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\title[Spherical designs for quaternions and modular forms]{Spherical designs for finite quaternionic unit groups and their applications to modular forms}

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\author{\firstname{Masatake} \lastname{Hirao}}
\curraddr{Department of Information Science and Technology\\
	Tokyo University of Science\\
	2641 Yamazaki\\ Noda-shi\\ Chiba 278-8510\\
	Japan (current)}
\address{Department of Information Science and Technology\\
	Aichi Prefectural University\\
	Nagakute-city\\ Aichi\\ 480-1198\\
	Japan (former)}
\email{hirao@rs.tus.ac.jp}

\author{\firstname{Hiroshi} \lastname{Nozaki}}
\address{Department of Mathematics Education\\ 
	Aichi University of Education\\
	1 Hirosawa\\ Igaya-cho\\ Kariya\\ Aichi 448-8542\\
	Japan}
\email{hnozaki@auecc.aichi-edu.ac.jp}

\author{\firstname{Koji} \lastname{Tasaka}}
\address{Department of Mathematics\\
	Kindai University\\
	Osaka 577-8502\\
	Japan\\
	Research Institute for Mathematical Sciences\\
	Kyoto University\\
	Kyoto 606-8502\\ 
	Japan}
\email{tasaka@math.kindai.ac.jp,\ tasaka@kurims.kyoto-u.ac.jp}


\thanks{This work is partially supported by JSPS KAKENHI Grant Number 19K03445, 20K03736, 23K03034, 24K06688 and 24K06871.}
\CDRGrant[JSPS KAKENHI]{19K03445}
\CDRGrant[JSPS KAKENHI]{20K03736}
\CDRGrant[JSPS KAKENHI]{23K03034}
\CDRGrant[JSPS KAKENHI]{24K06688}
\CDRGrant[JSPS KAKENHI]{24K06871}

% Keywords and phrases:
\keywords{Finite quaternionic unit group, spherical designs of harmonic index, harmonic strength, Linear programming bound, uniqueness, spherical theta functions, modular forms}
  
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\begin{document}

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%% Abstracts must be placed before \maketitle
\begin{abstract}
For a finite subset $X$ of the $d$-dimensional unit sphere, the harmonic strength $T(X)$ of $X$ is the set of $\ell\in \mathbb{N}$ such that $\sum_{x\in X} P(x)=0$ for all harmonic polynomials $P$ of homogeneous degree $\ell$.
We will study three exceptional finite groups of unit quaternions, called the binary tetrahedral group $2T$ of order 24, the octahedral group $2O$ of order 48, and the icosahedral group $2I$ of order 120, which can be viewed as a subset of the 3-dimensional unit sphere.
For these three groups, we determine the harmonic strength and show the minimality and the uniqueness as spherical designs. 
In particular, the group $2O$ is unique as a minimal subset $X$ of the 3-dimensional unit sphere with $T(X)=\{22,14,10,6,4,2 \}\cup \mathbb{O}^+$, where $\mathbb{O}^+$ denotes the set of all positive odd integers.
This result provides the first characterization of $2O$ from the spherical design viewpoint.

For $G\in \{2T,2O,2I\}$, we consider the lattice $\mathcal{O}_{G}$ generated by $G$ over $R_G$ on which the group $G$ acts by multiplication, where $R_{2T}=\mathbb{Z},\ R_{2O}=\mathbb{Z}[\sqrt{2}],\ R_{2I}=\mathbb{Z}[(1+\sqrt{5})/2]$ are the rings of integers. 
We introduce the spherical theta function $\theta_{G,P}(z)$ attached to the lattice $\mathcal{O}_G$ and a harmonic polynomial $P$ of degree $\ell$ and prove that they are modular forms. 
By applying our results on the characterization of $G$ as a spherical design, we determine the cases in which the $\mathbb{C}$-vector space spanned by all $\theta_{G,P}(z)$ of harmonic polynomials $P$ of homogeneous degree $\ell$ has dimension zero--without relying on the theory of modular forms.
\end{abstract}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Background}
A finite non-empty subset $X$ of the unit sphere $\mathbb{S}^{d-1}$ of the $d$-dimensional Euclidean space $ \R^d$ is a \emph{spherical $t$-design} if it holds that
\begin{equation*}\label{eq:spherical_design}
\frac{1}{|X|} \sum_{\boldsymbol{x}\in X} P(\boldsymbol{x}) = \frac{1}{|\mathbb{S}^{d-1}|} \int_{\boldsymbol{x} \in \mathbb{S}^{d-1}}P(\boldsymbol{x})d\sigma(\boldsymbol{x}) \quad \mbox{for all $P\in \R[x_1,\ldots,x_d]$ of degree $\le t$},
\end{equation*}
where $\sigma$ is the surface measure on $\mathbb{S}^{d-1}$ and $|\mathbb{S}^{d-1}|=\int_{\boldsymbol{x} \in \mathbb{S}^{d-1}}d\sigma(\boldsymbol{x})$.
This concept was first introduced by Delsarte--Goethals--Seidel in \cite{DelsarteGoethalsSeidel77}.
Since then, the problem of determining the maximum value of $t$ for a given finite subset $X$ of the unit sphere, referred to as the \emph{strength} of $X$, and the characterization of such subsets as spherical designs have been studied in connection with various algebraic structures~\cite{BannaiBannai09,BBTZ17}.

We wish to study these problems in the context of spherical designs of harmonic index, which is a slight generalization of spherical $t$-designs.
For $X\subset \R^d$, we define the \emph{harmonic strength} $T(X)$ of $X$ by
\begin{equation}\label{eq:T(X)}
T(X):=\left\{\ell \in \N \ \middle|\  \sum_{\boldsymbol{x}\in X} P(\boldsymbol{x})=0,\ \forall P\in {\rm Harm}_\ell (\R^d)\right\},
\end{equation}
where ${\rm Harm}_\ell (\R^d)$ denotes the $\R$-vector space of harmonic polynomials of homogeneous degree $\ell$.
Note that $X$ is a spherical $t$-design if and only if $1,2,\ldots,t\in T(X)$.
A finite subset $X\subset \mathbb{S}^{d-1}$ such that $T\subset T(X)$ is called a \emph{spherical $T$-design}.
This concept was introduced by Delsarte--Seidel \cite{DS89} as a spherical analogue of the design in association schemes \cite[Section 3.4]{Delsarte:PHD},
and has been actively studied in the past decade (cf.~\cite{BannaiOkudaTagami15,HiraoNozakiTasaka,Miezaki13,MisawaMunemasaSawa,OkudaYu16,Pandey,ZhuBannaiBannaiKimYu17}).

Our fundamental problems are as follows.


\begin{question}\label{Q1}
\begin{itemize}
\item[(i)] Given $X\subset \R^d$, determine $T(X)$.
\item[(ii)] Given $T\subset \N$, characterize finite subsets $X\subset \R^d$ such that $T\subset T(X)$.
\end{itemize}
\end{question}


As an illustrative example of addressing Problem \ref{Q1}, consider a finite subset $X\subset \mathbb{S}^{d-1}$ that is \emph{antipodal}, meaning that $-\boldsymbol{x}\in X$ for all $\boldsymbol{x}\in X$.
In this case, the harmonic index $T(X)$ contains the set $\mathbb{O}^+$ of all positive odd integers, providing a partial solution to (i).
Regarding (ii), a finite subset $X\subset \mathbb{S}^{d-1}$ must necessarily be antipodal if $\mathbb{O}^+\subset T(X)$ (cf.~\cite{MisawaMunemasaSawa}).

Remark that Problem \ref{Q1} (i) is closely connected to the theory of modular forms through spherical theta functions, which was first pointed out by Venkov \cite{Venkov84}.
Regarding Problem \ref{Q1} (ii), we note that for given $t,d\in \N$, the existence of spherical $t$-designs was shown in \cite{SZ84}. 
Bondarenko--Radchenko--Viazovska \cite{BondarenkoRadchenkoViazovska} further showed that for each $N\ge c_d t^{d-1}$, there exists a spherical $t$-design in $\mathbb{S}^{d-1}$ consisting of $N$ points, where $c_d$ is a constant depending only on $d$.

In this paper, we mainly consider a set of certain quaternions as $X$ in Problem \ref{Q1}.
Let us briefly describe it.
Denote by $\mathbb{H}$ the $\R$-algebra of (Hamilton's) \emph{quaternion} with generators $\{i,j\}$ and the relations $i^2=j^2=-1$ and $ij=-ji$.
Taking $\{1,i,j,ij\}$ as an $\R$-basis of $\mathbb{H}$, we will identify $\mathbb{H}$ with $\R^4$ by expressing each quaternion $x\in \mathbb{H}$ as $x=x_1+x_2i+x_3j +x_4 ij$, and then associating it with the vector $\boldsymbol{x}=(x_1,x_2,x_3,x_4)\in \R^4$. 
The subgroup $\mathbb{H}_1:=\{x\in \mathbb{H}\mid N(x)=1\}$ of $\mathbb{H}^\times$, where $N(x)=x_1^2+x_2^2+x_3^2+x_4^2$ is the \emph{norm} of $x$, is then regarded as the unit sphere $ \mathbb{S}^3$ in $\R^4$.

We will start from the following finite subgroups of $\mathbb{H}_1$ (cf.~\cite[Section 3.5]{ConwaySmith}): the \emph{binary tetrahedral} (resp.~\emph{binary octahedral} and \emph{binary icosahedral}) group $2T$ (resp.~$2O$ and $2I$) of order 24 (resp.~48 and 120). 
The group $2I$ is also referred to as the icosian group in \cite[p.207]{ConwaySloan}.
The quotient groups $2T/\{\pm1\},2O/\{\pm1\}$ and $2I/\{\pm1\}$ are isomorphic to $A_4,S_4$ and $A_5$, respectively, which are known as exceptional finite subgroups of the rotation group ${\rm SO}(3)$ of $\R^3$ (cf.~\cite[Proposition 11.5.2]{Voight2021}).
As a subset of $\R^4$, each $G\in \{2T,2O,2I\}$ is contained in $ F_G^4$, where $F_{2T}:=\Q,\ F_{2O}:=\Q\big(\sqrt{2}\big)$ and $F_{2I}:=\Q\big(\sqrt{5}\big)$ are number fields.

For $G\in \{2T,2O,2I\}$, let $\mathcal{O}_G:=\langle G\rangle_{R_G}$ be the $R_G$-algebra generated by $G$, where $R_G$ is the ring of integers of the number field $F_G$.
The ring $\mathcal{O}_G$ becomes a maximal $R_G$-order of the quaternion algebra over $F_G$ (cf.~\cite[Chapter 11]{Voight2021}).
In particular, the $\Z$-order $\mathcal{O}_{2T}$ is known as the \emph{Hurwitz order}.
Table \ref{table:orders} below gives a basis of $\mathcal{O}_G$.

\begin{table}[ht] 
\caption{Orders}
\label{table:orders}
\begin{tabular}{|l|c|c|c|} \hline
Group $G$ & $2T$ & $2O$ & $2I$\\\hline 
Order $|G|$ & 24 & 48 & 120\\\hline
Number field $F_G$ & $\Q$ & $\Q(\sqrt{2})$ & $\Q(\sqrt{5})$\\\hline
Ring of integers $R_G$ & $\Z$ & $\Z[\sqrt{2}]$ & $\Z[\tau]$\\\hline
$R_G$-order $\mathcal{O}_G$ & $\langle 2T\rangle_\Z$ & $\langle 2O\rangle_{\Z[\sqrt{2}]}$ & $\langle 2I\rangle_{\Z[\tau]}$\\\hline
Basis of $\mathcal{O}_G$ & $\{1, i, j , \omega\}$ & $\{1,\alpha,\beta,\alpha\beta\}$ & $\{1 , i, \zeta, i\zeta\}$\\\hline
\end{tabular}
\end{table}

Here we have set
\begin{equation}\label{eq:O_basis_elements}
\begin{aligned}
\tau=\frac{1+\sqrt{5}}{2},\ \omega=\frac{-1+i+j+ij}{2},\ \alpha=\frac{1+i}{\sqrt{2}},\  \beta=\frac{1+j}{\sqrt{2}},\ \zeta=\frac{\tau+\tau^{-1}i+j}{2}.
\end{aligned}
\end{equation}

For a positive integer $m$, we let
\[ \mathcal{O}_{G,m} := \{x\in \mathcal{O}_G\mid \iota_G(N(x))=m\},\]
where $\iota_G:R_G\rightarrow \Z$ is a certain projection defined in \eqref{eq:def_iota}.
A key point is that the composition $\iota_G(N(x))$ defines a positive definite integral quadratic form $\mathcal{Q}_G$ (see the proof of Proposition \ref{prop:counting} for the explicit descriptions).
As such, the theory of modular forms developed by Hecke and Schoeneberg becomes applicable.
Using this framework, we will show that $|\mathcal{O}_{G,m}|>0$ for all $m\in\N$ in Proposition \ref{prop:counting}.
Note that when $G=2T$, the subset $\mathcal{O}_{G,m}\subset \R^4$ for $G= 2T$ coincides with the $m$-shell of the Hurwitz order; that is, the set of lattice points on the sphere with radius $\sqrt{m}$ in $\R^4$.
In contrast, when $G\neq 2T$, the subset $\mathcal{O}_{G,m}\subset \R^4$ does not necessarily lie on a sphere in $\R^4$. 



