0)=(i-{\mathfrak{e}}_{n-1})/2$ and
$\max((i-1-{\mathfrak{e}}_{n-1})/2,0)=(i-1-{\mathfrak{e}}_{n-1})/2$, and by
(E), we easily see that we have
${\mathbb{B}}_{i.1}+{\mathbb{B}}_{i,2}={\mathfrak{e}}_{n-1}+1$. Moreover,
since we have ${\mathfrak{e}}_{n-1}+1\leq i\leq{\mathfrak{e}}_{n}/2$, we have

$(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-2)/2<(i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-1)/2\leq(i-{\mathfrak{e}}_{n}/2-3)/2<0,$

and hence ${\mathbb{B}}_{i.3}+{\mathbb{B}}_{i,4}=0$. This proves (D). Assume
that $i\leq{\mathfrak{e}}_{n-1}$. Then,
$\max((i-{\mathfrak{e}}_{n-1})/2,0)=\max((i-1-{\mathfrak{e}}_{n-1})/2,0)=0$,
and by (E), we have ${\mathbb{B}}_{i.1}+{\mathbb{B}}_{i,2}=i+1$. Assume that
$i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-2\geq 0$. Then, by (E), we have
${\mathbb{B}}_{i,3}+{\mathbb{B}}_{i.4}=i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}$,
and hence
${\mathbb{B}}_{i}=2i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}+1\leq{\mathfrak{e}}_{n-1}+1$,
which proves (D). Assume that $i-{\mathfrak{e}}_{n}+{\mathfrak{e}}_{n-1}-1\leq
0$. Then, we have ${\mathbb{B}}_{i,3}+{\mathbb{B}}_{i,4}\leq 1$. Moreover,
since we have

$i\leq{\mathfrak{e}}_{n}-{\mathfrak{e}}_{n-1}+1\leq{\mathfrak{e}}_{n-1},$

and ${\mathfrak{e}}_{n}/2$ is an integer, we have
${\mathfrak{e}}_{n}/2\leq{\mathfrak{e}}_{n-1}-1$. Hence we have

${\mathbb{B}}_{i}\leq
i+2\leq{\mathfrak{e}}_{n}/2+2\leq{\mathfrak{e}}_{n-1}+1.$

This completes (D), and hence (C).

Let $n$ be odd and assume that $\varepsilon_{n-1}\not=0$. Then, by Theorem
5.6, (2.1), for $0\leq i\leq{\mathfrak{e}}_{n}/2$, $a_{l}(q^{1/2})$ is given
by

$\displaystyle a_{i}(q^{1/2})$
$\displaystyle=\sum_{\max(i-{\mathfrak{e}}_{n-1},0)\leq j\leq
i}b_{i-j}q^{(i-j)/2}$ $\displaystyle-\varepsilon_{n}\sum_{0\leq
j\leq{\mathfrak{e}}_{n-1}+i-{\mathfrak{e}}_{n}-1}b_{{\mathfrak{e}}_{n}-i-2j+2}q^{({\mathfrak{e}}_{n}-i+j+1)/2}\varepsilon_{n-1}^{j}.$

Then the assertion can be proved in the same manner as above. Finally let $n$
be odd and assume that $\varepsilon_{n-1}=0$. Then, we have

$\displaystyle{\mathcal{G}}(H;q^{1/2},X)=\sum_{i=0}^{{\mathfrak{e}}_{n-1}}b_{i}(q^{1/2}X)^{i}+\varepsilon_{n}X^{{\mathfrak{e}}_{n}}\sum_{i=0}^{{\mathfrak{e}}_{n-1}}b_{i}(q^{1/2}X^{-1})^{i}.$

Then, for $i\leq{\mathfrak{e}}_{n}/2$, $a_{i}(q^{1/2})$ is given by

$\displaystyle
a_{i}(q^{1/2})=q^{l/2}b_{i}+\varepsilon_{n}q^{({\mathfrak{e}}_{n}-i)/2}b_{{\mathfrak{e}}_{n}-i}.$

Since ${\mathfrak{e}}_{n-1}$ is odd, we have ${\mathfrak{e}}_{n-1}+1\geq 2$.
Thus the assertion can be proved by the induction assumption. This completes
the induction.

