text:
\section{Introduction}

An elliptic normal curve of degree $N$ is an elliptic curve $E$ with an immersion in a projective space $\P^{N-1}$ as a degree~$N$ curve that is contained in no hyperplane.   Suppose $N\ge 4$, and the base field $K$ is a field of characteristic not dividing~$N$.
By the Riemann-Roch theorem, $E/K$ can be realized as an elliptic normal curve of degree~$N$ in $\P^{N-1}_{K}$ by means of a complete linear system $|D|$ with any effective divisor $D$ of degree~$N$.  In particular, we may take $D=N\cdot O$, where $O$ is the origin of the group structure of~$E$.  Moreover, it is known that such an $E$ is defined by a system of $N(N-3)/2$ quadratic equations.

Classically, over the complex number field $\C$, Bianchi \cite{Bianchi} first wrote down equations for the case $N=5$, where elliptic quintic in $\P^{4}$ is given by a set of five quadratic equations:
\[
\left\{\renewcommand{\arraystretch}{1.33}\begin{array}{l}
x_{0}^{2} + \phi x_{2}x_{3}-\phi^{-1}x_{1}x_{4} = 0, \\
x_{1}^{2} + \phi x_{3}x_{4}-\phi^{-1}x_{2}x_{0} = 0, \\
x_{2}^{2} + \phi x_{4}x_{0}-\phi^{-1}x_{3}x_{1} = 0, \\
x_{3}^{2} + \phi x_{0}x_{1}-\phi^{-1}x_{4}x_{2} = 0, \\
x_{4}^{2} + \phi x_{1}x_{2}-\phi^{-1}x_{0}x_{3} = 0.
\end{array}\right.
\]
Here, $\phi$ is a parametrizing modular function of level 5.
Klein \cite{Klein} generalized it to obtain such quadratic equations for general odd integers $N$, using Riemann's sigma functions.  What is significant is that their formulas describe the universal family of elliptic curves with level $N$ structure. In fact, V\'elu \cite{Velu:1978} realized it as schemes over $\operatorname{Spec}\Z[1/N]$.  By looking at the origin $O\in E$, we also obtain a model of modular curve associated with the principal congruence
subgroup $\Gamma(N)$.

In the case where $N$ is even, Hurwitz \cite{Hurwitz} obtained similar formulas following Klein's method, although it does not give the same kind of universal family.  The purpose of this paper is to study this case in detail.  We take a slightly different immersion from that of Hurwitz, and obtain the universal family of elliptic curves associated with a certain congruence subgroup of level~$2N$.

Let us describe it in more detail. Consider the immersion $E\to \P^{N-1}$ using the complete linear system $|N\cdot O|$.  For a $N$-torsion point $T\in E[N]$, let us denote by $\tau_{T}$ the translation-by-$T$ map $P\mapsto P+T$.  It is easy to see that $\tau_{T}$ can be lifted to an automorphism of the ambient space $\P^{N-1}$.  By choosing a basis $(S,T)$ of $E[N]$, we can find a coordinate system such that the translations $\tau_{S}$ and $\tau_{T}$ are expressed in simple forms similar to Bianchi's equations above (see Proposition~\ref{prop:proj-coord-odd} for detail).

In case $N$ is odd, the choice of such a coordinate system is unique once we fix a primitive $N$th root of unity $\zeta_{N}$, and the pair $(S,T)$ such that the Weil pairing $e_{N}(S,T)=\zeta_{N}$.  Thus, we obtain an immersion of the family of elliptic curves with level $N$ structure to a single $\P^{N-1}$, and by associating the origin of the elliptic curves, we obtain a morphism from the modular curve $X(N)$ to $\P^{N-1}$.

In case $N$ is even, the situation is more complicated.  In this case, the coordinate system mentioned above is not unique; we need to specify a further structure than a level $N$ structure.  As a consequence we will obtain the universal elliptic curve corresponding to a slightly smaller group than $\Gamma(N)$, which we will denote by $\Gamma^{(N)}(2N)$. (See \S3 for detail.)
In addition, we write down the quadratic equations satisfied by the image of $E$ in this particular coordinate system in the case of $K=\C$ using a certain type of theta functions.
More precisely, we define the theta functions denoted by $\theta_{k}^{(N)}(z,\tau)$ (see Definition~\ref{def:theta-N}) that become the coordinate functions of the coordinate system described above, and we obtain quadratic equations of the image using the relations between theta functions coming from Jacobi's identity \eqref{eq:Jacobi4} or \eqref{eq:Jacobi2}.  It should be noted that it is essentially the addition formula of the elliptic curve.

In \S\S\ref{sec:level-4}--\ref{sec:level-8} we work out in detail the cases $N=4,6$, and $8$.  There we show the explicit equations of the modular curve and the universal curve over~it.

\subsection*{Acknowledgements}
Kaneko was supported in part by JSPS KAKENHI Grant Numbers JP23340010, JP15K13428, JP16H06336.
Kuwata was supported by JSPS KAKENHI Grant Numbers JP23540028, JP26400023, and by the Chuo University Grant for Special Research.

\section{Preliminaries}

Let $E$ be an elliptic curve defined over $K$.  Throughout this section we fix a positive integer $N$ greater than~$3$, and suppose that the base field $K$ is a field of characteristic not dividing $N$.  Let $K_{s}$ be a separable closure of $K$, and let $\Gamma_{K}=\operatorname{Gal}(K_{s}/K)$.
Denote by $E(K)$ the group of $K$-rational points of~$E$, and $E[N]=\{P\in E(K_{s})\mid NP=O\}$ the subgroup of $N$-torsion points.

In this section we give a summary of necessary facts about the Weil pairings and central extensions of $E[N]$.  We follow V\'elu \cite{Velu:1978} faithfully, but we included here for the convenience of the reader.

\subsection{The Weil pairing}

We first fix some notation.
\begin{itemize}
\item $K(E)^{\times}$ : the multiplicative group of the function field $K(E)$ of $E$.
\item $\operatorname{Div}_{K}(E)$ : the group of $K$-divisors of $E$; i.e., the group of formal $\Z$-linear combination $D=\sum_{P\in E(K_{s})} n_{P}\{P\}$%
\footnote{For a point $P\in E(K_{s})$, we denote by $\{P\}$ the base of the formal sums associated with~$P$.  Thus, the sum $\{P\}+\{Q\}$ is a formal sum, while the sum in $\{P+Q\}$ means the addition in the elliptic curve~$E$.}%
 such that $n_{P}\in \Z$, $n_{P}=0$ for all but finitely many $P\in E(K_{s})$, and $n_{\, \action{\sigma}\!P}=n_{P}$ for all $\sigma\in \Gamma_{K}=\operatorname{Gal}(K_{s}/K)$.
\item $\operatorname{Div}_{K}^{0}(E)$ : the kernel of $\deg:\operatorname{Div}_{K}(E)\to \Z$.
\item $\operatorname{Nul}_{K}(E)$ : the kernel of the homomorphism $\operatornamewithlimits{\mathrm{Sum}}:\operatorname{Div}_{K}(E)\to E(K)$ defined by $\sum_{P}n_{P}\{P\}\mapsto\operatornamewithlimits{\mathrm{Sum}}\limits_{P}\,n_{P}P$, where $\operatornamewithlimits{\mathrm{Sum}}$ means the addition in~$E$.
\item $P_{K}(E) = \operatorname{Div}_{K}^{0}(E)\cap \operatorname{Nul}_{K}(E)$.
\end{itemize}

We have an obvious homomorphism $\div:K(E)^{\times}\to P_{K}(E)$, and the following is fundamental.

\begin{theorem}[Abel-Jacobi]\label{th:able}
The following sequence is exact.
\begin{equation}\label{eq:Abel}
1 \longrightarrow K^{\times} \longrightarrow K(E)^{\times}
\overset{\div}{\longrightarrow} P_{K}(E) \longrightarrow 0.
\qedhere
\end{equation}
\end{theorem}