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Main results}

Let us state our results on Problem \ref{Q1} (i).
For each $G\in\{2T,2O,2I\}$, we will show in Proposition \ref{prop:O_strength} that 
\begin{equation}\label{eq:G_acts_on_O_{G,m}}
T(G)\subset T\big(\mathcal{O}_{G,m}\big), \quad \forall m\in \N.
\end{equation}
This implies that determining the harmonic strength $T(G)$ gives a partial answer to Problem \ref{Q1} (i) for the case $X=\mathcal{O}_{G,m}$.
The first main result of this paper is a complete description of $T(G)$.



 
\begin{theo}\label{thm:harmonic_strength}
The harmonic strength of the finite quaternionic groups $2T, 2O$ and $2I$ are given by 
\begin{align*}
&{\rm (i)}\ T(2T) = \{ 10,4,2\}\cup \mathbb{O}^+, \\ 
&{\rm (ii)}\ T(2O) = \{22,14,10,6,4,2 \}\cup \mathbb{O}^+, \\
&{\rm (iii)}\ T(2I) =  \{ 58,46,38,34,28,26,22,18,16,14,10,8,6,4,2\}\cup \mathbb{O}^+,
\end{align*}
where $\mathbb{O}^+$ denotes the set of all positive odd integers.
\end{theo}

Let us turn to Problem \ref{Q1} (ii).
We first use the \emph{linear programming method} established in \cite{DelsarteGoethalsSeidel77} to obtain a lower bound for the cardinality of finite subsets of $\mathbb{S}^3$ whose harmonic strength includes (more precisely, contains a subset of) $T(G)$.
Our result shows that each $G\in\{2T,2O,2I\}$ is the `minimal' subset of $\mathbb{S}^3$ among those whose harmonic strength contains $T(G)$.  
\begin{theo}\label{thm:LP_bound}
Let $X$ be an antipodal finite subset of $\mathbb{S}^3$.
\begin{itemize}
\item[(i)] If $ T(X)\supset \{10,4,2\} $, then $|X|\ge 24$.
\item[(ii)] If $  T(X)\supset \{14,10,6,4,2\}$, then $|X|\ge 48$.
\item[(iii)] If $T(X)\supset \{10,8,6,4,2\}$, then $|X|\ge 120$.
\end{itemize}
\end{theo}

In the case of spherical $t$-designs, the linear programming method provides a so-called Fisher type bound: if $X\subset \mathbb{S}^{d-1}$ is a spherical $t$-design, then $|X|\ge b_{d,t}$, where $b_{d,t}= \binom{d+e-1}{e}+\binom{d+e-2}{e-1}$ if $t=2e$ and $b_{d,t}= 2\binom{d+e-1}{e}$ if $t=2e+1$.
However, it is not known whether the lower bound obtained by this method provides the maximum lower bound of spherical $t$-designs for a given $t$.
Circumstantially, a tight design, meaning that a spherical $t$-design $X$ that attains the lower bound $|X|=b_{d,t}$, is known to have some good extremal properties \cite{BannaiSloane81,CK07}, if it exists, and its classification has therefore been studied extensively.

In a similar vein, the study of classification of `tight' spherical $T$-designs has attracted a lot of attention (although no general formulation of tightness is known).
It was started in \cite{BannaiOkudaTagami15} for the case $T=\{t\}$.
The case $t = 4$ was also investigated in \cite{OkudaYu16}. 
Zhu et al.~\cite{ZhuBannaiBannaiKimYu17} obtained the classification 
of tight spherical $\{t\}$-designs for $t=6,8$, 
as well as the asymptotic non-existence of tight spherical $\{2e\}$-designs for $e \geq 3$. 
They also studied the existence problem for tight spherical $T$-designs for some $T$, including the case $T = \{8, 4\}$. 

Our classification problem is based on the fact that, for an antipodal spherical $T$-design $X$ of $\mathbb{S}^{d-1}$ with $N$ points, 
its orthogonal transformation $g(X):=\{g(\boldsymbol{x})\mid \boldsymbol{x}\in X\}$ for $g\in O(\R^d)$ is again an antipodal spherical $T$-design of $\mathbb{S}^{d-1}$ with $N$ points, where $O(\R^d):=\{g:\R^d\rightarrow \R^d:\mbox{linear map}\mid\langle g (\boldsymbol{x}),g(\boldsymbol{y})\rangle=\langle\boldsymbol{x},\boldsymbol{y}\rangle, \ \forall \boldsymbol{x},\boldsymbol{y}\in \R^d\}$ is the orthogonal transformation group. 
We say $X$ is {\itshape unique} as an antipodal spherical $T$-design of $\mathbb{S}^{d-1}$ with $N$ points if every antipodal spherical $T$-design of $\mathbb{S}^{d-1}$ with $N$ points is an orthogonal transformation of $X$.

\begin{theo}\label{thm:uniqueness}
\begin{itemize}
\item[(i)] The group $2T$ is unique as an antipodal spherical $\{10,4,2\}$-design of $\mathbb{S}^3$ with $24$ points.
\item[(ii)] The group $2O$ is unique as an antipodal spherical $\{14,10,6,4,2\}$-design of $\mathbb{S}^3$ with $48$ points.
\item[(iii)] The group $2I$ is unique as an antipodal spherical $\{10,8,6,4,2\}$-design of $\mathbb{S}^3$ with $120$ points.
\end{itemize}
\end{theo}

Theorem \ref{thm:uniqueness} is referred to as a uniqueness theorem in the study of the classification of spherical designs.
Remark that Theorem \ref{thm:uniqueness} (i) is obtained from Theorem 1.1 in our previous work \cite{HiraoNozakiTasaka}; we have shown the uniqueness of the normalized $D_4$ root system $\frac{1}{\sqrt{2}}{\bf D}_4$, all permutations of $\frac{1}{\sqrt{2}}(\pm1,\pm1,0,0)$, as an antipodal spherical $\{10,4,2\}$-design of $\mathbb{S}^3$ with 24 points. 
Indeed, the set $2T$ is an orthogonal transformation of $\frac{1}{\sqrt{2}}{\bf D}_4$ (see Remark \ref{rem:D_4}).
Theorem \ref{thm:uniqueness} (iii) is also a consequence of 
the uniqueness of the 600-cell as a spherical $11$-design of $\mathbb{S}^3$ with 120 points shown in \cite{BoyvalenkovDanev01}.
In this paper, we will prove Theorem \ref{thm:uniqueness} (ii) only.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Spherical theta functions}
Let $G\in\{2T,2O,2I\}$. 
We will also consider a connection with the \emph{spherical theta function} $\theta_{G,P}(z) $ defined for $P\in {\rm Harm}_\ell(\R^4)$ by
\begin{equation}\label{eq:theta} 
\theta_{G,P}(z) := \sum_{x\in \mathcal{O}_{G}} P(x) q^{\iota_G(N(x))} \quad (q=e^{2\pi iz}).
\end{equation}
This function is holomorphic on the complex upper half-plane and becomes an elliptic modular form of weight $2+\ell$ for $\Gamma_0(2)$ when $G=2T$, of weight $4+\ell$ for $\Gamma_0(2)$ when $G=2O$, and of weight $4+\ell$ for ${\rm SL}_2(\Z)$ when $G=2I$, as shown by Hecke and Schoeneberg (cf.~\cite[\S3.1]{Ebeling}, \cite[\S4.9]{Miyake}, \cite[\S6]{Ogg}).
The spherical theta function can be written as
\begin{equation*}\label{eq:q_expansion}
\theta_{G,P}(z) = \sum_{m\ge0} \bigg(\sum_{x\in\mathcal{O}_{G,m}} P(x)\bigg)q^{m}.
\end{equation*} 
From this, we see that, for $\ell\in \N$, the condition $\ell \in T\big(\mathcal{O}_{G,m}\big)$ is equivalent to the statement that the coefficient of $q^{m}$ vanishes in each of modular forms $\theta_1,\ldots,\theta_v$, where $\theta_1,\ldots,\theta_v$ form a basis of the $\C$-vector space 
\begin{equation*}\label{eq:def_V} 
\Theta(G,\ell):= \langle \theta_{G,P}(z) \mid P\in {\rm Harm}_\ell(\R^4)\rangle_\C,
\end{equation*}
which is a subspace of the $\C$-vector space of modular forms.
In particular, if $\dim_\C \Theta(G,\ell)=0$, then $\ell \in T(\mathcal{O}_{G,m})$ holds for all $m\in \N$.
This idea was employed in \cite{Pache05, Venkov84} to determine which degrees are included in the harmonic strength of the shells of a lattice. In fact, by applying the theory of modular forms to estimate the dimension of the space of spherical theta functions of a lattice, they partially determined the harmonic strength of the shells of several important lattices.

In our case, we obtain the following relationship between the harmonic strength $T(G)$ and the dimension of the space $\Theta(G,\ell)$.

\begin{theo}\label{thm:dim=0}
For $G\in\{2T,2O,2I\}$, we have that $\dim_\C \Theta(G,\ell)=0$ if and only if $\ell \in T(G)$.
\end{theo}

It is worth emphasizing that, combining Theorems \ref{thm:harmonic_strength} and \ref{thm:dim=0}, we can determine all $\ell\in \N$ for which the space $\Theta(G,\ell)$ is zero-dimensional.
Our proof does not rely on the theory of modular forms.

We remark that the problem whether the space of modular forms is generated by spherical theta functions is a version of the basis problem as raised by Eichler \cite{Eichler} (see also \cite{Martin} and references therein). 
While they use all classes of lattices with the same genus, we focus on the spherical theta functions associated with a single lattice, from the viewpoint of spherical designs.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Contents}
In Section 2, we give explicit descriptions of the groups $2T$, $2O$, and $2I$.
We then recall the notion of spherical designs and present a method, using Gegenbauer polynomials and distance distributions, to determine whether $\ell \in T(X)$ for a given finite subset $X \subset \mathbb{S}^{d-1}$.
Section 3 is devoted to the proofs of our main results: Theorems \ref{thm:harmonic_strength}, \ref{thm:LP_bound}, and \ref{thm:uniqueness}.
In Section 4, we first describe the quadratic form $\mathcal{Q}_G$ obtained from $\iota_G(N(x))$ and compute the set $\mathcal{O}_{G,1}$ to prove that $|\mathcal{O}_{G,m}| > 0$ for all $m \in \mathbb{N}$, using the theory of modular forms.
Theorem \ref{thm:dim=0} is proved based on the computation of $\mathcal{O}_{G,1}$.
Additional data on $\Theta(G, \ell)$ such as a dimension conjecture will be given.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Preliminaries}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Exceptional quaternionic unit groups}
We provide a brief overview of quaternions from the books \cite{ConwaySmith,Coxeter}.
Let $\mathbb{H}=\{x_1+x_2i+x_3j +x_4k \mid x_1,\ldots,x_4\in \R\}$ be the $\R$-algebra of quaternions, where $1,i,j,k$ are an $\R$-basis such that $i^2=j^2=-1$ and $ij=k=-ji$.
Its (non-commutative) multiplication is given for $x=x_1+x_2i+x_3j+x_4k$ and $y = y_1+y_2i +y_3j+ y_4k$ by
\begin{equation}\label{eq:product}
\begin{aligned}
xy&= (x_1 y_1 - x_2 y_2 - x_3 y_3 - x_4 y_4)+( x_2 y_1 + x_1 y_2 - x_4 y_3 + x_3 y_4)i\\
&+( x_3 y_1 + x_4 y_2 + x_1 y_3 - x_2 y_4)j+ ( x_4 y_1 - x_3 y_2 + x_2 y_3 + x_1 y_4)k.
\end{aligned}
\end{equation}
For $x=x_1+x_2i+x_3j+x_4k\in \mathbb{H}$, denote by $\bar{x}=x_1-x_2i-x_3j- x_4k$ the \emph{conjugate} of $x$ and by 
$N(x)=x\bar{x}=
x_1^2+x_2^2+x_3^2+x_4^2$
the (reduced) \emph{norm} of $x$.
The norm map $N: \mathbb{H}^\times\rightarrow \R_{>0}^\times$ is a surjective group homomorphism.