∎

######  Theorem 5.8.

Let $H$ be as in Theorem 5.7. Let $s\in{\mathbb{C}}$ such that
$\mathrm{Re}(s)\leq r_{0}$. Then

$|{\mathcal{F}}(H;q^{1/2},q^{s})|\leq
q^{{\mathfrak{e}}_{n}r_{0}/2}\prod_{i=1}^{n}({\mathfrak{e}}_{i}+1)\prod_{i=1}^{n-1}q^{e_{i}/2}.$

###### Proof.

We have

$\displaystyle{\mathcal{F}}(H;q^{1/2},q^{s})=\sum_{i=-{\mathfrak{e}}_{n}/2}^{-1}a_{i+{\mathfrak{e}}_{n}/2}(q^{1/2})q^{si}+\sum_{i=0}^{{\mathfrak{e}}_{n}/2}a_{i}(q^{1/2})q^{si}.$

Thus the assertion follows from Theorem 5.7. ∎

######  Corollary 5.9.

Let the notation be as above. Let $n$ be even. Then

$|{\mathcal{F}}(H;q^{1/2},q^{s})|\leq
q^{r_{0}{\mathfrak{e}}_{n}/2}q^{(n-1){\mathfrak{e}}_{n}/4}\prod_{i=1}^{n}({\mathfrak{e}}_{i}+1).$

###### Proof.

By definition,

$\displaystyle\sum_{i=1}^{n-1}{\mathfrak{e}}_{i}\leq\sum_{i=1}^{n-1}(n-i)a_{i}.$

For $\underline{a}=(a_{1},\ldots,a_{n})$, put
$|\underline{a}|=a_{1}+\cdots+a_{n}$. First suppose that $|\underline{a}|$ is
even, then, by definition, we have ${\mathfrak{e}}_{n}=|\underline{a}|$, and
hence we have

$\displaystyle
2\sum_{i=1}^{n-1}(n-i)a_{i}-(n-1){\mathfrak{e}}_{n}=\sum_{i=1}^{n}(n-2i+1)a_{i}$
$\displaystyle=\sum_{i=1}^{n/2}(n-2i+1)(a_{i}-a_{n-i+1})\leq 0.$

This implies that we have

$\displaystyle\sum_{i=1}^{n-1}(n-i)a_{i}\leq\frac{n-1}{2}{\mathfrak{e}}_{n},$

and hence

$\displaystyle\sum_{i=1}^{n-1}{\mathfrak{e}}_{i}\leq\frac{n-1}{2}{\mathfrak{e}}_{n}.$

Next suppose that $|\underline{a}|$ is odd, then, again by definition, we have
${\mathfrak{e}}_{n}=|\underline{a}|-1$, and $a_{n}\geq a_{1}+1$. Hence we have

$\displaystyle
2\sum_{i=1}^{n-1}(n-i)a_{i}-(n-1){\mathfrak{e}}_{n}=\sum_{i=1}^{n}(n-2i+1)a_{i}+(n-1)$
$\displaystyle=\sum_{i=2}^{n/2}(n-2i+1)(a_{i}-a_{n-i+1})+(n-1)(a_{1}+1-a_{n})\leq
0.$

Thus the assertion can be proved in the same manner as above.

∎

######  Theorem 5.10.

Let $q=\\#({\mathfrak{o}}/{\mathfrak{p}})$. Let
$B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\mathrm{nd}}$. Then there is a naive
$\mathrm{EGK}$ datum
$H=(a_{1},\ldots,a_{n};\varepsilon_{1},\ldots,\varepsilon_{n})$ of length $n$
such that $(a_{1},\ldots,a_{n})$ is the Gross-Keating invariant of $B$ and

${\mathcal{F}}(H;q^{1/2},X)=\widetilde{F}_{{\mathfrak{p}}}(B,X).$

###### Proof.

The assertion follows from [9, Theorem 4.3], [9, Corollary 5.1], [10, Theorem.
1.1], and [10, Proposition 4.5]. ∎

By Corollary 5.9 and Theorems 5.8 and 5.10, we immediately have the following
theorem.