Let $G\subset E[N](K)$ be a cyclic group of order~$N$.  We consider \eqref{eq:Abel} as the exact sequence of $G$-modules.
Define an action of $G$ on various groups as follows:
\begin{itemize}
\item trivially on $K^{\times}$, $\Z$, and $E(K)$.
\item by translation on $K(E)$; i.e., if $f\in K(E)$ and $T\in G$, define $\action{T}\!f(X)=f(X-T)$, where $X$ is a generic point of~$E$.
\item by translation on $\operatorname{Div}_{K}(E)$; i.e., if $D}%{\mathfrak{d}=\sum_{P}n_{P}\{P\}$ and $T\in G$, define
\[
\action{T}\!D}%{\mathfrak{d}=\sum_{P}n_{P}\{P+T\}.
\]
\end{itemize}
Clearly, the action of $G$ on $\operatorname{Div}_{K}(E)$ induces actions on $\operatorname{Div}_{K}^{0}(E)$ and $P_{K}(E)$.
The Weil pairing is obtained essentially as the connecting homomorphism $\delta$ of the long exact sequence of the group cohomology obtained from \eqref{eq:Abel}:
\begin{equation}\label{eq:long-exact-seq}
0 \longrightarrow {K(E)^{\times}}^{G} \overset{\div}\longrightarrow P_{K}(E)^{G}
\overset{\delta}{\longrightarrow} H^{1}(G,K^{\times}).
\end{equation}

\begin{theorem}\label{thm:weil-pairing}
\begin{enumerate}
\item
The homomorphism $\delta$ in \eqref{eq:long-exact-seq} induces  a $\Gamma_{K}$-isomor\-phism $\Psi:G'=E[N]/G \overset{\simeq}{\longrightarrow}\operatorname{Hom}(G,\mu_{N})$.
\item
For $S'\in G'$ and $T\in G$, define $e_{G}(S',T)$ by
\[
e_{G}(S',T)=\Psi(S')(T).
\]
Then $e_{G}$ is a $\Gamma_{K}$ compatible nondegenerate bilinear form $G'\times G\to \mu_{N}$.
\item
For $S'\in G'$, choose $S\in E[N]$ such that $S\bmod G=S'$, and define $D}%{\mathfrak{d}_{S}=\{S\}-\{O\}\in \operatorname{Div}_{K_{s}}^{0}(E)$.  Choose $f_{S}\in K_{s}(E)^{\times}$ such that
\[
\div f_{S} = \sum_{T\in G}\action{T}\!D}%{\mathfrak{d}_{S}
= \sum_{T\in G}\bigl(\{S+T\}-\{T\}\big).
\]
Then, we have%
\[
e_{G}(S',T)=\action{T}\!f_{S}/f_{S}=f_{S}(X-T)/f_{S}(X).
\]
\end{enumerate}
\end{theorem}

\begin{proof}
Since $G\subset E[N](K)$ by assumption, we have $\mu_{N}\in K^{\times}$ and thus $H^{1}(G,K^{\times})\simeq \operatorname{Hom}(G,\mu_{N})$.  The proof is a standard diagram chasing.  See V\'elu \cite[Ch.~1]{Velu:1978} for detail.
\end{proof}

The usual Weil pairing $e_{N}:E[N]\times E[N]\to \mu_{N}$ is defined in a similar manner.  The relation between $e_{G}$ and $e_{N}$ is given by
\[
e_{N}(S,T)=e_{G}(\phi(S),T) \quad \text{for $S, T\in E[N]$},
\]
where $\phi$ is the isogeny $E\to E/G$.

\subsection{Central extension $E[N](D}%{\mathfrak{d})$}
A central extension of a group $G$ is a short exact sequence of groups
$1\to A\to H\to G\to 1$ such that $A$ is in the center of the group $H$.
Here, we consider central extensions of $E[N]$ by $K^{\times}$, that is, short exact sequences of groups
\[
1 \longrightarrow K^{\times} \longrightarrow H
\longrightarrow E[N] \longrightarrow 0.
\]
(Note that the operation of $K^{\times}$ is noted multiplicatively, while that of $E[N]$ is noted additively.)

For a divisor $D}%{\mathfrak{d}\in\operatorname{Div}_{K}(E)$, we can construct a central extension.

\begin{definition}\label{def:extension}
Let $D}%{\mathfrak{d}\in \operatorname{Div}_{K}(E)$ be a divisor of degree divisible by~$N$.  Define
\[
E[N](D}%{\mathfrak{d})=\{(T,f)\in E[N]\times K(E)^{\times}\mid \div f =\action{T}\!D}%{\mathfrak{d}-D}%{\mathfrak{d}\},
\]
with a group operation on $E[N](D}%{\mathfrak{d})$ given by
\[
(S,g)(T,f)=(S+T,g\cdot\action{S}\!f).
\]
\end{definition}

Indeed, we have a natural inclusion $K^{\times}\to E[N](D}%{\mathfrak{d})$ given by $c\mapsto (O,c)$, and an exact sequence
\[
1 \longrightarrow K^{\times} \longrightarrow E[N](D}%{\mathfrak{d})
\longrightarrow E[N] \longrightarrow 0.
\]
From now on, we identify $(O,c)\in E[N](D}%{\mathfrak{d})$ with $c\in K^{\times}$ and consider $K^{\times}$ as a subgroup in $E[N](D}%{\mathfrak{d})$ (to ease the notation).

\begin{remark}\label{rem:Heisenberg}
If $N$ is odd, $E[N](D}%{\mathfrak{d})$ is essentially the Heisenberg group attached to $E[N]$.
\end{remark}

\begin{lemma}\label{lem:lin-equiv}
If two divisors $D}%{\mathfrak{d}_{1}$ and $D}%{\mathfrak{d}_{2}$ are linearly equivalent, then the extensions $E[N](D}%{\mathfrak{d}_{1})$ and $E[N](D}%{\mathfrak{d}_{2})$ are isomorphic.
\end{lemma}

\begin{proof} Straightforward.
\end{proof}

In general $E[N](D}%{\mathfrak{d})$ is not commutative and each element may be of infinite order.
For $S,T\in E[N]$, we denote the commutator and the $N$th power by
\begin{align}\label{eq:<T,T'>}
&\<S,T\> =(S,g)(T,f)(S,g)^{-1}(T,f)^{-1}
=(g/\action{T}\!g)\cdot(\action{S}\!f/f)
\\
&v(T)=(T,f)^{N}=f\cdot\action{T}f\cdot\action{2T}f\cdot\dots\cdot\action{(N-1)T}f.
\end{align}
Recall that we identify $(O,c)$ with $c$.  It is easy to see that $\<S,T\>$ and $v(T)$ are independent of the choice of $f$ and $g$.

\begin{proposition}\label{prop:<T,T'>}

Let $D}%{\mathfrak{d}$ be a divisor of degree divisible by $N$, and let $S,T\in E[N]$.  Then, we have
\begin{enumerate}
\item
$\<S,T\>$ is a bilinear form on $E[N]$ with its value in $\mu_{N}$, and
\[
\<S,T\> = e_{N}(T,S)^{\frac{\degD}%{\mathfrak{d}}{N}}.
\]
\item
$v(T)$ is determined modulo ${K^{\times}}^{N}$
\end{enumerate}\qedhere
\end{proposition}

\begin{proof}
V\'elu \cite[Def.~2.2 and Prop.~2.3]{Velu:1978}.
\end{proof}

The case where $D}%{\mathfrak{d}=N\{O\}$ is of particular interest.   In this case, the results depend on the parity of~$N$.

\begin{proposition}\label{prop:E[N](NO)}
Let $D}%{\mathfrak{d}$ be a divisor linearly equivalent to $N\{O\}$. Let $T$ be in $E[N]$, $\varphi$ a function in $K_{s}(E)$ such that $\div\varphi=N\{T\}-N\{O\}$.
\begin{enumerate}
\item
If $N$ is odd, $v$ is trivial; i.e., $v(T)\equiv 1 \bmod {K^{\times}}^{N}$.
\item
Let $\tilde T$ be a point in $E[2N]$ such that $2\tilde T=T$.  Then, we have $v(T)\equiv \varphi(\tilde T)^{N} \bmod {K^{\times}}^{N}$.  As a consequence,
the homomorphism $v^{2}$ is trivial.
\end{enumerate}
\end{proposition}

\begin{proof}
V\'elu \cite[Prop.~2.6]{Velu:1978}.
\end{proof}

\section{Projective immersion of an elliptic curve associated with a cyclic subgroup}