The set of unit quaternions $\mathbb{H}_1=\{\varepsilon \in \mathbb{H}\mid N(\varepsilon)=1\}$ forms a subgroup of $\mathbb{H}^\times$.
Let $\varepsilon \in \mathbb{H}_1$. 
We note that $\varepsilon^{-1}=\bar{\varepsilon}$.
For $x\in {\rm Im}\ \mathbb{H}:=\{x_2i+x_3j+x_4k\mid x_2,x_3,x_4\in \R\}$, we see that $\varepsilon x \varepsilon^{-1} \in {\rm Im}\, \mathbb{H}$.
The matrix $Z_\varepsilon$ defined by the coefficient matrix $(\varepsilon i\varepsilon^{-1},\varepsilon j \varepsilon^{-1},\varepsilon k \varepsilon^{-1})=(i,j,k) Z_\varepsilon$ is a rotation matrix, i.e. $Z_\varepsilon\in {\rm SO}(3):=\{A\in {\rm O}(3)\mid  \det A=1\}$.
From this, the map
\[ \rho:  \mathbb{H}_1 \rightarrow {\rm SO}(3), \quad \varepsilon \mapsto Z_\varepsilon\]
is a homomorphism.
Moreover, the map $\rho$ is surjective.
Its kernel is $\{\pm 1\}$ (\cite[Corollary 2.4.21]{Voight2021}).
Namely, we obtain
\[\mathbb{H}_1/\{\pm1\}\cong {\rm SO}(3).\]
Thus, for any finite subgroup $G$ of $\mathbb{H}_1$ with $-1\in G$, we have the embedding $G/\{\pm 1\}\rightarrow {\rm SO}(3)$.
Here we recall that every finite subgroup of ${\rm SO}(3)$ is completely classified as follows (cf.~\cite[Proposition 11.5.2]{Voight2021}): 
\begin{itemize}
\item[(i)] a cyclic group; 
\item[(ii)] a dihedral group;
\item[(iii)] the tetrahedral group $A_4$ of order 12;
\item[(iv)] the octahedral group $S_4$ of order 24; or
\item[(v)] the icosahedral group $A_5$ of order 60.
\end{itemize}



Finite subgroups of $\mathbb{H}_1$ whose images under the map $\rho$ are isomorphic to the cases (iii), (iv) and (v) above are called the binary tetrahedral group $2T$ of order 24, the binary octahedral group $2O$ of order 48 and the binary icosahedral group $2I$ (of order 120), respectively.
Their generators can be found in \cite[Chapter 6.5]{Coxeter}.
We follow the definition given in \cite[Section 3.5]{ConwaySmith}.

The representations of the groups $2T$, $2O$, and $2I$ we use are as follows.
Denote by $Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ the \emph{quaternion group} of order 8.
For a subgroup $G\subset \mathbb{H}^\times$, we write 
\[xG=\{x\varepsilon \mid \varepsilon \in G\}\]
for the $G$-orbit of $x\in \mathbb{H}$.
Note that $|xG|=|G|$ for $x\in \mathbb{H}^\times $.
With these notations, the exceptional quaternionic unit groups $2T$, $2O$ and $2I$ can be written as follows:
\begin{equation}\label{eq:def_exceptional_groups}
\begin{aligned}
2T&=Q_8 \cup \omega\,Q_8 \cup \omega^2\,Q_8,\\
2O&=2T \cup \alpha\, 2T,\\
2I&=2T \cup \zeta\, 2T\cup \zeta^2\, 2T \cup \zeta^3\, 2T \cup \zeta^4\, 2T,
\end{aligned}
\end{equation}
where $\omega,\alpha,\zeta$ are defined in \eqref{eq:O_basis_elements}.
Note that these elements are units: $\omega^3=1$ and $\alpha^4=\zeta^5=-1$.
The description given in \eqref{eq:def_exceptional_groups} coincides with that in \cite[Section 3.5]{ConwaySmith}, which can be verified using symbolic computations performed with the help of the computer software Mathematica.

For the latter purpose, we recall that the multiplication by a quaternion yields an orthogonal transformation.

\begin{lemm}\label{lem:quaternion_orthogonal}
For $x\in \mathbb{H}^\times$, the $\R$-linear map 
\[ m_x:\R^4\rightarrow \R^4, \ y\mapsto \frac{x}{\sqrt{N(x)}} y\] 
is an orthogonal transformation, i.e. $m_x\in O(\R^4)$.
Here we are identifying $\mathbb{H}$ with $\R^4$ as $\R$-vector spaces by taking $(y_1,y_2,y_3,y_4)\in \R^4$ for $y=y_1+y_2i+y_3j+y_4k\in \mathbb{H}$. 
\end{lemm}
\begin{proof}
For $x=x_1+x_2i+x_3j+x_4k\in \mathbb{H}^\times$, let
\begin{equation}\label{eq:M_x}
M_x:=\frac{1}{\sqrt{N(x)}} \begin{pmatrix}x_1&x_2&x_3&x_4 \\ -x_2&x_1&x_4 &-x_3\\ -x_3&-x_4 & x_1 & x_2\\ -x_4 & x_3 & -x_2 & x_1\end{pmatrix}.
\end{equation}
From \eqref{eq:product} we obtain
\[\frac{1}{\sqrt{N(x)}}xy=yM_x ,\quad (\forall y\in \mathbb{H}),\]
where $yM_x$ is the matrix product under the identification $\mathbb{H}$ with $\R^4$.
Since $M_x$ is an orthogonal matrix, we obtain the desired result.
\end{proof}

Note that the map
\[\mathbb{H}_1 \longrightarrow O(4),\ x\longmapsto M_x\]
is an injective group homomorphism, where $O(4)$ is the orthogonal group.


\begin{rema}\label{rem:D_4}
In our previous paper \cite{HiraoNozakiTasaka}, we have defined the $D_4$ root system ${\bf D}_4$ as the set of all permutations of $(\pm1,\pm1,0,0)$.
Using the identification of $\mathbb{H}$ with $\R^4$, one can check that $\frac{1}{\sqrt{2}}{\bf D}_4=\alpha 2T=m_\alpha(2T)$, where $\alpha$ is given in \eqref{eq:O_basis_elements}.
From Lemma \ref{lem:quaternion_orthogonal}, we see that the set $\frac{1}{\sqrt{2}}{\bf D}_4$ is an orthogonal transformation of $2T$.
\end{rema}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Spherical design}
The concept of spherical $T$-designs introduced by Delsarte and Seidel \cite{DS89} applies to finite subsets of the unit sphere ${\mathbb S}^{d-1}: = \{ \boldsymbol{x} = (x_1, \ldots, x_d) \in \R^{d} \mid \langle \boldsymbol{x}, \boldsymbol{x} \rangle = 1 \}$ in the $d$-dimensional Euclidean space $\R^d$, where for $\boldsymbol{x},\boldsymbol{y}\in \R^d$, its inner product is denoted by
\[ \langle \boldsymbol{x}, \boldsymbol{y} \rangle := \sum_{a = 1}^{d} x_ay_a.\]
Under the identification $\mathbb{H}$ with $\R^4$, for $x\in \mathbb{H}$ we have $N(x)=\langle \boldsymbol{x}, \boldsymbol{x} \rangle $, and hence, $\mathbb{S}^3=\mathbb{H}_1$.

For an integer $\ell\ge0$, denote by ${\rm Hom}_{\ell}(\R^d)$ the $\R$-vector subspace of $\R[x_1, \ldots, x_d]$ consisting of all homogeneous polynomials of degree $\ell$ in $d$ variables.
We write
\[\Delta:= \frac{\partial^2}{\partial x_1^2}+\cdots +\frac{\partial^2}{\partial x_d^2}\]
for the Laplacian operator.
Let 
\[{\rm Harm}_{\ell}(\R^d) :=\{P\in {\rm Hom}_{\ell}(\R^d)\mid \Delta P=0\}\]
be the $\R$-vector space of harmonic polynomials of homogeneous degree $\ell$.
It is well known (see e.g. Theorem 3.2 in \cite{DelsarteGoethalsSeidel77}) that $\dim_\R {\rm Harm}_\ell(\R^d)= \binom{\ell+d-1 }{\ell} - \binom{\ell+d-3}{\ell -2}$ holds for $\ell,d\in \N$ with $\ell\ge2$.

\begin{defi}\label{def:T-design}
For a subset $T$ of $\N$, a non-empty finite subset $X$ of $\mathbb{S}^{d-1}$ is called a {\itshape spherical $T$-design}
if for any $\ell \in T$ and $P \in {\rm Harm}_\ell (\R^d)$ we have that
\[\sum_{\boldsymbol{x} \in X} P(\boldsymbol{x}) = 0.\]
\end{defi}


Alternatively, a finite subset $X\subset \mathbb{S}^{d-1}$ is a spherical $T$-design if $T\subset T(X)$, where $T(X)$ is the harmonic strength of $X$ defined in \eqref{eq:T(X)}.



For a subset $X$ of $\R^d$, we write $-X:=\{-\boldsymbol{x}\mid \boldsymbol{x}\in X\}$. 
A set $X$ is said to be \emph{antipodal} if we have $-X=X$. 
If a finite subset $X\subset \mathbb{S}^{d-1}$ is antipodal, then the harmonic strength $T(X)$ contains all positive odd integers.
Conversely, for a finite subset $X\subset \mathbb{S}^{d-1}$, if $T(X)$ contains all positive odd integers, then $X$ is antipodal (cf.~\cite{MisawaMunemasaSawa}).

For an antipodal finite subset $X$ of $\R^d$, a subset $X'\subset X$ is called \emph{a half set of $X$} if $X$ is a disjoint union of $X'$ and $-X'$; $X'\sqcup (-X')=X$.
It follows that
\begin{equation}\label{eq:antipodal_cardinality}
|X|=2|X'|.
\end{equation}
For any antipodal finite subset $X$ of $\mathbb{S}^{d-1}$ (note that $\boldsymbol{0}\not\in X$), a half set of $X$ always exists, but not unique.
The harmonic strength of the half set satisfies $T(X') \cap 2\Z = \{2\ell \in \N\mid 2\ell\in T(X)\}$ (see e.g. \cite[Lemma 2.2]{HiraoNozakiTasaka}).
It should be noted that $T(X')$ may contain odd integers (see \cite{BannaiZhaoZhuZhu18}).


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Computing the harmonic strength}
We now explain our method to compute the harmonic strength defined in \eqref{eq:T(X)}.
We use the Gegenbauer (or ultraspherical) polynomials, which form a class of orthogonal polynomials (cf.~\cite[p.302]{AndrewsAskeyRoy} and \cite[(4.7.23)]{S75}).
For $\lambda>0$, the Gegenbauer polynomial $C_\ell^\lambda(s)\in \R[s]$ of degree $\ell$ is defined by 
the coefficient of $u^\ell$ in the formal power series $\Phi^{\lambda}(s;u):=(1-2su+u^2)^{-\lambda}$:
\[
\Phi^{\lambda}(s;u)=\sum_{\ell \geq 0}C_\ell^\lambda(s)u^\ell.
\]
It is also obtained by the recurrence relation $\ell C_\ell^\lambda(s) = 2(\ell+\lambda-1)s C_{\ell-1}^\lambda(s) -(\ell+2\lambda-2)C_{\ell-2}^\lambda(s)$ for $\ell\ge2$ with $C_0^\lambda(s)=1,\ C_1^\lambda(s)=2\lambda s$.
For example, the case $\ell=2$ is $C_2^{\lambda}(s)=-\lambda +2\lambda (1+\lambda) s^2$.
It follows that $C_\ell^\lambda(1) = \frac{2\lambda (2\lambda+1)(2\lambda+2)\cdots (2\lambda+\ell-1)}{\ell!}$.

\begin{lemm}\label{lem:gegenbauer}
Let $X$ be a finite subset of $\mathbb{S}^{d-1}$.
For $\ell\in \N$, we have that
\[\sum_{\boldsymbol{x},\boldsymbol{y}\in X} C_{\ell}^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y}\rangle ) =0\]
if and only if $\ell\in T(X)$, where $T(X)$ is the harmonic strength of $X$ defined in \eqref{eq:T(X)}.
\end{lemm}
\begin{proof}
Write $c_{\ell,d}:=\dim_\R {\rm Harm}_\ell(\R^d)$.
Let $\{S_{\ell,n}^{(d)}\mid n=1,2,\ldots,c_{\ell,d} \}$ be an orthonormal basis of ${\rm Harm}_\ell(\R^d)$ with respect to the inner product defined for 
$F_1,F_2\in {\rm Harm}_\ell(\R^d)$ by 
\[( F_1,F_2) :=\frac{1}{|\mathbb{S}^{d-1}|} \int_{\boldsymbol{x}\in \mathbb{S}^{d-1}} F_1(\boldsymbol{x})F_2(\boldsymbol{x})d\sigma(\boldsymbol{x}).\]