######  Theorem 5.11.

Let $B\in{\mathcal{H}}_{n}({\mathfrak{o}})^{\mathrm{nd}}$ with $n$ even and
$s\in{\mathbb{C}}$ such that $\mathrm{Re}(s)\leq r_{0}$. Let
${\mathfrak{e}}_{i}={\mathfrak{e}}_{i}(B)$ be that in Section 4. Then we have
the following estimates.

  * (1)

$\displaystyle|\widetilde{F}_{\mathfrak{p}}(B,\alpha)|\leq
q^{r_{0}{\mathfrak{e}}_{n}/2}q^{(n-1){\mathfrak{e}}_{n}/4}\prod_{i=1}^{n}({\mathfrak{e}}_{i}+1).$

  * (2)

$\displaystyle|\widetilde{F}_{\mathfrak{p}}(B,\alpha)|\leq
q^{r_{0}{\mathfrak{e}}_{n}/2}\prod_{i=1}^{n-1}q^{e_{i}/2}.\prod_{i=1}^{n}({\mathfrak{e}}_{i}+1).$

##  6\. Proofs of Theorem 3.1 and 3.2

In this section we prove our main result.

######  Lemma 6.1.

Let $\kappa\in{1\over 2}+{\mathbb{Z}}$ such that $\kappa\geq 13/2$. Let
$g(z)=\sum_{m=1}^{\infty}c_{g}(h){\bf e}(mz)\in S_{\kappa}(\varGamma_{0}(4))$.
Then, for any fundamental discriminant $D$ we have

$\displaystyle c_{g}(|D|)\ll_{\varepsilon,g}|D|^{\kappa/2-1/3+\varepsilon}\
(\varepsilon>0).$

###### Proof.

It is known that $D$ can be expressed as $D=2^{s}D^{\prime}$, with $s=0,2$ or
$3$ and $D^{\prime}$ a square free odd integer. Suppose that $s=0$. Then, by
[5, Corollary 1.3], we have

$\displaystyle c_{g}(|D|)\ll_{\varepsilon,g}|D|^{\kappa/2-1/3+\varepsilon}\
(\varepsilon>0).$

Suppose that $s\geq 2$. Let $T_{\kappa,1}^{4}(4)$ be the operator on
$S_{\kappa}(\varGamma_{0}(4))$ in [18, page 450]. Let

$\widetilde{g}(z):=g|T_{\kappa,1}^{4}(4)(z)=\sum_{m=1}^{\infty}b(m){\bf
e}(mz).$

Then, by Theorem [18, Theorem 1.7], we have
$c_{g}(|D|)=b(2^{s-2}|D^{\prime}|)$. Since $\widetilde{g}$ belongs to
$S_{\kappa}(\varGamma_{0}(4))$ and $2^{s-2}|D^{\prime}|$ is square free, again
by [5, Corollary 1.3], we have

$\displaystyle
c_{g}(|D|)=b(2^{s-2}|D^{\prime}|)\ll_{\varepsilon,\widetilde{g}}|2^{s-2}D^{\prime}|^{\kappa/2-1/3+\varepsilon}\ll_{\varepsilon,g}|D|^{\kappa/2-1/3+\varepsilon}\
(\varepsilon>0).$

This completes the proof. ∎

######  Remark 6.2.

For $g\in S_{\kappa}^{+}(\varGamma_{0}(4))$, $g|T_{\kappa,1}^{4}(4)$ does not
necessarily belong to $S_{\kappa}^{+}(\varGamma_{0}(4))$.

For an element $C\in{\mathcal{H}}_{r}({\mathbb{Z}})_{>0}$, put

$\boldsymbol{\Delta}(C)=\begin{cases}{\mathfrak{f}}_{C}^{2}&\text{ if }r\text{
is even}\\\ 2^{r-1}\det C&\text{ if }r\text{ is odd.}\end{cases}$

Let $B=(b_{ij})\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. For each prime number
$p$, we denote by
$\mathrm{GK}(B)^{(p)}=\underline{a}^{(p)}=(a_{1}^{(p)},\ldots,a_{n}^{(p)})$
the Gross-Keating invariant of $B$ viewing $B$ as an element of
${\mathcal{H}}_{n}({\mathbb{Z}}_{p})$, and for $r=1,\ldots,n$ put