In this section we assume $E[N]\subset E(K)$.
Let $D}%{\mathfrak{d}\in \operatorname{Div}_{K}(E)$ be a divisor.  If $\deg D}%{\mathfrak{d}\ge 3$, it is well known that the complete linear system $|D}%{\mathfrak{d}|$ gives an immersion $E\hookrightarrow |D}%{\mathfrak{d}|$ in a projective space.  Here, we take a closer look at this fact.

\begin{definition}\label{def:L(D)}
For $D}%{\mathfrak{d}\in \operatorname{Div}_{K}(E)$ and nonnegative integer $d$, define
\begin{align*}
&\L^{d}(D}%{\mathfrak{d}) =\{ h\in K(E)^{\times}\mid \div h + dD}%{\mathfrak{d}\geq 0\} \cup \{0\}
\\
&\L^{*}%{\bullet}(D}%{\mathfrak{d})=\textstyle{\bigoplus_{d\ge0}}\,\L^{d}(D}%{\mathfrak{d}),
\\
&\mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})=\operatorname{Sym}^{*}%{\bullet}\L^{1}(D}%{\mathfrak{d}).
\end{align*}
\end{definition}
By the Riemann-Roch theorem, $\L^{d}(D}%{\mathfrak{d})$ is a $K$-vector space of dimension $d\deg D$, and the space $\L^{*}%{\bullet}(D}%{\mathfrak{d})$ is equipped naturally with a structure of graded algebra by the multiplication of functions.
There is a canonical homomorphism of graded algebras $\mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})\to \L^{*}%{\bullet}(D}%{\mathfrak{d})$, and let $\mathscr{I}^{*}%{\bullet}(D}%{\mathfrak{d})$ be its kernel. We have the exact sequence
\[
0\longrightarrow \mathscr{I}^{*}%{\bullet}(D}%{\mathfrak{d}) \longrightarrow \mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})
\longrightarrow \L^{*}%{\bullet}(D}%{\mathfrak{d}).
\]

\begin{proposition}\label{prop:proj-normality}
Let $D$ be an effective divisor of degree $N\geq 4$.
\begin{enumerate}
\item There is an exact sequence
\[
0\longrightarrow \mathscr{I}^{*}%{\bullet}(D}%{\mathfrak{d}) \longrightarrow \mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})
\longrightarrow \L^{*}%{\bullet}(D}%{\mathfrak{d}) \longrightarrow 0.
\]
\item
$\dim\mathscr{I}^{2}(D}%{\mathfrak{d})=N(N-3)/2$.
\item
$\mathscr{I}^{*}%{\bullet}(D}%{\mathfrak{d})$ is generated by $\mathscr{I}^{2}(D}%{\mathfrak{d})$, i.e., $\mathscr{I}^{*}%{\bullet}(D}%{\mathfrak{d})=\mathscr{I}^{2}(D}%{\mathfrak{d})\cdot \mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})$.
\item
$E\hookrightarrow \operatorname{Proj} \mathscr{S}^{*}%{\bullet}(D}%{\mathfrak{d})\simeq \P^{N-1}$ is the scheme-theoretic intersection of the quadrics which contain it.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) \ The right exactness of this sequence is one of the equivalent definition of projective normality.  For proof see V\'elu \cite[Th.~3.3]{Velu:1978}.  See also Hartshorne \cite[Ex.~IV.4.2]{Hartshorne:AG} and Mumford \cite[p.~55]{Mumford:1970a}. (2) follows immediately from (1) as $\dim \mathscr{S}^{2}(D}%{\mathfrak{d})=N(N+1)/2$ and $\dim \L^{2}(D}%{\mathfrak{d})=2N$.

(3) \ V\'elu \cite[Th.~3.9]{Velu:1978}.  (4) follows from (3).
\end{proof}

The central extension $E[N](D}%{\mathfrak{d})$ has a representation in $\L^{*}%{\bullet}(D}%{\mathfrak{d})$.  More precisely, we have
\begin{proposition}\label{prop:L(D)-rep}
\begin{enumerate}
\item
For $h\in \L^{d}(D}%{\mathfrak{d})$ and $(T,f)\in E[N](D}%{\mathfrak{d})$, the function $\action{T}h\cdot f^{d}$ is in $\L^{d}(D}%{\mathfrak{d})$.
\item
If we denote by $\tau_{d}(T,f)$ the automorphism $h\mapsto \action{T}h\cdot f^{d}$ of $\L^{d}(D}%{\mathfrak{d})$, the map $\tau_{d}:(T,f)\mapsto \tau_{d}(T,f)$ is a representation of $E[N](D}%{\mathfrak{d})$ in $\L^{d}(D}%{\mathfrak{d})$.  In addition, the representations $\tau_{d}$ extend to a representation $\tau$ of $E[N](D}%{\mathfrak{d})$ in the group of automorphism of the graded ring $\L^{*}%{\bullet}(D}%{\mathfrak{d})$.
\item
If $D}%{\mathfrak{d}$ is an effective divisor of degree $\geq 2$ and $d\geq 1$, the kernel of the representation $\tau_{d}$ consists of the elements $(O,f)$, where $f$ is a constant satisfying $f^{d}=1$. In particular, $\tau_{1}$ is faithful.
\end{enumerate}
\end{proposition}

\begin{proof}
V\'elu \cite[Prop.~2.13]{Velu:1978}.
\end{proof}

We call $\tau$ the canonical representation of $E[N](D}%{\mathfrak{d})$.

\begin{lemma}\label{lem:G(dd)-lin-equiv}
If $D}%{\mathfrak{d}_{1}$ and $D}%{\mathfrak{d}_{2}$ are two linearly equivalent effective divisors, the canonical representations of the extensions $E[N](D}%{\mathfrak{d}_{1})$ and $E[N](D}%{\mathfrak{d}_{2})$ are isomorphic. Moreover, if $\sigma:E[N]\to E[N](D}%{\mathfrak{d}_{1})$ is a section \textup{(}as sets\textup{)} of $E[N](D}%{\mathfrak{d}_{1})\to E[N]$, then the map $\mu\sigma$ is also a section of $E[N](D}%{\mathfrak{d}_{2})\to E[N]$, and the associated $2$-cocycles are equal.
\end{lemma}

\begin{proof}
Let $\varphi$ be a function such that $\div\varphi=D}%{\mathfrak{d}_{2}-D}%{\mathfrak{d}_{1}$. Define the maps
\(
\nu: E[N](D}%{\mathfrak{d}_{1})\to E[N](D}%{\mathfrak{d}_{2})\) and \
\mu: \L^{*}%{\bullet}(D}%{\mathfrak{d}_{1})\to \L^{*}%{\bullet}(D}%{\mathfrak{d}_{2}),
\)
by
\[
\nu\bigl((T,f)\bigr)=\bigl(T,f\cdot\action{T}\!\varphi/\varphi\bigr)
 \quad\text{ and }\quad
\mu(h)=h/\varphi^{d}
\]
for any $(T,f)\in E[N](D}%{\mathfrak{d})$ and $h\in \L^{d}(D}%{\mathfrak{d})$.
Then it is easy to show that they are isomorphisms, and we have
\[
\tau_{d}\bigl(\nu\bigl((T,f)\bigr)\bigr)\mu(h)
=\mu\bigl(\tau_{d}(T,f)h\bigr).
\]
The last statement is also straight forward.
\end{proof}

Let $C\subset E[N]$ be a cyclic subgroup of order $N$.  Suppose $D}%{\mathfrak{d}$ is an effective divisor of degree $N$ invariant under the translations by $C$.  Then, we may choose a particular coordinate system of $\P^{N-1}$ so that the immersion $E\hookrightarrow |D}%{\mathfrak{d}|\simeq \P^{N-1}$ is expressed in a simple way.  To do so, we consider the decomposition of $\L^{1}(D}%{\mathfrak{d})$ into eigenspaces.