From the addition theorem \cite[Theorem 9.6.3]{AndrewsAskeyRoy}, for $\ell\in \N$ and $\boldsymbol{x},\boldsymbol{y}\in \mathbb{S}^{d-1}$, one has
\[ \sum_{n=1}^{c_{\ell,d}}S_{\ell,n}^{(d)}(\boldsymbol{x})S_{\ell,n}^{(d)}(\boldsymbol{y}) = \frac{c_{\ell,d}}{C_\ell^{(d-2)/2}(1)} C_\ell^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y}\rangle).\]
Let $Z_n := \sum_{\boldsymbol{x}\in X}S_{\ell,n}^{(d)}(\boldsymbol{x})\in \R$.
Then we have
\[\sum_{n=1}^{c_{\ell,d}}Z_n^2= \sum_{\boldsymbol{x},\boldsymbol{y}\in X}\sum_{n=1}^{c_{\ell,d}}S_{\ell,n}^{(d)}(\boldsymbol{x})S_{\ell,n}^{(d)}(\boldsymbol{y}) = \frac{c_{\ell,d}}{C_\ell^{(d-2)/2}(1)}  \sum_{\boldsymbol{x},\boldsymbol{y}\in X} C_{\ell}^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y}\rangle )  .\]
Thus, the result follows from the definition: $\ell \in T(X)$ if and only if $Z_n=0$ holds for all $n=1,2,\ldots, c_{\ell,d}$.
\end{proof}


\begin{prop} \label{prop:gene}
Let $d\geq 3$.
For a finite subset $X$ of $\mathbb{S}^{d-1}$ and $\ell\in \N$, we have that $\ell \in T(X)$ if and only if 
the coefficient of $u^\ell$ in $\sum_{\boldsymbol{x},\boldsymbol{y} \in X}\Phi^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y} \rangle;u)$ is zero. 
\end{prop}
\begin{proof}
    Since 
\begin{align*}
\sum_{\boldsymbol{x},\boldsymbol{y} \in X}\Phi^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y} \rangle;u)
&=
\sum_{\boldsymbol{x},\boldsymbol{y} \in X} \sum_{\ell\geq 0} C_\ell^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y}  \rangle) u^\ell   \\
&=
 \sum_{\ell\geq 0} \left(\sum_{\boldsymbol{x},\boldsymbol{y} \in X}C_\ell^{(d-2)/2}(\langle \boldsymbol{x}, \boldsymbol{y} \rangle)\right) u^\ell, 
\end{align*}
the result follows from Lemma \ref{lem:gegenbauer}.
\end{proof}

The power series $\sum_{\boldsymbol{x},\boldsymbol{y} \in X}\Phi^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y} \rangle;u)\in \R[[u]]$ as in Proposition \ref{prop:gene} can be computed from the \emph{distance distribution} $R_X:=\{(s, A_s(X))\mid s \in A(X)\cup\{1\}\}$ of $X$, where for $s\in \R$, we write
\begin{equation}\label{eq:d_a}
A_s(X):=|\{(\boldsymbol{x},\boldsymbol{y})\in X\times X\mid \langle \boldsymbol{x},\boldsymbol{y}\rangle=s\}|
\end{equation}
and 
\begin{equation}\label{eq:A(X)}
A(X):=\{\langle \boldsymbol{x},\boldsymbol{y}\rangle \mid \boldsymbol{x},\boldsymbol{y}\in X, \ \boldsymbol{x}\neq\boldsymbol{y}\}
\end{equation}
denotes the set of inner products of two distinct elements in $X\subset \mathbb{S}^{d-1}$.
Note that $A_1(X)=|X|$.
With this, one has
\begin{equation*}\label{eq:Phi_A(X)}
\sum_{\boldsymbol{x},\boldsymbol{y} \in X} \Phi^{(d-2)/2}(\langle \boldsymbol{x},\boldsymbol{y} \rangle;u)=\sum_{s\in A(X)\cup\{1\}}A_s(X) \Phi^{(d-2)/2}(s;u).
\end{equation*}
Our strategy is to express the above right-hand side in terms of simple rational functions in $u$ and check the conditions under which the coefficient of $u^\ell$ vanishes.

Below, we only use the case $d=4$. In this case, we simply write
\[\Phi(s;u):=\Phi^{1}(s;u)=\frac{1}{1-2su+u^2}.\]



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proofs of main results}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem \ref{thm:harmonic_strength}}
To prove Theorem \ref{thm:harmonic_strength}, we use the theory of invariant polynomials.
Let ${\rm Hom}_{\ell}(\C^2)$ denote the $\C$-vector subspace of $\C[z_1, z_2]$ consisting of all homogeneous polynomials of degree $\ell$ in two variables.
Let us recall an action of the group $\mathbb{H}_1$ on ${\rm Hom}_\ell (\C^2)$.

We first note that a quaternion $x \in \mathbb{H}$ can be written as $x=z_1+z_2j $ for some $z_1,z_2\in \C$:
\[x = x_1 + x_2 i + x_3 j + x_4 k = (x_1 + x_2 i) + (x_3 + x_4 i) j.\]
For $x=z_1+z_2j\in \mathbb{H}$, we set 
\[ C_x := \begin{pmatrix} z_1&z_2\\ -\overline{z_2}&\overline{z_1}\end{pmatrix},\]
where $\overline{z}$ denotes the complex conjugate for $z\in \C$.
The matrix $C_x$ lies in the special unitary group $SU(2)=\{A\in U(2)\mid \det A=1\}$ if $x\in \mathbb{H}_1$.
Moreover, the map
\begin{equation}\label{eq:map_C_x}
\mathbb{H}_1\longrightarrow SU(2),\ \varepsilon\longmapsto C_{\varepsilon}
\end{equation}
is an injective group homomorphism.
This shows that for $\varepsilon\in \mathbb{H}_1$ and $ P\in {\rm Hom}_{\ell}(\C^2)$, setting
\[ (\varepsilon P)(z_1,z_2):=P\big( (z_1,z_2)C_\varepsilon\big)\]
we obtain an action of $\mathbb{H}_1$ on the space ${\rm Hom}_{\ell}(\C^2)$.

Let $G$ be a finite subgroup of $\mathbb{H}_1$, which can be viewed as a subgroup of $SU(2)$ via \eqref{eq:map_C_x}.
Denote by ${\rm Hom}_{\ell}(\C^2)^G=\{P\in {\rm Hom}_{\ell}(\C^2)\mid  \varepsilon P=P, \ \forall \varepsilon \in G\}$ the $\C$-vector space of $G$-invariant homogeneous polynomials, and by
\[\Psi_G(u):=\sum_{\ell\geq 0} \dim_\C {\rm Hom}_\ell (\C^2)^G u^\ell\]
the \emph{Molien series} (or Hilbert--Poincar\'{e} series) of $G$.
It is well known (cf.~\cite[Theorem 4.3.2]{Smith}) that the Molien series $\Psi_G(u)$ can be calculated by
\[  \Psi_G(u)= \frac{1}{|G|} \sum_{\varepsilon \in G} \frac{1}{\det (I-uC_\varepsilon)}.\]


We now prove that, for a finite subgroup $G$ of $\mathbb{H}_1$, the formal power series $(1/|G|^2)\sum_{x,y \in G} \Phi(\langle x,y \rangle;u)$ coincides with the Molien series $\Psi_G(u)$ of $G$. 

\begin{lemm}\label{lem:SU(2)}
Let $G$ be a finite subgroup of $\mathbb{H}_1$. Then we have  
\[\Psi_G(u)= \frac{1}{|G|^2}\sum_{x,y \in G} \Phi(\langle x,y \rangle; u),\]
where the inner product $\langle x,y \rangle$ is given by $\langle x_1+x_2 i+x_3 j +x_4 k,y_1+y_2i+y_3 j+y_4 k \rangle=x_1y_1+x_2y_2+x_3y_3+x_4y_4$ via the identification $\mathbb{H}$ with $\R^4$.
\end{lemm}
\begin{proof}
From Lemma \ref{lem:quaternion_orthogonal}, for $x,y\in \mathbb{H}_1$, we have $\langle x,y\rangle = \langle 1,x^{-1}y\rangle$.
Thus, it follows that 
\begin{align*}
\frac{1}{|G|^2}\sum_{x,y \in G} \Phi(\langle x,y \rangle; u)&=
\frac{1}{|G|}\sum_{x \in G} \frac{1}{1-2\langle 1,x \rangle u+u^2}    \\
&=\frac{1}{|G|}\sum_{x\in G} \frac{1}{1-2x_1 u+u^2} \\
&= \frac{1}{|G|}\sum_{x \in G}\frac{1}{\det (I-u C_x)},
\end{align*}
where $\det (I-u C_x)=1-2x_1 u+u^2$ is easily verified. 
\end{proof}

For a finite subgroup $G\subset \mathbb{H}_1$ viewed as a finite subgroup of $SU(2)$, the Molien series $\Psi_G(u)$ is explicitly calculated as follows \cite[Section 4.5]{Sbook}: 
\begin{table}[ht]
    \centering
    \caption{Molien series of $G$}
    \label{tab:molien}
    \begin{tabular}{|c|c|} \hline
    $G$ & $\Psi_G(u)$
    \\ \hline \hline 
       $C_n$  &  $\displaystyle{\frac{1-u^{2n}}{(1-u^2)(1-u^n)^2}=\frac{1+u^{n}}{(1-u^2)(1-u^n)}}$\\ \hline 
       $D_{2n}$ &$\displaystyle{\frac{1-u^{4n+4}}{(1-u^4)(1-u^{2n})(1-u^{2n+2})}=\frac{1+u^{2n+2}}{(1-u^4)(1-u^{2n})}}$  \\ \hline
       $2T$ & $\displaystyle{\frac{1-u^{24}}{(1-u^6)(1-u^8)(1-u^{12})}=\frac{1+u^{12}}{(1-u^6)(1-u^8)}}$\\ \hline 
       $2O$ & $\displaystyle{\frac{1-u^{36}}{(1-u^8)(1-u^{12})(1-u^{18})}=\frac{1+u^{18}}{(1-u^8)(1-u^{12})}}$\\ \hline 
       $2I$ & $\displaystyle{\frac{1-u^{60}}{(1-u^{12})(1-u^{20})(1-u^{30})}=\frac{1+u^{30}}{(1-u^{12})(1-u^{20})}}$\\ \hline 
    \end{tabular}
\end{table}

Here $C_n$ and $D_{2n}$ are the cyclic and dihedral group of order $n$ and $4n$, respectively.
A quaternionic representation is given as follows:
\[C_{n}:=\{e^{2m\pi i/n}\mid m=0,1,\ldots, n-1\}, \quad D_{2n}:=C_{2n} \cup C_{2n}',\]
where we set $e^{i\theta}=\cos \theta + i \sin \theta$ and $C_{2n}':=\{e^{m\pi i/n}j \mid m=0,1\ldots, 2n -1\}$. 

\begin{proof}[Proof of Theorem \ref{thm:harmonic_strength}]
       We compute the generating series in Proposition \ref{prop:gene}.
       As a prototype, we give the proof for the case $X=2T$ (Theorem \ref{thm:harmonic_strength} (i)); the proofs for the other cases are similar.

    From Lemma \ref{lem:SU(2)} and Table \ref{tab:molien}, one has
    \begin{align*}
        \frac{1}{|2T|^2}\sum_{x,y \in 2T} \Phi(\langle x, y \rangle;u)&=\frac{1+u^{12}}{(1-u^6) (1-u^8)}.
    \end{align*}
    
    We now determine where the coefficient of $u^\ell$ in the above power series is zero.
    Let $B_{\ell}=\{(a,b)\in \Z_{\ge0}^2\mid 3a+4b=\ell\}$.
    Then we have
    \[ \frac{1}{(1-u^6)(1-u^8)} = \sum_{a,b\ge0}u^{6a+8b}=\sum_{\ell\ge0}|B_\ell|u^{2\ell}.\]
    We will show that $|B_\ell| =0$ if and only if $\ell=1,2,5$, 
    from which by Proposition \ref{prop:gene} the result 
    $T(2T) = \{ 10,4,2\}\cup \mathbb{O}^{+}$ 
    follows.
    For $\ell\in \N$, suppose $|B_\ell|>0$. Then there exists $(a,b)\in  \Z_{\ge0}^2$ such that $\ell= 3a+4b=3(a+b)+b$.
    Therefore, $|B_\ell|>0$ holds if and only if $\ell \in S_0\cup S_1\cup S_2$ holds, where, for $b\in\{0,1,2\}$, we let $S_b=\{3a+b\in \Z \mid a\ge b \}$.
    Since $\N\setminus \big(S_0\cup S_1\cup S_2\big)=\{1,2,5\}$, we have done.