${\mathfrak{e}}_{r,B}^{(p)}=\begin{cases}2[(a_{1}^{(p)}+\cdots+a_{r}^{(p)})/2]&\text{
if }r\text{ is even}\\\ a_{1}^{(p)}+\cdots+a_{r}^{(p)}&\text{ if }r\text{ is
odd},\end{cases}$

and in particular put ${\mathfrak{e}}_{B}^{(p)}={\mathfrak{e}}_{n,B}^{(p)}$.
Moreover, for $r\leq n$, put

$\mathscr{E}_{r}(B)=\mathrm{GCD}_{X\in M_{nr}({\mathbb{Z}})\atop
B[X]>0}\boldsymbol{\Delta}(B[X]),$ $D_{r}(B)=\mathrm{GCD}_{X\in
M_{nr}({\mathbb{Z}})\atop B[X]>0}2^{2[r/2]}\det B[X].$

Clearly ${\mathscr{E}}_{r}(B)$ divides $D_{r}(B)$,

######  Lemma 6.3.

Let $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$ with $n$ even.

  * (1)

For any positive integer $r\leq n$, the product
$\prod_{p|{\mathfrak{f}}_{B}}p^{{\mathfrak{e}}_{r,B}^{(p)}}$ divides
${\mathscr{E}}_{r}(B)$. In particular, we have

$\prod_{p{\mathfrak{f}}_{B}}p^{{\mathfrak{e}}_{n,B}^{(p)}}={\mathfrak{f}}_{B}^{2}.$

  * (2)

For any positive integer $r\leq n$, $G_{r}(B)=D_{r}(B)$.

###### Proof.

The assertion (1) follows from Theorems 4.1, 4.2. We prove (2). By Lemma 4.3,
$G_{r}(B)$ divides $D_{r}(B)$. To prove that $D_{r}(B)$ divides $G_{r}(B)$,
for each prime number $p$, we denote by ${\bf g}_{r}(B)^{(p)}$ and ${\bf
d}_{r}(B)^{(p)}$ the quantities ${\bf g}_{r}(B)$ and ${\bf d}_{r}(B)$ defined
in Section 3, respectively, viewing $B$ as an element of
${\mathcal{H}}_{n}({\mathbb{Z}}_{p})^{\rm nd}$. Let $X\in
M_{nr}({\mathbb{Z}}_{p})$ and suppose that $\det B[X]\not=0$. Then we can take
$X_{0}\in M_{nr}({\mathbb{Z}})$ such that $X_{0}\equiv X\text{ mod
}p^{e}M_{nr}({\mathbb{Z}}_{p})$ with $e>\mathrm{ord}_{p}(\det B[X])$. Then, by
definition, we have

$\mathrm{ord}_{p}(D_{r}(B))\leq\mathrm{ord}_{p}(\det
B[X_{0}])=\mathrm{ord}_{p}(\det B[X]).$

By Theorem 4.4, this implies that

$\mathrm{ord}_{p}(D_{r}(B))\leq{\bf d}_{r}(B)^{(p)}={\bf g}_{r}(B)^{(p)},$

and hence $D_{r}(B)$ divides $\prod_{p}p^{{\bf g}_{r}(B)^{(p)}}=G_{r}(B)$.
This proves the assertion (2). ∎

The following theorem is a refined version of Theorem 3.1.

######  Theorem 6.4.

Let the notation and the assumption be as in Theorem 3.1. Let
$B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. The we have the following
estimates.

  * (1)

We have

$\displaystyle|c_{I_{n}(h)}(B)|\leq|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{f}}_{B}^{k-1}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})).$

  * (2)

We have

$\displaystyle|c_{I_{n}(h)}(B)|\leq|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{f}}_{B}^{k-(n+1)/2}\prod_{i=1}^{n-1}{\mathscr{E}}_{i}(B)^{1/2}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})),$

and in particular we have

$\displaystyle|c_{I_{n}(h)}(B)|\leq|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{f}}_{B}^{k-(n+1)/2}\prod_{i=1}^{n-1}G_{i}(B)^{1/2}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})).$