\begin{definition}\label{def:L(D,chi)}
For any character $\chi\in \operatorname{Hom}(C,\mu_{N})$ and a positive integer~$d$, define
\[
\L^{d}(D}%{\mathfrak{d},\chi) = \{h\in \L^{d}(D}%{\mathfrak{d})\mid
\action{U}h=\chi(U)\cdot h \text{ for all $U$ in $C$} \}.
\]
Then, define
\[
\L^{d}(D}%{\mathfrak{d}) = \textstyle{\bigoplus_{\chi}}\L^{d}(D}%{\mathfrak{d},\chi),
\quad
\text{and}
\quad
\L^{*}%{\bullet}(D}%{\mathfrak{d})=\textstyle{\bigoplus_{d,\chi}}\L^{d}(D}%{\mathfrak{d},\chi).
\]
\end{definition}

If $h\in \L^{d}(D}%{\mathfrak{d},\chi)$ and $h'\in \L^{d'}(D}%{\mathfrak{d},\chi')$, then the function $hh'$ belongs to $\L^{d+d'}\bigl(D}%{\mathfrak{d},\chi\chi')$.  As a consequence, $\L^{*}%{\bullet}(D}%{\mathfrak{d})$ has a structure of graded algebra by $\mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}\times \operatorname{Hom}(C,\mu_{N})$.

\begin{definition}\label{def:char_S}
Let $C=\<T\>\subset E[N]$ be the cyclic subgroup generated by $T$.  For any $S\in E[N]$, define the character $\chi_{S}\in \operatorname{Hom}(C,\mu_{N})$ by
\[
\chi_{S}(U)=e_{N}(S,U) \quad\text{for all $U\in C$}.
\]
\end{definition}

\begin{proposition}\label{prop:L(D)}
\begin{enumerate}
\item
The space $\L^{1}(D}%{\mathfrak{d},\chi)$ is $1$-dimensional.
\item
If $(S,f_{S})$ is in $E[N](D}%{\mathfrak{d})$, the function $f_{S}$ is a base of $\L^{1}(D}%{\mathfrak{d},\chi_{S})$.
\item
The map $\tau_{d}(S,f_{S})$ from $\L^{d}(D}%{\mathfrak{d},\chi)$ to $\L^{d}(D}%{\mathfrak{d},\chi\chi_{S}^{d})$ is an isomorphism for all $d\ge 0$.
\end{enumerate}
\end{proposition}

\begin{proof}
Let $(S,f_{S})$ be in $E[N](D}%{\mathfrak{d})$. Since $\div f_{S} + D}%{\mathfrak{d}=\action{S}D}%{\mathfrak{d}\ge0$, $f_{S}$ is in $\L^{1}(D}%{\mathfrak{d})$.
For $U\in C$, we have $(U,1)\in E[N](D}%{\mathfrak{d})$.  By \eqref{eq:<T,T'>}, we have $\<U,S\>={\action{U}\!f_{S}}/{f_{S}}$, and by Proposition~\ref{prop:<T,T'>}, we have $\<U,S\>=e_{N}(S,U)=\chi_{S}(U)$.  Thus, we have $\action{U}\!f_{S}=\chi_{S}(U)f_{S}$, which shows $g\in \L(D}%{\mathfrak{d},\chi_{S})$.

If $h\in \L^{d}(D}%{\mathfrak{d},\chi)$, then
\[
\action{U}(\action{S}h\cdot f_{S}^{d}) = \action{U+S}h\cdot(\action{U}\!f_{S})^{d}=\chi(U)\cdot\action{S}h\cdot(\action{U}\!f_{S})^{d}
=\chi(U)\chi_{S}(U)^{d}(\action{S}h\cdot f_{S}^{d}).
\]
This implies $\tau_{d}(S,f_{S})$ sends $\L^{d}(D}%{\mathfrak{d},\chi)$ to $\L^{d}(D}%{\mathfrak{d},\chi\chi_{S}^{d})$.  As a consequence, $\tau_{d}$ permutes among the spaces $\L^{d}(D}%{\mathfrak{d},\chi)$ transitively as long as $d$ is relatively prime to~$N$.  In that case, the spaces $\L^{d}(D}%{\mathfrak{d},\chi)$ have the same dimension independent of~$\chi$.

By the Riemann-Roch theorem, $\dim \L^{1}(D}%{\mathfrak{d})=\degD}%{\mathfrak{d}=N$.  On the other hand
\begin{align*}
\dim \L^{1}(D}%{\mathfrak{d}) &= \sum_{\chi}\dim \L^{1}(D}%{\mathfrak{d},\chi)
= \#\operatorname{Hom}(C,\mu_{N}) \cdot\dim \L^{1}(D}%{\mathfrak{d},\chi_{0}) \\
&= N \dim \L^{1}(D}%{\mathfrak{d},\chi_{0}),
\end{align*}
where $\chi_{0}$ is the trivial character.  Thus, we conclude that $\dim \L^{1}(D}%{\mathfrak{d},\chi)=1$ for any $\chi$, and any nonzero function $h$ in $\L^{1}(D}%{\mathfrak{d},\chi_{S})$ is its base.
\end{proof}

\begin{proposition}\label{prop:I^2}
For any character $\chi\in \operatorname{Hom}(C,\mu_{N})$, the sequence
\[
0 \longrightarrow \mathscr{I}^{2}(D}%{\mathfrak{d},\chi)
\longrightarrow \mathscr{S}^{2}(D}%{\mathfrak{d},\chi)
\longrightarrow \L^{2}(D}%{\mathfrak{d},\chi)
\longrightarrow 0
\]
is exact.  Moreover, we have
\[
\dim \mathscr{I}^{2}(D}%{\mathfrak{d},\chi) = \frac{N-4+r_{\chi}}{2},
\]
where $r_{\chi}$ is a number of $\chi'\in \operatorname{Hom}(C,\mu_{N})$ such that $\chi'^{2}=\chi$.
\end{proposition}

\begin{proof}
V\'elu \cite[Cor.~3.6]{Velu:1978}.
\end{proof}

\section{Canonical coordinate system}

In this section we consider the projective immersion $E\hookrightarrow \bigl|N\{O\}\bigr|\simeq \P^{N-1}$.  We show that, by fixing a certain level structure on $E$, we can choose a unique coordinate system with prescribed properties.  The situation differs depending on the parity of $N$.  The odd case is well known, but we include here for the comparison with the even case.

\subsection{Odd case}\label{subsec:4.1}

Throughout this subsection we assume $N$ to be a positive odd integer.  This section serves as a prototype for the even case, and all the material in this section is written in V\'elue \cite{Velu:1978}.  See also Fisher \cite{Fisher:thesis}, \cite{Fisher:2001} for geometric treatment.

Consider the action on the pair of points in $E[N]$ by an element $\sltwo(a,b;c,d)\in M_{2}(\Z)$ given by
\[
(S',T')=(S,T)\sltwo(a,b;c,d)=(aS+cT,bS+dT).
\]
Recall that if the pair $(S, T)$ is a basis of $E[N]$ satisfying $e_{N}(S,T)=\zeta$, then the pair $(S',T')=(aS+cT,bS+dT)$ is once again a basis of $E[N]$ satisfying $e_{N}(S',T')=\zeta$ if and only if $\sltwo(a,b;c,d)\in SL_{2}(\Z)$.  The kernel of this action is given by
\[
\Gamma(N) = \left\{\left.
\sltwo(a,b;c,d)\in SL_{2}(\Z) \,\right|\,
\sltwo(a,b;c,d)\equiv \sltwo(1,0;0,1)\bmod N\right\},
\]
which is the principal congruence subgroup of level $N$ in $SL_{2}(\Z)$.

\begin{definition}[$\Gamma(N)$ structure]\label{def:level-Gamma(N)}
Fix a primitive $N$-th root of unity $\zeta$. A \emph{$\Gamma(N)$-structure} on an elliptic curve $E$ is a pair $(S, T)$ of points in $E[N]$ satisfying $e_{N}(S,T)=\zeta$.
\end{definition}

From now on we fix a primitive $N$th root of unity $\zeta$ in $K_{s}$, and a $\Gamma(N)$-structure $(S,T)$.  Define the divisor $D}%{\mathfrak{d}_{T}$ by
\[
D}%{\mathfrak{d}_{T}=\sum_{n=0}^{N-1}\{nT\}.
\]
Clearly, $D}%{\mathfrak{d}_{T}$ is invariant under the translation by~the cyclic group $C=\<T\>$.