 By the same argument as for $X=2T$, we can find all degree $\ell$ at which the coefficient of $u^\ell$ becomes zero in the power series given in Table \ref{tab:molien}.
    The detailed verification is left to the reader. 
\end{proof}


The harmonic strengths of the groups $C_n$ and $D_{2n}$ can also be obtained from Lemma~\ref{lem:SU(2)} and Table~\ref{tab:molien}.
We state the results without providing proofs.

\begin{theo}\label{thm:harmonic_strength_Dih}
The harmonic strengths of the cyclic group $C_n$ and the dihedral group $D_{2n}$ are given by
\[
T(C_n)= 
\begin{cases}
    \mathbb{O}^+ & \text{if $n$ is even,} \\ % (n\geq 2),  
    \{\ell \in \mathbb{O}^+ \mid \ell < n\} & \text{if $n$ is odd $(n \geqslant 2)$},
\end{cases}
\]
and 
\[T(D_{2n}) =\{2\ell \in \N \mid 0<\ell < n, \ \ell \equiv 1\bmod 2  \} \cup 
\mathbb{O}^{+}\qquad  (n\geq 2). 
\]
\end{theo}


For subgroups $G$ of $\mathbb{H}_1$, the formal power series $\sum_{x,y \in G} \Phi(\langle x,y \rangle; u)$ can be written explicitly as shown in Lemma~\ref{lem:SU(2)}.
In general, the crucial point in this approach is that the series $\sum_{x,y \in G} \Phi(\langle x,y \rangle; u)$ can be rewritten in a sufficiently simple form to determine the non-vanishing of its coefficients.
However, finding such a simple form is generally difficult, as the coefficients of this series may not be integers.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem \ref{thm:LP_bound}}
In order to give a lower bound on the cardinality of a spherical $T$-design, we recall a useful inequality due to Delsarte--Goethals--Seidel \cite{DelsarteGoethalsSeidel77}.

Let 
\[ Q_\ell^{(d)}(s):=\frac{d+2\ell-2}{d-2} C_{\ell}^{(d-2)/2}(s)\]
be the scaled Gegenbauer polynomial used in \cite{DelsarteGoethalsSeidel77}.
With this normalization, we have $Q_\ell^{(d)}(1)=\dim_\R {\rm Harm}_\ell(\R^d)$.
As examples, one has
\begin{equation}\label{eq:Q_l^{d}}
\begin{aligned}
Q_2^{(d)}(s) &=\frac{1}{2} (d+2) \left(d s^2-1\right),\quad Q_3^{(d)}(s) =\frac{1}{6} d (d+4) s \left((d+2) s^2-3\right),\\
Q_4^{(4)}(s) &=5\left( 16 s^4-12 s^2+1\right),\quad Q_6^{(4)}(s) =7 \left(64 s^6-80 s^4+24 s^2-1\right).
\end{aligned}
\end{equation}


Hereafter, we fix $d=4$.
Recall that the polynomials $Q_\ell(s)=Q_\ell^{(4)}(s)$ form a basis of the polynomial algebra $\R[s]$, and every polynomial $F(s)\in \R[s]$ of degree $r$ can be written uniquely in the form
\begin{equation}
\label{eq:F}
F(s) = \sum_{\ell = 0}^{r} f_\ell Q_{\ell} (s),
\end{equation}
called the \emph{Gegenbauer expansion} of $F(s)$.
Note that, since the polynomials $Q_\ell(s)$ are the orthogonal polynomials on the closed interval $[-1, 1]$ with respect to the inner product of the weight function $(1 - s^2)^{1/2}$, the coefficient $f_\ell\in \R$ given in \eqref{eq:F} can be computed from
\[
f_\ell = \frac{ \int_{-1}^{1} F(s) Q_{ \ell} (s) (1 - s^2)^{\frac12} \; ds}{ \int_{-1}^{1} Q_\ell(s)^2(1 - s^2)^{\frac12} \; ds}.
\]


\begin{lemm}\label{lem:LP_bound}
Let $X\subset \mathbb{S}^3$ be a spherical $T$-design and $F(s)\in \R[s]$ of degree $r$.
Suppose that the Gegenbauer expansion \eqref{eq:F} of $F(s)$ satisfies 
 $f_\ell \leq 0 $
 for all $\ell \not\in T$ (we call $F(s)$ a test function).
Then we have the inequality
\[ f_0|X|^2-F(1)|X| \ge \sum_{s\in A(X)} F(s)A_s(X),\]
where $A_s(X)$ and $A(X)$ are defined in \eqref{eq:d_a} and \eqref{eq:A(X)}, respectively. 
\end{lemm}
\begin{proof}
This is an immediate consequence of \cite[Lemma 3.1]{HiraoNozakiTasaka} (see also \cite[Corollary 3.8]{DelsarteGoethalsSeidel77}). 
\end{proof}

We are now in a position to show Theorem \ref{thm:LP_bound}.

\begin{proof}[Proof of Theorem \ref{thm:LP_bound}]
(i) This was shown in \cite[Theorem 3.2]{HiraoNozakiTasaka}, but we repeat the proof.
Let $X'$ be a half set of an antipodal spherical $\{10,4,2\}$-design $X$ of $\mathbb{S}^3$.
From \eqref{eq:antipodal_cardinality} it suffices to show that $|X'|\ge12$.
Let 
\[ F_{2T}(s):= \frac{1}{11264} Q_{10}(s) + \frac{1}{2560} Q_{4} (s) + 
\frac{1}{768} Q_{2} (s) + \frac{3}{1024}.\]
Applying this to Lemma \ref{lem:LP_bound}, we obtain
\begin{equation}\label{eq:F_{2T}_ineq} 
 \frac{3}{1024} |X'|^2 - F_{2T}(1)|X'| \ge \sum_{s\in A(X')} F_{2T}(s)A_s(X').
\end{equation}
Note that the polynomial $F_{2T}$ has the following factorization:
\begin{equation}\label{eq:F_{2T}}
F_{2T}(s)= s^2 \left(s + \frac{1}{2} \right)^2 \left(s - \frac{1}{2} \right)^2 \left( \left(s^2 - \frac78\right)^2 + \frac34\right).
\end{equation}
It follows from \eqref{eq:F_{2T}} that $F_{2T}(s)\ge0$ for any $s\in [-1,1]$.
Since $A(X')\subset (-1,1)$, we obtain the inequality $\sum_{s\in A(X')} F_{2T}(s)A_s(X')\ge0$, and hence,
\[ |X'| \ge \frac{1024}{3} F_{2T}(1)=12.\]
 
(ii) Let $X'$ be a half set of an antipodal spherical $\{14,10,6,4,2\}$-design $X$ of $\mathbb{S}^3$.
Consider
\begin{align*}
F_{2O}(s) &:= \frac{1}{245760} Q_{14}(s) + \frac{1}{135168} Q_{10}(s) + \frac{1}{114688} Q_6(s) + \frac{1}{49152} Q_4(s) \\ 
   &\quad + \frac{1}{147456 }Q_2(s) + \frac{1}{8192}.
\end{align*}
With this, Lemma \ref{lem:LP_bound} gives the inequality 
\[\frac{1}{8192} |X'|^2 - F_{2O}(1)|X'| \ge \sum_{s\in A(X')} F_{2O}(s)A_s(X').\]
From another expression
\begin{equation}\label{eq:F_{2O}}
F_{2O}(s)=s^2\left( s+\frac12\right)^2  \left( s-\frac12\right)^2 \left(  s^2-\frac12\right)^2 \left(\left(s^2- \frac78 \right)^2 +\frac14 \right),
\end{equation}
we see that $F_{2O}(s)\ge0$ for any $s\in [-1,1]$.
Therefore, one gets the desired inequality 
\[ |X'| \ge 8192 F_{2O}(1)= 24.\]

(iii) 
Let $X'$ be a half set of an antipodal spherical $\{10,8,6,4,2\}$-design $X$ of $\mathbb{S}^3$.
Applying 
\begin{equation}\label{eq:F_{2I}}
\begin{aligned}
F_{2I}(s) &:= 
- \frac{1}{1114112} Q_{16}(s)
- \frac{11}{4915200} Q_{14}(s)
+ \frac{21}{1802240} Q_{10}(s)
+ \frac{17}{491520} Q_{8}(s) \\
&\quad + \frac{11}{163840} Q_{6}(s)
+ \frac{177}{1638400} Q_4(s)
+ \frac{149}{983040} Q_2(s)
+ \frac{3}{16384} \\
&= s^2 \left(s - \frac{\tau^{-1}}{2} \right)^2
\left(s + \frac{\tau^{-1}}{2} \right)^2
\left(s - \frac{\tau}{2} \right)^2
\left(s + \frac{\tau}{2} \right)^2 \left(s^2 - \frac14\right)^2
\left( \frac{6}{5} - s^2 \right)
\end{aligned}
\end{equation}
to Lemma \ref{lem:LP_bound}, we obtain 
$\frac{3}{16384} |X'|^2 - F_{2I}(1)|X'|\ge 0$ (note that the coefficients of $Q_{16}$ and $Q_{14}$ in the Gegenbauer expansion of $F_{2I}$ are negative).
Hence, it holds that
\[ |X'|\ge  \frac{16384}{3} F_{2I}(1) = 60. \qedhere\]
\end{proof}

\begin{rema}
By finding a test function of degree $17$, which is different from our $F_{2I}(s)$, Andreev \cite[Theorem 1]{Andreev} proved that every spherical $11$-design $X\subset \mathbb{S}^3$ satisfies $|X|\ge120$ (see also \cite[Theorem 6]{BoyvalenkovDanev01}).
\end{rema}


From the above proofs, we are led to the following definition.

\begin{defi}
\begin{itemize}
\item[(i)] An antipodal spherical $\{10,4,2\}$-design $X$ of $\mathbb{S}^3$ is said to be \emph{minimal} when $|X|=24$.
\item[(ii)] An antipodal spherical $\{14,10,6,4,2\}$-design $X$ of $\mathbb{S}^3$ is said to be \emph{minimal} when $|X|=48$.
\item[(iii)] An antipodal spherical $\{10,8,6,4,2\}$-design $X$ of $\mathbb{S}^3$ is said to be \emph{minimal} when $|X|=120$.
\end{itemize}
\end{defi}

Using the expressions \eqref{eq:F_{2T}}, \eqref{eq:F_{2O}} and \eqref{eq:F_{2I}} of our test functions, one can control the set $A(X)$ of inner products for the above minimal antipodal spherical $T$-designs $X$. 

\begin{coro}\label{cor:angles}
\begin{itemize}
\item[(i)] An antipodal spherical $\{10,4,2\}$-design $X$ of $\mathbb{S}^3$ is minimal if and only if
\[A(X) \subset \left\{-1,\pm \frac12,0\right\}.\]
\item[(ii)] An antipodal spherical $\{14,10,6,4,2\}$-design $X$ of $\mathbb{S}^3$ is minimal if and only if 
\[A(X)\subset \left\{-1,\pm \frac{1}{\sqrt{2}},\pm \frac12,0\right\}.\]
\item[(iii)] An antipodal spherical $\{10,8,6,4,2\}$-design $X$ of $\mathbb{S}^3$ is minimal if and only if
\[A(X)\subset \left\{-1,\pm \frac{\tau^{-1}}{2},\pm \frac{\tau}{2}, \pm \frac12,0\right\}.\]
\end{itemize}
\end{coro}
\begin{proof}
For a half set $X'$ of an antipodal spherical $\{10,4,2\}$-design $X$ of $\mathbb{S}^3$, it follows from \eqref{eq:F_{2T}_ineq} that the equality $|X'|=12$ holds if and only if $F_{2T}(s)=0$ for all $s\in A(X')$.
Thus, the result follows from the expression in \eqref{eq:F_{2T}}.
The other two cases are obtained in much the same way as above and are therefore omitted.
\end{proof}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem \ref{thm:uniqueness}}

In this subsection, we show the uniqueness of $X=2O$ as a minimal antipodal spherical $\{14,10,6,4,2\}$-design, whose cardinality is $48$. 

\begin{proof}[Proof of Theorem \ref{thm:uniqueness} (ii)] 
Let 
\[ A:= \left\{-1,\pm \frac{1}{\sqrt{2}},\pm \frac12,0\right\}.\]
From Corollary \ref{cor:angles} (ii), the set of the inner products satisfies $A(X)\subset A$ (namely, $X$ is an $A$-code).
Fix an arbitrary point $\boldsymbol{x}_0 \in X$, and we may suppose $\boldsymbol{x}_0=(1,0,0,0)$. 
For $s\in \R$, let 
\[ X_s: =\{ \boldsymbol{x} \in X \mid \langle \boldsymbol{x}, \boldsymbol{x}_0 \rangle =s \}.  \]
Note that $X_{-s}=-X_s$, since $X$ is antipodal.
Each point $\boldsymbol{x}$ of $X_{s}$ has the form $\boldsymbol{x}=(s,x_2,x_3,x_4)$. For such $\boldsymbol{x}$ with $s\neq \pm 1$, we write $\widetilde{\boldsymbol{x}}=\sqrt{1/(1-s^2)} (x_2,x_3,x_4)$ which is on the unit sphere $\mathbb{S}^2$.
For $s \in A(X)\setminus\{-1\}$, the set 
 \[\widetilde{X}_{s}:=\{\widetilde{\boldsymbol{x}} \in \mathbb{S}^2 \mid \boldsymbol{x} \in X_{s}\} \] 
is called the derived code with respect to $\boldsymbol{x}_0$. 
By \cite[Theorem 8.2]{DelsarteGoethalsSeidel77}, each derived code
$\widetilde{X}_s$ is a spherical $3$-design on $\mathbb{S}^2$, as $X$ is a spherical $7$-design and an $A$-code.