###### Proof.

From now on we write $\widetilde{F}_{p}(B,X)$ instead of
$\widetilde{F}_{\mathfrak{p}}(B,X)$ if ${\mathfrak{p}}=(p)$. We have

$\displaystyle
c_{I_{n}(h)}(B)=c_{h}(|{\mathfrak{d}}_{B}|){\mathfrak{f}}_{B}^{k-(n+1)/2}\prod_{p|{\mathfrak{f}}_{B}}\widetilde{F}_{p}(B,\alpha_{p}).$

We have $|\alpha_{p}|=1$ for any prime number $p$. Hence, by Theorem 5.11 (1)
and Lemma 6.3, we have

$\displaystyle|\prod_{p|{\mathfrak{f}}_{B}}\widetilde{F}_{p}(B,\alpha_{p})|$
$\displaystyle\leq\prod_{p|{\mathfrak{f}}_{B}}p^{(n-1){\mathfrak{e}}_{B}^{(p)}/4}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)}))$
$\displaystyle={\mathfrak{f}}_{B}^{(n-1)/2}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})).$

This proves the assertion (1). Similarly, by Theorem 5.11 (2), we have

$\displaystyle|\prod_{p|{\mathfrak{f}}_{B}}\widetilde{F}_{p}(B,\alpha_{p})|$
$\displaystyle\leq{\mathfrak{f}}_{B}^{k-(n+1)/2}\prod_{i=1}^{n-1}\prod_{p|{\mathfrak{f}}_{B}}p^{{\mathfrak{e}}_{i,B}^{(p)}/2}\prod_{i=1}^{n}\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})).$

By Lemma 6.3 (1) and the remark before it, we have

$\prod_{p|{\mathfrak{f}}_{B}}p^{{\mathfrak{e}}_{i,B}^{(p)}/2}\leq{\mathscr{E}}_{i}(B)^{1/2}\leq
D_{i}(B)^{1/2}=G_{i}(B)^{1/2}.$

This proves the assertion (2).

∎

Proofs of Theorems 3.1 and 3.2

By Theorem 4.1, we have
${\mathfrak{e}}_{B}^{(p)}=2\mathrm{ord}_{p}({\mathfrak{f}}_{B})$ and
$e_{i,B}\leq\mathrm{ord}_{p}(\det(2B))$ for any $i=1,\ldots,n$. Hence we have

$\prod_{p|{\mathfrak{f}}_{B}}(1+{\mathfrak{e}}_{i,B}^{(p)})\leq d(\det(2B))$

for any $i=1,\ldots,n$, where $d(a)$ is the number of positive divisors of $a$
for $a\in{\mathbb{Z}}_{>0}$. Hence, by Theorem 6.4 (1), we have

$\displaystyle|c_{I_{n}(h)}(B)|$
$\displaystyle\leq|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{f}}_{B}^{k-1}d(\det(2B))^{n}$
$\displaystyle=|c_{h}(|{\mathfrak{d}}_{B}|)|{\mathfrak{d}}_{B}|^{-k/2+1/2}d(\det(2B))^{n}\det(2B)^{(k-1)/2}$

By Lemma 6.1 we have

$\displaystyle
c_{h}(|{\mathfrak{d}}_{B}|)\ll_{\varepsilon,h}|{\mathfrak{d}}_{B}|^{k/2-n/4-1/12+\varepsilon}\quad(\varepsilon>0)$

Hence we have

$\displaystyle c_{I_{n}(h)}(B)$
$\displaystyle\ll_{\varepsilon,I_{n}(h)}|{\mathfrak{d}}_{B}|^{-n/4+5/12+\varepsilon}d(\det(2B))^{n}(\det(2B))^{(k-1)/2},$

We have $|{\mathfrak{d}}_{B}|^{\varepsilon}\leq\det(2B)^{\varepsilon}$ and

$\displaystyle
d(\det(2B))^{n}\ll_{\varepsilon}(\det(2B))^{\varepsilon}\quad(\varepsilon>0)$

for any $B\in{\mathcal{H}}_{n}({\mathbb{Z}})_{>0}$. Thus we complete the proof
of Theorem 3.1. Similarly, we can prove Theorem 3.2.

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