Since $\operatornamewithlimits{\mathrm{Sum}}_{n=0}^{N-1}nT=O$, $D}%{\mathfrak{d}_{T}$ is linearly equivalent to~$N\{O\}$.  Choose $f_{S,T}$ such that
\[
\div f_{S,T}=\action{S}D}%{\mathfrak{d}_{T}-D}%{\mathfrak{d}_{T}.
\]
Since $(S,f_{S,T})^{N}\equiv 1 \bmod {K^{\times}}^{N}$\! by Proposition~\ref{prop:E[N](NO)} a), by multiplying a suitable constant if necessary, we may assume $(S,f_{S,T})^{N}=1$.  Define functions $X_{0},X_{1},\dots,$ $X_{N-1}\in K(E)$ indexed by the elements of $k\in\Z/N\Z$ by
\begin{align*}
&X_{0} = 1,
 \\
&X_{k}
=\tau(S,f_{S,T})X_{k-1}
\quad \text{for $1\le k \le N-1$.}
\end{align*}

\begin{proposition}\label{prop:X_k-odd}
\begin{enumerate}
\item
$X_{k}$ is a basis of $\L^{1}(D}%{\mathfrak{d}_{T},\chi_{S}^{k})$ for $k\in \Z/N\Z$.
\item
The actions of $(T,1)\in E[N](D}%{\mathfrak{d}_{T})$ is given by
\[
\tau(T,1)X_{k}=\action{T}X_{k}=e_{N}(S,T)^{k}X_{k}.
\]
\item
$X_{k}(-P)=X_{-k}(P)$ for any $P\in E$.
\end{enumerate}
\end{proposition}

Interpreting the above proposition geometrically, we have

\begin{proposition}\label{prop:proj-coord-odd}
Let $E$ be an elliptic curve, and $E\hookrightarrow\P^{N-1}$ be an immersion as an elliptic normal curve of degree~$N$ via the complete linear system $|N\{O\}|$.  Choose a primitive $N$th root of unity $\zeta$ and a $\Gamma(N)$-structure $(S,T)$. Then, there exists a unique coordinate system of $\P^{N-1}$ such that the translation maps $\tau_{S}$, $\tau_{T}$, and the multiplication-by-$(-1)$ map $[-1]$ are given by the following elements $M_{T}$, $M_{S}$ and $M_{[-1]}$ in $PGL_{N}(K(\zeta))$, respectively.
\begin{equation}\label{eq:matrices}
\begin{gathered}
M_{S}=
\setlength{\arraycolsep}{3pt}
\renewcommand{\arraystretch}{0.9}
\left[\begin{array}{*5c|c}
0 &  &  \cdots &  &  0 &  1\\\hline
1 &   &   &   &   &  0 \\
  & 1 &   &   &   &   \\
  &  & \ddots  &  &  & \vdots   \\
  &   &   & 1 &   &   \\
  &   &   &   & 1 & 0
\end{array}\right],
\quad
M_{T}=
\renewcommand{\arraystretch}{1.1}
\left[\begin{array}{*5c}
1 &   &   &   &   \\
  & \zeta &   &   &   \\
  &   & \zeta^{2} &   &   \\
  &   &   & \ddots &   \\
  &   &   &   & \zeta^{N-1}
\end{array}\right],
\\
\setlength{\arraycolsep}{3pt}
\renewcommand{\arraystretch}{0.9}
M_{[-1]}=\left[\begin{array}{c|*5c}
1 & 0 &   & \cdots  &   & 0 \\\hline
0 &   &   &   &   & 1 \\
  &   &   &   & 1 &  \\
\vdots  &   &   & \scalebox{-1}[1]{$\ddots$} &   &   \\
  &   & 1 &   &    &   \\
0 & 1 &   &   &   &
\end{array}\right]
\end{gathered}
\end{equation}
\end{proposition}

\begin{remark}
When we use the square bracket $[\quad ]$ for matrix, we mean it is a class in $PGL$ or $PSL$.
\end{remark}

\begin{proposition}\label{prop:quad-eqs-odd}
The notation being as above, we have the following.
\begin{align*}
&\mathscr{I}^{2}(D}%{\mathfrak{d}_{T})
= \textstyle{\bigoplus_{k\in\Z/N\Z}}\mathscr{I}^{2}(D}%{\mathfrak{d}_{T},\chi_{S}^{k}),\\
&\mathscr{I}^{2}(D}%{\mathfrak{d}_{T},\chi_{S}^{k})=\mathscr{I}^{2}(D}%{\mathfrak{d}_{T})\cap \<X_{i}X_{j}\mid i+j\equiv k \mod N\>,
\\
&\dim \mathscr{I}^{2}(D}%{\mathfrak{d}_{T},\chi_{S}^{k}) =(N-3)/2, \quad
\dim \mathscr{I}^{2}(D}%{\mathfrak{d}_{T}) =N(N-3)/2
\end{align*}
In other words, $E\hookrightarrow \P^{N-1}$ is defined by $N(N-3)/2$ quadratic equations.
\end{proposition}

\subsection{Even case}\label{subsec:4.2}

In this subsection we assume $N$ to be a positive even integer.
A critical difference in this case is that $\operatornamewithlimits{\mathrm{Sum}}_{n=0}^{N-1}nT=\frac{N}{2}T\neq O$, and thus $D}%{\mathfrak{d}_{T}=\sum_{n=0}^{N-1}\{nT\}$ is \emph{not} linearly equivalent to~$N\{O\}$.  To define a divisor linearly equivalent to $N\{O\}$, we use a $\tilde T$ such that $2\tilde T=T$.

First we fix a primitive $N$-th root of unity $\zeta$ and a $\Gamma(N)$-structure $(S,T)$, then we consider $\Gamma(2N)$-structures on $E$ in relation to this fixed $\Gamma(N)$-structure.  To do so, we first choose a primitive $2N$th root of unity $\tilde\zeta$ such that $\tilde\zeta^{2}=\zeta$.

\begin{definition}\label{def:level-2N}
Suppose $(S,T)$ is a $\Gamma(N)$-structure on $E$ with $e_{N}(S,T)=\zeta=\tilde \zeta^{2}$.
\begin{enumerate}
\item
We say that a $\Gamma(2N)$-structure $(\tilde S,\tilde T)$ on $E$ is \emph{above} $(S,T)$ if
\[
2\tilde S=S, \quad 2\tilde T=T, \quad\text{and}\quad
e_{2N}(\tilde S,\tilde T)=\tilde\zeta.
\]
\item
We say that two $\Gamma(2N)$-structures $(\tilde S,\tilde T)$ and $(\tilde S',\tilde T')$ are \emph{similar} if $(\tilde S',\tilde T')$ equals either
\[
(\tilde S,\tilde T) \quad\text{or}\quad
(\tilde S,\tilde T)\sltwo(1+N,0;0,1+N).
\]
\end{enumerate}
\end{definition}

\begin{lemma}\label{lem:level-N-2N}
For a given $\Gamma(N)$-structure $(S,T)$, there are eight $\Gamma(2N)$-structures $(\tilde S,\tilde T)$ above $(S,T)$.  Up to similarity, these eight are classified into four classes.  If $(\tilde S,\tilde T)$ is one of them, then the following four represent the four different classes\/:
\[
(\tilde S,\tilde T), \quad
(\tilde S,\tilde T)\sltwo(1,N;0,1),\quad
(\tilde S,\tilde T)\sltwo(1,0;N,1),\quad
(\tilde S,\tilde T)\sltwo(1,N;N,1).
\]
\end{lemma}