Notice that $X$ is distance invariant from 
Theorem 7.4 in \cite{DelsarteGoethalsSeidel77}, namely $ |X_s |$ does not depend on the choice of the initial $\boldsymbol{x}_0 \in X$.
Thus, we have $A_s(X)=|X_s|\times |X|$, where $A_s(X)$ is defined in \eqref{eq:d_a}.
From Lemma \ref{lem:gegenbauer}, one obtains the following system of equations in $|X_s|$: 
\[\sum_{s \in A\cup\{1\} }|X_s| Q_\ell^{(4)}(s) = 0\]
for each $\ell \in \{2, 4, 6\}$ (see \eqref{eq:Q_l^{d}} for the polynomials $Q_\ell^{(4)}(s)$). Together with the conditions $|X_{\pm 1} |=1$ and $|X_s| = |X_{-s}|$, this system of equations is solved as 
\[ |X_0| = 18, \qquad |X_{\pm 1/\sqrt{2}}| = 6, \qquad |X_{\pm 1/2}| = 8.\]
Since a spherical 3-design on $\mathbb{S}^2$ with $6$ points (namely, a tight 3-design) is unique and is a cross polytope (the regular octahedron) up to orthogonal transformations (cf.~\cite{BB09,BZ20}), we see that $\widetilde{X}_{\pm 1/\sqrt{2}}$ is an orthogonal transformation of $\{(\pm 1,0,0),(0,\pm 1,0),(0,0,\pm 1)\}$.
By taking a suitable orthogonal transformation, we may express
\begin{align*}
    X_1&=\{(1,0,0,0)\}, \qquad 
    X_{-1}=\{(-1,0,0,0)\}, \\ 
    X_{1/\sqrt{2}}&= \left\{\left(\frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}},0 ,0\right),
    \left(\frac{1}{\sqrt{2}},0,\pm \frac{1}{\sqrt{2}},0 \right),
    \left(\frac{1}{\sqrt{2}},0,0,\pm \frac{1}{\sqrt{2}}\right)
    \right\}, \\
     X_{-1/\sqrt{2}}&= \left\{\left(-\frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{2}},0 ,0\right),
    \left(-\frac{1}{\sqrt{2}},0,\pm \frac{1}{\sqrt{2}},0 \right),
    \left(-\frac{1}{\sqrt{2}},0,0,\pm \frac{1}{\sqrt{2}}\right)
    \right\}. 
\end{align*}

We now compute the coordinates of $X_{\pm1/2}$.
From the comment below Definition 8.1 in \cite{DelsarteGoethalsSeidel77}, one has
\[ A(\widetilde{X}_{1/2}) \subset \left\{\frac{4}{3}s -\frac{1}{3} \ \middle| \ s \in A \right \}\cap [-1,1)=\left\{\pm \frac{1}{3}, \frac{2\sqrt{2}-1}{3},-1 \right\}.\]
 Since $\widetilde{X}_{1/2}$ is a spherical $3$-design, it follows from Lemma \ref{lem:gegenbauer} that 
 \begin{equation} \label{eq:0}
 \sum_{\widetilde{\boldsymbol{x}}, \widetilde{\boldsymbol{y}} \in \widetilde{X}_{1/2}} \langle \widetilde{\boldsymbol{x}}, \widetilde{\boldsymbol{y}} \rangle =0.     
 \end{equation}
If \( (2\sqrt{2}-1)/3 \) appears as an inner product in \eqref{eq:0}, then the left-hand side of \eqref{eq:0} is never zero.
This implies that $A(\widetilde{X}_{1/2}) \subset \left\{\pm \frac{1}{3},-1 \right\}$. 
Thus by \cite[Theorem 7.4]{DelsarteGoethalsSeidel77}, the set $Y=\widetilde{X}_{1/2}$ is distance invariant, namely, $|\{\boldsymbol{y}\in Y\mid \langle \boldsymbol{y},\boldsymbol{y}_0\rangle=s\}|$ does not depend on the choice of $\boldsymbol{y}_0\in Y$.
Fix $\boldsymbol{y}_0\in Y$ and let $Y_s=\{\boldsymbol{y}\in Y\mid \langle \boldsymbol{y},\boldsymbol{y}_0\rangle=s\}$.
Note that $|Y_1|=1$.
From Lemma \ref{lem:gegenbauer}, one has
\[\sum_{s\in \{\pm 1,\pm \frac13 \} }|Y_s| Q_\ell^{(3)}(s) = 0, \quad \ell=1,2,3.\]
This system of equations has the following solutions:
\[ |Y_{\pm 1/3}|=3, \qquad |Y_{-1}|=1.\]
Therefore, $Y$ is the cube in $\mathbb{S}^2$ (cf.~\cite{BB09,BZ20}). 
From \cite[Lemma 1]{BoyvalenkovDanev01}, for $\boldsymbol{y} \in Y$ and $\widetilde{\boldsymbol{x}} \in \widetilde{X}_{1/\sqrt{2}}$, we see that
\[ \langle \boldsymbol{y}, \widetilde{\boldsymbol{x}} \rangle \in 
\left\{\sqrt{\frac{8}{3}} s -\frac{1}{\sqrt{3}}\ \middle| \ s \in A\right\} \cap[-1,1)=\left\{\pm \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}}-{\frac{1}{\sqrt{3}}}\right\}.\]
Using \cite[Corollary 2]{BoyvalenkovDanev01}, we compute the distance distribution of $Y$ with respect to $\widetilde{\boldsymbol{x}} \in \widetilde{X}_{1/\sqrt{2}}$: 
\[ \left|Y_{1/\sqrt{3}}^{\widetilde{\boldsymbol{x}}}\right|= 4, \qquad \left|Y_{-1/\sqrt{3}}^{\widetilde{\boldsymbol{x}}}\right|=4, \qquad \left|Y_{\sqrt{2/3}-1/\sqrt{3}}^{\widetilde{\boldsymbol{x}}}\right|=0,  \]
where $Y_s^{\widetilde{\boldsymbol{x}}}=\{\boldsymbol{\boldsymbol{y}} \in Y \mid \langle \boldsymbol{y}, \widetilde{\boldsymbol{x}} \rangle = s \}$. This implies that for each $\boldsymbol{x} \in X_{1/\sqrt{2}}$, we have that
\[ |\{\boldsymbol{y} \in X_{1/2} \mid \langle \boldsymbol{y} ,\boldsymbol{x} \rangle = 1/\sqrt{2}\}|=4, 
\qquad 
|\{ \boldsymbol{y} \in X_{1/2} \mid \langle \boldsymbol{y}, \boldsymbol{x}\rangle = 0\}|=4. \]
Using the above expression of $X_{1/\sqrt{2}}$, we can uniquely determine 
\[ X_{1/2}=\left\{\left(\frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}, \pm \frac{1}{2}\right ) \right\}, \]
where we take all choices of signs. 
$X_{-1/2}$ is obtained by $-X_{1/2}$.


Finally we determine $\boldsymbol{x}=(0,x_2,x_3,x_4)\in X_0$. 
The inner products between elements in $X_0$ and in $X_{1/\sqrt{2}}$ belong to $A$, so $\pm \frac{x_j}{\sqrt{2}} \in A$ for 
$j=2,3,4$. 
This together with $x_2^2+x_3^2+x_4^2=1$ shows that $x_j \in \{0, \pm 1, \pm 1/\sqrt{2}\}$.
Therefore, $X_0$ should be a subset of 
\begin{align*}
\Big\{ &(0,\pm 1, 0,0), (0,0,\pm 1, 0),(0,0,0,\pm 1), \\ 
&\left(0,\pm \tfrac{1}{\sqrt{2}},\pm \tfrac{1}{\sqrt{2}},0  \right),\left(0,\pm \tfrac{1}{\sqrt{2}},0,\pm \tfrac{1}{\sqrt{2}}  \right),\left(0,0,\pm \tfrac{1}{\sqrt{2}},\pm \tfrac{1}{\sqrt{2}}  \right) \Big\}.
\end{align*}
This set is size 18, and hence, coincides with $X_0$. 


From the coordinates of $X_{1/\sqrt{2}}$, we can uniquely determine the coordinates of $X_s$ for each $s \in A(X)$. This implies the uniqueness of $2O$ as desired. 
\end{proof}





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Application to modular forms}\label{sec:order}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Orders and modular forms}
Let $F$ be a totally real number field and $R$ its ring of integers.
The $F$-vector space 
\[D(F):=\{x_1+x_2i+x_3j+x_4k\mid x_1,\ldots,x_4\in F\}\]
forms an $F$-algebra, called the \emph{quaternion algebra over $F$}.
A subring $O\subset D(F)$ is an \emph{$R$-order} if there exists a basis $\{b_1,b_2,b_3,b_4\}$ of $D(F)$ such that $O=Rb_1\oplus R b_2\oplus Rb_3\oplus R b_4$. 
Note that the set $N(O)$ of all norm values of $O$ is a subset of $R$ (cf.~\cite[Chapter 10]{Voight2021}).
An $R$-order of $D$ is said to be \emph{maximal} if it is not properly contained in another $R$-order of $D$.