\begin{proof}
There are sixteen pairs $(\tilde S,\tilde T)$ satisfying $2\tilde S=S$ and $2\tilde T=T$. Choose one pair $(\tilde S,\tilde T)$.  Replacing $\tilde S$ by $\tilde S+\frac{N}{2}S$ if necessary, we may assume $e_{2N}(\tilde S,\tilde T)=\tilde\zeta$.  Thus, we obtain at least one $\Gamma(2N)$-structure above $(S,T)$.  All the pairs $(\tilde S',\tilde T')$ satisfying $2\tilde S'=S$ and $2\tilde T'=T$ are given by
\[
(\tilde S,\tilde T) +(\tfrac{N}{2}S,\tfrac{N}{2}T)\sltwo(\epsilon_{1},\epsilon_{3};\epsilon_{2},\epsilon_{4}),
\quad \epsilon_{j}=0, \text{or } 1 \ (j=1,\dots,4).
\]
Now, using properties of the Weil pairings, we have
\begin{multline*}
e_{2N}(\tilde S+\epsilon_{1}\tfrac{N}{2}S+\epsilon_{2}\tfrac{N}{2}T,
\tilde T+\epsilon_{3}\tfrac{N}{2}S+\epsilon_{4}\tfrac{N}{2}T)
\\
=e_{2N}(\tilde S,\tilde T) e_{2N}(\tilde S,\tfrac{N}{2}T)^{\epsilon_{4}}e_{2N}(\tfrac{N}{2}S,\tilde T)^{\epsilon_{1}}
=(-1)^{\epsilon_{1}+\epsilon_{4}}e_{2N}(\tilde S,\tilde T).
\end{multline*}
Thus, the Weil pairing coincide if and only if $\epsilon_{1}=\epsilon_{4}$.  This implies that there are eight different $\Gamma(2N)$-structures above $(S,T)$.  It is also easy to see that the four pairs above represent the different classes up to similarity.
\end{proof}

In view of Lemma~\ref{lem:level-N-2N}, we define a subgroup $\Gamma^{(N)}(2N)$ of $\Gamma(N)$ by
\[
\Gamma^{(N)}(2N) =\left\{\left.\sltwo(a,b;c,d) \in SL_{2}(\Z)\,\right|\,
a\equiv d\equiv 1 \bmod
, b\equiv c\equiv 0\bmod 2N\right\}.
\]
It is easy to see that
\[
\Gamma(N)\vartriangleright\Gamma^{(N)}(2N)\vartriangleright\Gamma(2N).
\]
Since $N$ is even, we have
\begin{gather*}
\Gamma(N)/\Gamma^{(N)}(2N)
=\left\<\sltwo(1,N;0,1), \sltwo(1,0;N,1)\right\>
\simeq \Z/2\Z\times\Z/2\Z,
\\
\Gamma^{(N)}(2N)/\Gamma(2N)
=\left\<\sltwo(1+N,0;0,1+N)\right\>
\simeq \Z/2\Z.
\end{gather*}

\begin{remark}
If $N$ is odd, we have $\Gamma^{(N)}(2N)$ conincide with $\Gamma(2N)$, and $[\Gamma(N):\Gamma^{(N)}(2N)]=6$.
\end{remark}

\begin{definition}[$\Gamma^{(N)}(2N)$-structure]%
\label{def:Gamma^(N)(2N)-structure}
Fix a primitive $N$th root of unity $\zeta$, and choose $\tilde \zeta$ such that $\tilde\zeta^{2}=\zeta$.
Let $(S,T)$ be a $\Gamma(N)$-structure.
By a \emph{$\Gamma^{(N)}(2N)$-structure} we mean an equivalence class of $\Gamma(2N)$-structure $(\tilde S,\tilde T)$ above $(S,T)$ modulo similarity.
\end{definition}

We now fix a $\Gamma^{(N)}(2N)$-structure $(\tilde S, \tilde T)$ above $(S,T)$.
Define
\[
D}%{\mathfrak{d}_{\tilde T}=\sum\limits_{n=0}^{N-1}\bigl\{\tilde T+nT\bigr\}.
\]

\begin{lemma}\label{lem:D-tilde-T}
\begin{enumerate}
\item
The divisor $D}%{\mathfrak{d}_{\tilde T}$ is linearly equivalent to $N\{O\}$, and there is a function $\varphi_{\tilde T}$ such that $\div \varphi_{\tilde T}=D}%{\mathfrak{d}_{\tilde T}-N\{O\}$.
\item
Suppose $\tilde T'$ is another point satisfying $2\tilde T'=T$, and let $T_{2}=\tilde T'-\tilde T\in E[2]$.  Then we have
\[
D}%{\mathfrak{d}_{\tilde T'}=\action{T_{2}}D}%{\mathfrak{d}_{\tilde T}.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
(1) \
Since $\operatornamewithlimits{\mathrm{Sum}}_{E}D}%{\mathfrak{d}_{\tilde T}=N\tilde T+\frac{N}{2}T=O$, the assertion follows immediately from Able's Theorem (Theorem~\ref{th:able}).  (2) follows immediately from the definition.
\end{proof}

\begin{remark}
We may still use $D}%{\mathfrak{d}_{T}=\sum_{n=1}^{N-1}nT$ to obtain an immersion to $\P^{N-1}$ even though $\operatornamewithlimits{\mathrm{Sum}}_{E}D}%{\mathfrak{d}_{T}\neq O$.  This is how Hurwitz \cite{Hurwitz} and Stevens \cite{Stevens} studied.
\end{remark}

For a given $\Gamma^{(N)}(2N)$-structure $(\tilde S, \tilde T)$, there is a unique function $f_{\tilde S,\tilde T}$ satisfies the condition
\[
\div f_{\tilde S,\tilde T}=\action{S}\!D}%{\mathfrak{d}_{\tilde T} - D}%{\mathfrak{d}_{\tilde T},
\quad f_{\tilde S,\tilde T}(\tilde S)=1.
\]
In other words, $(S,f_{\tilde S,\tilde T})$ is in $E[N](D}%{\mathfrak{d}_{\tilde T})$.

\begin{definition}\label{def:X_k-even}
Define $X^{(\tilde S,\tilde T)}_{0},X^{(\tilde S,\tilde T)}_{1},\dots,X^{(\tilde S,\tilde T)}_{N-1}\in \L(D_{\tilde T})$ indexed by the elements of $\Z/N\Z$ by
\begin{align*}
&X^{(\tilde S,\tilde T)}_{0} = 1,
 \\
&X^{(\tilde S,\tilde T)}_{k}
=\tau(S,f_{\tilde S,\tilde T})X^{(\tilde S,\tilde T)}_{k-1}
\quad \text{for $1\le k \le N-1$.}
\end{align*}
If $(\tilde S,\tilde T)$ is understood, we write $X^{(\tilde S,\tilde T)}_{k}=X_{k}$ for simplicity.
\end{definition}

\begin{proposition}\label{prop:X_k-even}
\begin{enumerate}
\item
$X_{k}$ is well-defined, i.e.,
\[
\tau(S,f_{\tilde S,\tilde T})X_{N-1} = X_{0}
\]
\item
$X_{k}$ is a basis of $\L^{1}(D}%{\mathfrak{d}_{\tilde T},\chi_{S}^{k})$ for $k\in \Z/N\Z$.
\item
The actions of $(T,1)\in E[N](D}%{\mathfrak{d}_{\tilde T})$ is given by
\[
\tau(T,1)X_{k}=e_{N}(S,T)^{k}X_{k}.
\]
\end{enumerate}
\end{proposition}

\begin{proof}
(1) follows from the fact $(S,f_{\tilde S,\tilde T})^{N}=1$, which is by Proposition~\ref{prop:E[N](NO)} b).