Recall our orders from the introduction.
Let $F_{2T}=\Q,\ F_{2O}=\Q\big(\sqrt{2}\big)$ and $F_{2I}=\Q\big(\sqrt{5}\big)$.
From \eqref{eq:def_exceptional_groups}, we see that each $G\in\{2T,2O,2I\}$ is a subset of $F_G^4$.
Let $R_G$ be the ring of integers of $F_G$, i.e. $R_{2T}=\Z, R_{2O}=\Z[\sqrt{2}], R_{2I}=\Z[\tau]$ with $\tau=\frac{1+\sqrt{5}}{2}$, and $\mathcal{O}_G=\langle G\rangle_{R_G}$ the $R_G$-subalgebra of the quaternion algebra over $F_G$ generated by $G$.
Then, $\mathcal{O}_G$ is a maximal $R_G$-order (cf.~\cite[Chapter 11]{Voight2021}).
Its basis is given in Table \ref{table:orders}.
For $G\in \{2T,2O,2I\}$, we define the map $\iota_G:R_G\rightarrow\Z$ by
\begin{equation}\label{eq:def_iota}
\begin{aligned}
&\iota_{2T}(a)=a, \quad (a\in \Z=R_{2T}),\\
&\iota_{2O}(a+b\sqrt{2})=a, \quad (a+b\sqrt{2}\in \Z+\Z\sqrt{2}=R_{2O}),\\
&\iota_{2I}(a+b\tau)=a, \quad (a+b\tau\in \Z+\Z\tau=R_{2I}).
\end{aligned}
\end{equation}

For $G\in \{2T,2O,2I\}$, let
\[\mathcal{Q}_G(x):=\iota_G(N(x)).\]
For $m\in \N$, we study the following finite subset of $\R^4$:
\[ \mathcal{O}_{G,m}=\{ x\in \mathcal{O}_G\mid \mathcal{Q}_G(x)=m\}.\]
We give explicit formulas for $|\mathcal{O}_{G,m}|$.
For this, let $\sigma_k(m)=\sum_{d\mid m}d^k$ be the divisor function.
We set $\sigma_k(m/2)=0$ if $m$ is odd.

\begin{prop}\label{prop:counting}
\begin{itemize}
\item[(i)] For $m\in \N$, we have $|\mathcal{O}_{2T,m}| = 24\left(\sigma_1(m)-2\sigma_1\left(\frac{m}{2}\right)\right)>0$. 
Moreover, $\mathcal{O}_{2T,1}=2T$. 
\item[(ii)] For $m\in \N$, we have $|\mathcal{O}_{2O,m}| = 48\left(\sigma_3(m)+4\sigma_3\left(\frac{m}{2}\right)\right)>0$. 
Moreover, $\mathcal{O}_{2O,1}=2O$. 
\item[(iii)] For $m\in \N$, we have $|\mathcal{O}_{2I,m}| =  240\sigma_3(m)>0$. 
Moreover, $\mathcal{O}_{2I,1}=2I\cup \tau\, 2I$. 
\end{itemize}
\end{prop}
\begin{proof}
For each $G\in \{2T,2O,2I\}$, we first observe that $\mathcal{Q}_G(x)\in \Z$ for $x\in \mathcal{O}_G$ defines a positive definite integral quadratic form (for background on quadratic forms, see \cite[Chapter 6]{Ogg}).
This implies that the spherical theta function $\theta_{G,P}(z)=\sum_{x\in \mathcal{O}_G} P(x) q^{\mathcal{Q}_G(x)} \ (q=e^{2\pi i z})$ defined in \eqref{eq:theta}, is a modular form (cf.~\cite[\S4.9]{Miyake}, \cite[\S6]{Ogg}).
Then, using the theory of modular forms, we express the generating function $\theta_{G,1}(z)=\sum_{m\ge0}| \mathcal{O}_{G,m} |q^{m}$ in terms of the Eisenstein series $E_2(z)=1-24\sum_{m\ge1}\sigma_1(m)q^{m}$ and $E_4(z)=1+240\sum_{m\ge1}\sigma_3(m)q^{m}$.
For the theory of modular forms for the congruence subgroup, see \cite{DS}.
The quadratic form $\mathcal{Q}_G(x)$ is also used to give a description of $\mathcal{O}_{G,1}$ (we solve the Diophantine equation $\mathcal{Q}_G(x)=1$ with the aid of Mathematica).


For the Hurwitz order $\mathcal{O}_{2T}$, let $x=r_1+r_2 i+ r_3 j+ r_4 \omega\in \mathcal{O}_{2T}$ with $r_1,r_2,r_3,r_4\in \Z$.
Then one obtains
\[\mathcal{Q}_{2T}(x)= r_1^2+r_2^2+r_3^2+r_4^2-r_1r_4+r_2r_4+r_3r_4,\]
which defines a positive definite integral quadratic form of level 2 and discriminant 4.
Hence, the spherical theta function $\theta_{2T,P}(z)$ for $P\in {\rm Harm}_\ell(\R^4)$ is a modular form of weight $2+\ell$ for $\Gamma_0(2)$.
The space of modular forms of weight $2$ for $\Gamma_0(2)$ is 1-dimensional spanned by $2E_2(2z)-E_2(z)$.
Therefore, we have
\[\theta_{2T,1}(z)=2E_2(2z)-E_2(z) =1+24q+24q^2+96q^3+24q^4+\cdots. \]
By finding integral solutions to the equation $r_1^2+r_2^2+r_3^2+r_4^2-r_1r_4+r_2r_4+r_3r_4=1$, we obtain $\mathcal{O}_{2T,1}=2T$.
Below is an example of Mathematica code to compute the set $\mathcal{O}_{2T,1}$:
\begin{align*}
\texttt{Solve[}&\texttt{r1\string^2+r2\string^2+r3\string^2+r4\string^2-r1*r4+r2*r4+r3*r4==1,} \\
&\texttt{\{r1,r2,r3,r4\}, Integers]}
\end{align*}
For the case $\mathcal{O}_{2O}$, using the integral basis of $\Z[\sqrt{2}]$, we write $x=(y_1+z_1\sqrt{2})+(y_2+z_2\sqrt{2})\alpha +(y_3+z_3\sqrt{2})\beta+(y_4+z_4\sqrt{2})\alpha \beta \in \mathcal{O}_{2O}$ with $y_j,z_j\in \Z$.
Then we have
\begin{align*}
 \mathcal{Q}_{2O}(x) &= y_{1}^2+y_{1} y_{4}+2 y_{1} z_{2}+2 y_{1}
   z_{3}+y_{2}^2+y_{2} y_{3}+2 y_{2} z_{1}+2 y_{2}
   z_{4}+y_{3}^2+2 y_{3} z_{1}\\
 &+2 y_{3} z_{4}+y_{4}^2+2
   y_{4} z_{2}+2 y_{4} z_{3}+2 z_{1}^2+2 z_{1} z_{4}+2
   z_{2}^2+2 z_{2} z_{3}+2 z_{3}^2+2 z_{4}^2,
\end{align*}
which is positive definite.
The level and the discriminant of the above quadratic form are 2 and 16, respectively.
Hence, $\theta_{2O,P}(z)$ for $P\in {\rm Harm}_\ell(\R^4)$ is a modular form of weight $4+\ell$ for $\Gamma_0(2)$.
Note that the space of modular forms of weight $4$ for $\Gamma_0(2)$ is 2-dimensional.
Due to the Sturm bound \cite[Corollary 9.20]{Stein}, expressing $\theta_{2O,1}(z)$ in terms of the Eisenstein series requires the coefficient of $q$, namely, $|\mathcal{O}_{2O,1}|$.
Now, let the computer calculate integer solutions to $\mathcal{Q}_{2O}(x)=1$ to get an explicit description of $\mathcal{O}_{2O,1}$. 
The result is $\mathcal{O}_{2O,1}=2O$. 
Hence, $|\mathcal{O}_{2O,1}|=48$ and we get
\[ \theta_{2O,1}(z)=\frac15 E_4(z)+\frac45 E_4(2z) = 1+48q+624q^2+1344q^3+5232q^4+\cdots.\]



For the case $\mathcal{O}_{2I}$, we also write $x=(y_1+z_1\tau)+(y_2+z_2\tau)i +(y_3+z_3\tau)\zeta +(y_4+z_4\tau)i\zeta \in \mathcal{O}_{2I}$ with $y_j,z_j\in \Z$.
It holds that
\begin{align*} 
\mathcal{Q}_{2I}(x) &= y_{1}^2+y_{1} y_{4}+y_{1} z_{3}-y_{1}
   z_{4}+y_{2}^2-y_{2} y_{3}+y_{2} z_{3}+y_{2}
   z_{4}+y_{3}^2+y_{3} z_{1}\\
   &+y_{3} z_{2}+y_{4}^2-y_{4}
   z_{1}+y_{4} z_{2}+z_{1}^2+z_{1} z_{3}+z_{2}^2+z_{2}
   z_{4}+z_{3}^2+z_{4}^2.
   \end{align*}
This is a positive definite integral quadratic form of level 1 and discriminant 1.
Thus, $\theta_{2I,P}(z)$ for $P\in {\rm Harm}_\ell(\R^4)$ is a modular form of weight $4+\ell$ for ${\rm SL}_2(\Z)$.
Since the underlying vector space is 1-dimensional, we get
\[ \theta_{2I,1}(z)=E_4(z) = 1+240q+2160q^2+6720q^3+17520q^4+\cdots.\]
One can check that the set $\mathcal{O}_{2I,1}$ of integer solutions to the equation $\mathcal{Q}_{2I}(x)=1$ has 240 points. Moreover, we obtain $\mathcal{O}_{2I,1}=2I \cup \tau\, 2I$, meaning that the set $\mathcal{O}_{2I,1}$ consists of two copies of $2I$.
This completes the proof.
\end{proof}




\begin{rema}
As the formula $|\mathcal{O}_{2I,m}| = 240\sigma_3(m)$ suggests, the $\Z[\tau]$-order $\mathcal{O}_{2I}$ has a connection with the $E_8$ lattice (an even integral unimodular lattice in $\R^8$).
Let $\kappa_4 : D(\Q(\tau))\rightarrow \Q^8$ be the $\Q$-linear map defined by 
\[ \kappa_4\big( (a_1+b_1\tau)+(a_2+b_2\tau)i+(a_3+b_3\tau)j+(a_4+b_4\tau)k\big)=(a_1,b_1,a_2,b_2,a_3,b_3,a_4,b_4).\]
Then, the image $\sqrt{2}\,\kappa_4\big(\mathcal{O}_{2I}\big)$ is isometric to the $E_8$ lattice.
Furthermore, the map $\sqrt{2}\,\kappa_4$ gives a bijection from $\mathcal{O}_{2I,m}$ to the $2m$-shell of the $E_8$ lattice.
This is because for $x=(a_1+b_1\tau)+(a_2+b_2\tau)i+(a_3+b_3\tau)j+(a_4+b_4\tau)k\in\mathcal{O}_{2I}$, it holds that
\[\iota_{2I}(N(x))=a_1^2+b_1^2+\cdots+a_4^2+b_4^2=\langle \kappa_4(x),\kappa_4(x)\rangle,\] 
where we have used $\tau^2=\tau+1$.
For more details, see \cite[Chapter 8]{ConwaySloan}. 
\end{rema}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem \ref{thm:dim=0}}

We first prove \eqref{eq:G_acts_on_O_{G,m}}.
For this, recall that the multiplication by a quaternion yields an orthogonal transformation (Lemma \ref{lem:quaternion_orthogonal}).

\begin{prop}\label{prop:O_strength}
Let $G\in\{2T,2O,2I\}$. 
For all $m\in \N$, we have $T(G)\subset T\big(\mathcal{O}_{G,m}\big)$.
When $m=1$, the equality $T(G)=T\big(\mathcal{O}_{G,1}\big)$ holds.
\end{prop}
\begin{proof}
Let $m\in \N$.
For $\varepsilon\in G$ and $x\in \mathcal{O}_{G,m}$, since $N(x\varepsilon)=N(x)$, it holds that $x\varepsilon \in \mathcal{O}_{G,m}$.
This shows that $G$ acts on the set $\mathcal{O}_{G,m}$. 
Hence, we have a $G$-orbit decomposition 
\begin{equation}\label{eq:orbit_decomp} 
\mathcal{O}_{G,m} = \bigsqcup_{x\in S_m} xG
\end{equation}
of $\mathcal{O}_{G,m}$ as a disjoint union of $\{xG \mid x\in S_m\}$ for some $S_m\subset \mathcal{O}_{G,m}$.
By Lemma \ref{lem:quaternion_orthogonal}, the $G$-orbit $xG$ is an orthogonal transformation of $\sqrt{N(x)} G$.
It is well known that the orthogonal transformation group $O(\R^d)$ acts on the space ${\rm Harm}_\ell (\R^d)$ of homogeneous harmonic polynomials. In our case, for $P\in {\rm Harm}_\ell (\R^4)$ and $g\in O(\R^4)$, we have $gP\in {\rm Harm}_\ell (\R^4)$, where $(gP)(x):=P(g(x))$.
Therefore, for any $x\in S_m$, we see that $\ell \in T(G)$ if and only if $\ell \in T(xG)$ (note that since every $x \in \mathcal{O}_{G,m}$ is invertible in $\mathbb{H}^\times$, we have $|xG| = |G|$).
Finally, the inclusion $T(G)\subset T\big(\mathcal{O}_{G,m}\big)$ follows from the fact that $T(X_1\sqcup X_2)$ contains $ T(X_1)\cap T(X_2)$.

Let us turn to the case $m=1$.
From Proposition \ref{prop:counting}, for $G\in \{2T,2O\}$, we have $G=\mathcal{O}_{G,1}$.
Thus, $T(G)=T\big(\mathcal{O}_{G,1}\big)$.
For the case $G=2I$, since $T(2I)\subset T\big(\mathcal{O}_{2I,1}\big)$, it suffices to show $T(2I)\supset T\big(\mathcal{O}_{2I,1}\big)$.
Suppose $\ell\in T\big(\mathcal{O}_{2I,1}\big)$.
By Proposition \ref{prop:counting} (iii), for all $P\in {\rm Harm}_\ell(\R^4)$ one has
\[ 0=\sum_{x\in \mathcal{O}_{2I,1}} P(x)=\sum_{x\in 2I} P(x) + \sum_{x\in 2I} P(\tau x)=(1+\tau^\ell ) \sum_{x\in 2I} P( x).\]
Thus, $\ell \in T(2I)$.
This completes the proof.
\end{proof}

We are now in a position to prove Theorem \ref{thm:dim=0}.

\begin{proof}[Proof of Theorem \ref{thm:dim=0}]
We first prove that $\dim_\C \Theta(G,\ell)=0 \Rightarrow \ell \in T(G)$.
Suppose $\ell \not\in T(G)$.
Then, by Proposition \ref{prop:O_strength}, $\ell \not\in T\big(\mathcal{O}_{G,1}\big)$.
This shows that there exists $P\in {\rm Harm}_\ell(\R^4)$ such that $\theta_{G,P}(z)\neq0$.
Hence, $\dim_\C\Theta(G,\ell)\ge1$.