(2) and (3) follow immediately form Proposition~\ref{prop:L(D)} (1) and (3)\par
\end{proof}

\begin{proposition}\label{prop:X_k-2N-str}
\begin{enumerate}
\item
If $\tilde T'$ is another point such that $2\tilde T'=T$, then
\[
X_{k}^{(\tilde S,\tilde T')}
=\begin{cases}
X_{k}^{(\tilde S,\tilde T)} & \text{ if $e_{2}(\tilde T'-\tilde T,\frac{N}{2}T)=1$,}\\[\medskipamount]
X_{k+\frac{N}{2}}^{(\tilde S,\tilde T)}& \text{ if $e_{2}(\tilde T'-\tilde T,\frac{N}{2}T)=-1$.}
\end{cases}
\]
\item
If $\tilde S'$ is another point such that $2\tilde S'=S$, then
\[
X_{k}^{(\tilde S',\tilde T)}
=\begin{cases}
X_{k}^{(\tilde S,\tilde T)} & \text{ if $e_{2}(\tilde S'-\tilde S,\frac{N}{2}S)=1$,}\\[\smallskipamount]
(-1)^{k}X_{k}^{(\tilde S,\tilde T)} & \text{ if $e_{2}(\tilde S'-\tilde S,\frac{N}{2}S)=-1$.}
\end{cases}
\]
\end{enumerate}
\end{proposition}

\begin{proof}
See V\'elu \cite[Lemma~2.7]{Velu:1978}.
\end{proof}

\begin{proposition}\label{prop:level-2N}
Let $(\tilde S,\tilde T)$ be a $\Gamma^{(N)}(2N)$-struc\-ture above a $\Gamma(N)$-struc\-ture $(S,T)$, and
let $X_{k}^{(\tilde S,\tilde T)}$ $(k=0,\dots,N)$ be the functions defined in Definition~\ref{def:X_k-even}.
The functions $X_{k}^{(\tilde S,\tilde T)}$ are uniquly determined by the $\Gamma^{(N)}(2N)$-struc\-ture.  More precisely, if $(\tilde S',\tilde T')$ is another $\Gamma(2N)$-structure above $(S,T)$, then $X_{k}^{(\tilde S',\tilde T')}=X_{k}^{(\tilde S,\tilde T)}$ for all $k$ if and only if $(\tilde S',\tilde T')$ is similar to $(\tilde S,\tilde T)$.
\end{proposition}

\begin{proof}
This is an immediate consequence of Proposition~\ref{prop:X_k-2N-str}.
\end{proof}

\begin{proposition}\label{prop:proj-coord-even}
Let $E$ be an elliptic curve, and $E\hookrightarrow\P^{N-1}$ be an immersion as an elliptic normal curve of degree~$N$ via the complete linear system $|N\{O\}|$.  Choose a primitive $N$th root of unity $\zeta$, and $\tilde \zeta$ such that $\tilde\zeta^{2}=\zeta$.  Let $(\tilde S,\tilde T)$ be a $\Gamma^{(N)}(2N)$-structure above a $\Gamma(N)$-structure $(S,T)$. Then, there exists a unique coordinate system of $\P^{N-1}$ such that the translation maps $\tau_{S}$, $\tau_{T}$, and the multiplication-by-$(-1)$ map $[-1]$ are given by the matrices $M_{T}$, $M_{S}$ and $M_{[-1]}$ in \eqref{eq:matrices}, respectively.

For a given $\Gamma(N)$-structure $(S,T)$, there are four different choices of such coordinate systems related by the change of coordinates of $\P^{N-1}$ given by the transition matrices generated by the following two\/:
\[
M_{T}^{\frac{N}{2}}=
\left[
\begin{array}{*5r}
1 & 0 &  &   &   \\
0 & -1&  & O &  \\
  &   & \ddots &  & \\
  & O &  & 1 & 0 \\
  &   &  & 0& -1
\end{array}
\right],
\quad
\setlength{\arraycolsep}{3pt}
\renewcommand{\arraystretch}{0.8}
M_{S}^{\frac{N}{2}}=
\left[
\begin{array}{*3c|*3c}
& & & & &\\
& O & & & I_{\frac{N}{2}} & \\
& & & & &\\
\hline
& & & & &\\
& I_{\frac{N}{2}} & & & O  \\
& & & & &
\end{array}
\right],
\]
\end{proposition}

\begin{proof}
The existence of such a coordinate system follows from Proposition~\ref{prop:X_k-even}, and the uniqueness follows from Proposition~\ref{prop:level-2N}. The last part follows from Proposition~\ref{prop:X_k-2N-str}.
\end{proof}

\begin{proposition}\label{prop:quad-eqs-even}
The notation being as above, we have the following.
\begin{align*}
&\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T})
= \textstyle{\bigoplus_{k\in\Z/N\Z}}\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T},\chi_{S}^{k}),\\
&\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T},\chi_{S}^{k})=\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T})\cap \<X_{i}X_{j}\mid i+j\equiv k \mod N\>,
\\
&\dim \mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T},\chi_{S}^{k}) =
\begin{cases}
(N-2)/2 & \text{if $k\equiv 0\bmod 2$,}\\%[\medskipamount]
(N-4)/2 & \text{if $k\equiv 1\bmod 2$,}
\end{cases}
\\
&\dim \mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T}) =N(N-3)/2.
\end{align*}
\end{proposition}

To find a basis of $\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T})$, we need to find a basis of $\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T},\chi_{0})$ and $\mathscr{I}^{2}(D}%{\mathfrak{d}_{\tilde T},\chi_{S})$, which we will do for the case $K=\C$, the field of complex numbers using theta functions.

\section{Modular curves}

Suppose $N$ is even.  Fix a primitive $N$th root of unity $\zeta$, and fix $\tilde \zeta$ such that $\tilde\zeta^{2}=\zeta$.  Let $(E, (S,T), (\tilde S, \tilde T))$ be a triple such that $E$ is an elliptic curve, $(S,T)$ a $\Gamma(N)$-structure on $E$, and $(\tilde S,\tilde T)$ a $\Gamma^{(N)}(2N)$-structure above $(S,T)$. Such triples are parametrized by the modular curve corresponding to the modular group $\Gamma^{(N)}(2N)$.  We denote it by $Y^{(N)}(2N)$, and its compactification by $X^{(N)}(2N)$.

As we have seen in \S2.3, such a triple $(E, (S,T), (\tilde S, \tilde T))$ determines a unique immersion $\iota:E \hookrightarrow \P^{N-1}$.
If we choose two different $\Gamma^{(N)}(2N)$ structures $(\tilde S,\tilde T)$ and $(\tilde S',\tilde T')$ related by
\[
(\tilde S',\tilde T')=(\tilde S,\tilde T)\sltwo(a,b;c,d),
\quad \sltwo(a,b;c,d)\in SL_{2}(\Z),
\]
two different coordinate systems are related by $\rho\bigl(\sltwo(a,b;c,d)\bigr)$ of the change of coordinates of $\P^{N-1}$.  We thus have a representation
\[
\rho: SL_{2}(\Z) \to PGL_{n}(\bar K); \quad {\textstyle \sltwo(a,b;c,d) \mapsto \rho\left(\sltwo(a,b;c,d)\right)}.
\]
In the coordinate system through the immersion using $(\tilde S',\tilde T')$, the translations $\tau_{S}$ and $\tau_{T}$ are represented by the same matrices $M_{S}$ and $M_{T}$ as before.  Since $u\tilde S+v\tilde T=(du-bv)\tilde S'+(-cu+av)\tilde T'$, we have the following relation
\[
\rho\begin{pmatrix} a & b \\c & d\end{pmatrix}^{-1}
M_{S}^{u}M_{T}^{v}
\,\rho\sltwo(a,b;c,d)
=M_{S}^{du-bv}M_{T}^{-cu+av} \quad\text{ in } PGL_{N-1}(\bar K).
\]
Recall that $SL_{2}(\Z)$ is generated by
\[
A=\sltwo(0,-1;1,0)
\quad \text{and} \quad
B=\sltwo(1,1;0,1),
\]
and it is characterized by the relations among them:
\[
SL_{2}(\Z)=\<A,B\mid (AB)^{3}=A^{2}, B^{4}=I\>.
\]
The lemma below shows the candidates of $\rho(A)$ and $\rho(B)$.
\begin{lemma}\label{lem:rep}
{\upshape (1)} \
Let $R$ be an element of $PGL_{N-1}(\bar K)$ satisfying the relations
\[
R^{-1}M_{S}R = M_{T}^{-1}, \quad
R^{-1}M_{T}R = M_{S}, \quad
R^{-1}M_{[-1]}R = M_{[-1]},
\]
then $R$ is one of the following.
\begin{align*}
&A_{0}=\left[\renewcommand{\arraystretch}{1.25}
\begin{matrix}
1 & 1 & 1 &\cdots & 1 \\
1 & \zeta^{1} & \zeta^{2} &\cdots & \zeta^{(N-1)}\\
1 & \zeta^{2} & \zeta^{4} &\cdots & \zeta^{2(N-1)}\\
\vdots & \vdots & \vdots & \ddots &\vdots\\
1 & \zeta^{(N-1)} & \zeta^{2(N-1)} & \cdots & \zeta^{(N-1)^{2}}
\end{matrix}\right]
=\bigl[\zeta^{ij}\bigr]_{i,j=0}^{N-1},
\\
&A_{0}M_{S}^{\frac{N}{2}}, \quad
A_{0}M_{T}^{\frac{N}{2}}, \quad
A_{0}M_{S}^{\frac{N}{2}}M_{T}^{\frac{N}{2}}.
\end{align*}
\par{\upshape (2)} \
Let $R$ be an element of $PGL_{N}(\bar K)$ satisfying the relations
\[
R^{-1}M_{P}R = M_{P}, \quad
R^{-1}M_{Q}R = M_{P}^{-1}M_{Q}, \quad
R^{-1}M_{[-1]}R = M_{[-1]},
\]
then $R$ is one of the following.
\begin{align*}
&B_{0}
\setlength{\arraycolsep}{2pt}
\left[\begin{array}{cclclc}
1 &  &  &  &  &   \\
  & \tilde\zeta^{1(N-1)} &  &  &  &  \\
  &  & \ddots &  &  & \\
  &  &  & \tilde\zeta^{i(N-i)} &  &  \\
  &  &  &  & \ddots &  \\
  &  &  &  &  & \tilde\zeta^{(N-1)\cdot1}
\end{array}\right]
=\operatorname{Diag}\bigl[\tilde\zeta^{i(N-i)}\bigr]_{i=0}^{N-1},
\\
&B_{0}M_{S}^{\frac{N}{2}}, \quad
B_{0}M_{T}^{\frac{N}{2}}, \quad
B_{0}M_{S}^{\frac{N}{2}}M_{T}^{\frac{N}{2}}.
\end{align*}
where $\tilde\zeta$ is a primitive $2N$th root of unity with $\tilde\zeta^2=\zeta$.  In particular, $R$ is an element of order~$2N$.
\end{lemma}