Let us turn to the proof of $ \ell \in T(G)\Rightarrow \dim_\C \Theta(G,\ell)=0$.
For $\ell \in T(G)$, by Proposition \ref{prop:O_strength}, $\ell \in T\big(\mathcal{O}_{G,m}\big)$ holds for all $m\in\N$.
Hence, $\theta_{G,P}=0$ for all $P\in {\rm Harm}_\ell (\R^4)$.
Thus, $\dim_\C\Theta(G,\ell)=0$.
We have done.
\end{proof}

\begin{rema}\label{rem:HNT}
In our previous paper \cite[Section 6]{HiraoNozakiTasaka}, as an application of the uniqueness, we gave a decomposition of the $m$-shell of the $D_4$ lattice in terms of a disjoint union of the orthogonal transformations (up to scalar) of the $D_4$ root system.
We have the same structure: For any $m>0$, there exists a finite subset $U_m$ of the orthogonal transformation group $O(\R^4)$ such that 
\[\left\{ \frac{x}{\sqrt{N(x)}} \ \middle| \  x\in \mathcal{O}_{G,m}\right\}=\bigsqcup_{g\in U_m} g (G).\]
This is because the orbit $xG$ in \eqref{eq:orbit_decomp} is an orthogonal transformation of $\sqrt{N(x)} G$.
\end{rema}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{More on the space \texorpdfstring{$\Theta(G,\ell)$}{Theta(G, ell)}}

In this last subsection, we first give an upper bound of the dimension of the space $\Theta(G,\ell)$, by using the theory of invariant polynomials.
Then, numerical basis of the non-zero space $\Theta(G,\ell)$ are provided for some $\ell$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Invariant polynomials}
For a subgroup $G$ of $\mathbb{H}_1$, denote the $G$-invariant subspace of ${\rm Harm}_{\ell} (\R^4)$ by 
\[{\rm Harm}_\ell (\R^4)^G :=\{ P\in {\rm Harm}_\ell (\R^4) \mid \varepsilon P=P\ \mbox{for all}\ \varepsilon \in G\},\]
where we set $(\varepsilon P)(x):=P(x M_\varepsilon)$ for $x\in\R^4$ and $M_\varepsilon$ is the matrix defined in \eqref{eq:M_x}.
If we identify $\R^4$ with $\mathbb{H}$ as in Lemma \ref{lem:quaternion_orthogonal}, then $(\varepsilon P)(x)=P(\varepsilon x)$.

\begin{prop}\label{prop:upper_bound}
For $G\in \{2T,2O,2I\}$ and $\ell\in \N$, we have 
\[ \dim_\C \Theta(G,\ell) \le \dim_\R {\rm Harm}_\ell (\R^4)^G.\]
\end{prop}
\begin{proof}
From the representation theory of finite groups, we obtain
\[ {\rm Harm}_\ell (\R^4)= {\rm Harm}_\ell (\R^4)^G\oplus {\rm Span}_{\R}\{P- \varepsilon P \mid P\in {\rm Harm}_\ell (\R^4), \varepsilon\in G\}.\]
Then, the result follows from 
\begin{align*}
\theta_{G,{P-\varepsilon P}}(z)&=\sum_{x\in \mathcal{O}_G} \big( P(x)-P(\varepsilon x)\big) q^{\mathcal{Q}_G(x)}=\theta_{G,P}(z)-\sum_{x\in \varepsilon \mathcal{O}_G} P(x)q^{\mathcal{Q}_G(\varepsilon^{-1}x)}\\
&=\theta_{G,P}(z)-\sum_{x\in \mathcal{O}_G} P(x)q^{\mathcal{Q}_G(x)}=0,
\end{align*}
where we have used the fact that $G$ acts on $\mathcal{O}_G$ and that $N(\varepsilon x)=N(x)$ for $\varepsilon\in \mathbb{H}_1$ and $x\in \mathbb{H}$.
\end{proof}

The generating series of the dimension of ${\rm Harm}_\ell (\R^4)^G$ is called the \emph{harmonic Molien series of $G$}.
We denote it by
\[ \Psi_G^H(u):=\sum_{\ell\ge0}  \dim_\R {\rm Harm}_\ell (\R^4)^G u^\ell.\]
From \cite[Theorem 4.3.2]{Smith} and $ {\rm Hom}_\ell(\R^4)^G = {\rm Harm}_\ell (\R^4)^G\oplus  N(x)  {\rm Hom}_{\ell-2}(\R^4)^G$, we obtain
\[  \Psi_G^H(u)= \frac{1}{|G|} \sum_{\varepsilon \in G} \frac{1-u^2}{\det (I-uM_\varepsilon)}.\]
Using this, we can compute $\Psi_G^H(u)=\sum_{\ell\ge0}d_{G,\ell}u^\ell$ for $G\in \{2T,2O,2I\}$ as follows:
\begin{align*}
&\Psi_{2T}^H(u) = \frac{f_{2T}(u)-f_{2T}(u^{-1})u^{18}}{(1-u^4)^2(1-u^6)^2},\\
&\Psi_{2O}^H(u) = \frac{f_{2O}(u) - f_{2O}(u^{-1})u^{26}}{(1-u^6)^2(1-u^8)^2},\\ 
&\Psi_{2I}^H(u) = \frac{f_{2I}(u)-f_{2I}(u^{-1})u^{34}}{(1+u^2)^2(1-u^6)^2(1-u^{10})^2},
\end{align*}
where we set
\begin{align*}
f_{2T}(u)&=1 - 2 u^4 + 5 u^6 + 10 u^8,\\
f_{2O}(u)&=1 - 2 u^6 + 7 u^8 + 14 u^{12},\\
f_{2I}(u)&=1 + 2 u^2 +u^4 - 2 u^6 - 4 u^8 - 4 u^{10} + 10 u^{12} + 26 u^{14} +  18 u^{16}. 
\end{align*}

\begin{center}
\begin{tabular}{c|cccccccccccccccccccc}
$\ell$ & 2&4 &6 &8&10&12&14&16&18&20&22&24 \\ \hline
$d_{2T,\ell}$  &0 & 0 & 7 & 9 & 0 & 26 & 15 & 17 & 38 & 42 & 23 & 75\\ \hline
$d_{2O,\ell}$ &  0 & 0 & 0 & 9 & 0 & 13 & 0 & 17 & 19 & 21 & 0 & 50 \\ \hline
$d_{2I,\ell}$ &  0 & 0 & 0 & 0 & 0 & 13 & 0 & 0 & 0 & 21 & 0 & 25\\ \hline
\end{tabular}
\end{center}

We believe that the inequality in Proposition \ref{prop:upper_bound} becomes an equality only when $\dim_\C \Theta(G,\ell)=0$.
For the (conjectural) dimension of the space $\Theta(G,\ell) $, see the subsequent subsections.


\subsubsection{Case \texorpdfstring{$G=2T$}{G=2T}}
Let $\Lambda$ be the $D_4$ lattice generated by the $D_4$ root system.
This lattice is unique as an even integral lattice of level 2 (see e.g. \cite[Theorem 7.2]{HiraoNozakiTasaka}).
From Remark \ref{rem:D_4}, we see that $\Lambda=(1+i)\mathcal{O}_{2T}$.
Thus, the space $\Theta(2T,\ell)$ coincides with the $\C$-vector space spanned by the spherical theta function of the $D_4$ lattice.
It is generated by newforms of weight $2+\ell$ for $\Gamma_0(2)$ (see \cite[Remark 7.3]{HiraoNozakiTasaka}), namely, we have $\Theta(2T,\ell)=S_{2+\ell}^{\rm new}(2)$ the $\C$-vector space spanned by newforms of weight $2+\ell$ for $\Gamma_0(2)$.
Let $S_k(N)$ denote the $\C$-vector space spanned by cusp forms with real Fourier coefficients of weight $k$ for $\Gamma_0(N)$.
It is known that $S_k(1)$ and $S_k(2)$ have a basis whose Fourier coefficients are rational numbers.
Thus, from the dimension formula (cf.~\cite[Section 3.5]{DS}), we have $\sum_{k\ge0}\dim_\C S_k(1)u^k = \frac{u^{12}}{(1-u^4)(1-u^6)}$ and $\sum_{k\ge0}\dim_\C S_k(2)u^k = \frac{u^{8}}{(1-u^2)(1-u^4)}$.
Since $\dim_\C S_k^{\rm new}(2)= \dim_\C S_k(2)-2\dim_\C S_k(1)$, we obtain the dimension formula for $\Theta(2T,\ell)$ as follows:
\[ \sum_{\ell\ge0} \dim_\C \Theta(2T,\ell) u^\ell =\frac{1+u^{12}}{(1-u^6)(1-u^8)}.\] 

\subsubsection{Case \texorpdfstring{$G=2O$}{G=2O}}
From Theorems \ref{thm:harmonic_strength} and \ref{thm:dim=0}, we have $\dim_\C \Theta(2O,2\ell)=0$ if and only if $2\ell\in\{22,14,10,6,4,2\}$.
The first non-trivial space $\Theta(2O,8)$ is a subspace of cusp forms of weight 12 for $\Gamma_0(2)$.
One can observe that $\Theta(2O,8)$ is 1-dimensional spanned by
\[ \Delta(z)+64 \Delta(2z) = q+40q^2+252q^3-3008q^4+4830q^5+\cdots,\]
where $\Delta(z)=q\prod_{n\ge1}(1-q^n)^{24}$ is the unique cusp form of weight 12 for ${\rm SL}_2(\Z)$.
Moreover, we have
\begin{align*}
\Theta(2O,12)&=\C(q-3368q^2+58092q^3-268736q^4+\cdots),\\
\Theta(2O,16)&=\C(q+39880q^2+1279452q^3-1999808q^4-174409410q^5+\cdots),\\
\Theta(2O,18)&=\C(q-79024q^2+5687604q^3+103384576q^4-1438933770q^5+\cdots),\\
\Theta(2O,20)&=\C(q+200632q^2+57854412q^3-580869056q^4 + 12385840110 q^5 \\
&\phantom{=\C(q}\,- 70670680416 q^6+\cdots).
\end{align*}

We may expect that
\[ \sum_{\ell\ge0} \dim_\C \Theta(2O,\ell) u^\ell \stackrel{?}{=} 1+ \frac{u^8+u^{12}-u^{14}}{(1-u^6)(1-u^8)}=\frac{1+u^{18}}{(1-u^8)(1-u^{12})}.\] 


\subsubsection{Case \texorpdfstring{$G=2I$}{G=2I}}
Theorems \ref{thm:harmonic_strength} and \ref{thm:dim=0} imply that $\dim_\C \Theta(2I,2\ell)=0$ if and only if 
\[2\ell\in\{58,46,38,34,28,26,22,18,16,14,10,8,6,4,2\}.\]
The first nontrivial space $\Theta(2I,12) $ is 1-dimensional, generated by $ E_4(z)\Delta(z)$.
One can observe that
\begin{align*}
\Theta(2I,20) &=\C(41q + 719736q^2 - 26525268q^3 + 1272600128q^4+\cdots),\\
\Theta(2I,24) &=\C(23q + 2267928q^2 + 438932196q^3 +\cdots),\\
\Theta(2I,30) &=\C(51q + 107369712q^2 - 33500700684q^3 +\cdots),\\
\Theta(2I,32)&=\C(2207q + 11194301112q^2 + 953165204964q^3 +\cdots), \\
\Theta(2I,36)&=\C(107q + 3571780392q^2 - 1430566003836q^3 + \cdots).
\end{align*}

We may expect that
\[ \sum_{\ell\ge0} \dim_\C \Theta(2I,\ell) u^\ell \stackrel{?}{=}\frac{1+u^{30}}{(1-u^{12})(1-u^{20})}.\] 



% A longer acknowledgment at the end looks like this:

%\longthanks{We thank B.~Pascal for some interesting comments on an earlier version of this manuscript.}

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