\begin{proposition}\label{prop:rep}
There is a representation $\rho:SL_{2}(\Z)\to PGL_{N-1}(K(\tilde\zeta))$ satisfying
\[
\rho\sltwo(0,-1;1,0) = A_{0}
\quad \text{and} \quad
\rho\sltwo(1,1;0,1) = B_{0},
\]
where $A_{0}$ and $B_{0}$ are in Lemma~\ref{lem:rep}.  Its kernel equals $\Gamma^{(N)}(2N)$.
\end{proposition}

\begin{proof}
Since $SL_{2}(\Z)$ is characterized by the relation $(AB)^{3}=A^{2}$ and $A^{4}=I$, it suffices to verify that $(A_{0}A_{0})^{3}=A_{0}^{2}$ and $A_{0}^{4}=I$.  This can be done by straightforward calculations.

Since $\operatorname{ord}(B_{0})=2N$, the kernel of $\rho$ contains $\Gamma(2N)$ but does not contain $\Gamma(N)$.  Also, we have the relation
\[
\sltwo(1,N;0,1)
\sltwo(1,0;N-1,1)
\sltwo(1,N;0,1)
\sltwo(1,0;1,1)
\equiv
\sltwo(1+N,0;0,1+N) \bmod 2N,
\]
and a simple calculation shows that
\[
B_{0}^{N}(A_{0}^{-1}B_{0}^{-1}A_{0})B_{0}^{N}(A_{0}^{N-1}B_{0}A_{0})
=I_{N}.
\]
This implies $\ker \rho=\Gamma^{(N)}(2N)$.
\end{proof}

Let $\iota$ be the immersion associated with a triple $(E, (S,T), (\tilde S, \tilde T))$.  Then,
associating $\iota(O)\in \P^{N-1}$ to the triple $(E, (S,T), (\tilde S, \tilde T))$, we obtain a map $Y^{(N)}(2N)\to \P^{N-1}$. Since $\iota(O)$ is fixed by $M_{[-1]}$, it is contained in the linear subspace $H$ of $\P^{N-1}$ fixed by $M_{[-1]}$.  $H$ is given by the equations $X_{k}=X_{N-k}$, $k=1,\dots, \tfrac{N}{2}-1$.  Thus the coordinate of $\iota(O)$ is of the form
\begin{equation}\label{eq:iota(O)}
(a_{0}:a_{1}:\dots:a_{\frac{N}{2}-1}:a_{\frac{N}{2}}:a_{\frac{N}{2}-1}:\dots:a_{2}:a_{1}).
\end{equation}
Let $E_{Y^{(N)}(2N)}$ be the universal elliptic curve over $Y^{(N)}(2N)$, and $o$ the $0$-section $Y^{(N)}(2N)\to E_{Y^{(N)}(2N)}$.  We have the diagram
\[\begin{CD}
E_{Y^{(N)}(2N)} @>>> \P^{N-1} \\
@A o AA @AAA\\
Y^{(N)}(2N) @>>> H \simeq \P^{\frac{N}{2}}.
\end{CD}\]

The linear space $H$ is stable under the representation $\rho:SL_{2}(\Z)\to PGL_{N-1}(K(\tilde\zeta))$ in Proposition~\ref{prop:rep}.  Moreover, $\rho(-I_{2})=M_{[-1]}$ acts trivially on $H$, and $\ker\rho=\Gamma^{(N)}(2N)$.
Thus, $\rho$ induces a representation
\[
\bar\rho: SL_{2}(\Z)/(\pm\Gamma^{(N)}(2N)) \to PGL_{\frac{N}{2}+1}(K(\tilde\zeta)).
\]
Define the coordinates $(\bar X_{0}:\bar X_{1}:\dots:\bar X_{\frac{N}{2}})$  of $H$ by
\[
\renewcommand{\arraystretch}{1.2}
\left\{
\begin{array}{l}
\bar X_{0} = X_{0}, \\
\bar X_{1} = X_{1}+X_{N-1}, \\
\bar X_{2} = X_{2}+X_{N-2}, \\
\qquad \dots  \\
\bar X_{\frac{N}{2}-1} = X_{\frac{N}{2}-1}+X_{\frac{N}{2}+1},\\
\bar X_{\frac{N}{2}} = X_{\frac{N}{2}}.
\end{array}\right.
\]
Then, $\bar\rho$ is given by
\begin{gather*}
\bar\rho\sltwo(0,-1;1,0)
=\left[\renewcommand{\arraystretch}{1.25}
\setlength{\arraycolsep}{3pt}
\begin{matrix}
1 & 1 & 1 &\cdots & 1 \\
2 & \zeta+\zeta^{-1} & \zeta^{2}+\zeta^{-2} & \cdots & -2\\
2 & \zeta^{2}+\zeta^{-2} & \zeta^{4}+\zeta^{-4} &\cdots & -2\\
\vdots & \vdots & \vdots & \ddots &\vdots\\
1 & -1 & 1 & \cdots & (-1)^{\frac{N}{2}}
\end{matrix}\right], \\
\bar\rho\sltwo(1,1;0,1)
=\setlength{\arraycolsep}{1pt}
\left[\begin{array}{*5c}
1 &  &  &  & \\
  & \tilde\zeta^{1(N-1)} &  &  & \\
  &  & \tilde\zeta^{2(N-2)} &  & \\
  &  &  & \ddots &  \\
  &  &  &  & \tilde\zeta^{N^{2}/4}\\
\end{array}\right].
\end{gather*}
In particular, we have
\begin{gather*}
\bar\rho\sltwo(1,N;0,1)
=
\left[\addtolength{\arraycolsep}{0.5pt}\begin{array}{*5c}
1 &  &  &  &\\
  & -1 &  &  \\
  &  & 1 &  &   \\
  &  &  & \ddots  &\\
  &  &  &  &(-1)^{\frac{N}{2}}
\end{array}\right],
\\
\bar\rho\sltwo(1,0;N,1)
=\addtolength{\arraycolsep}{4pt}
\left[\begin{array}{*5c}
  &  &  &   & 1 \\
  &  &  & 1 &   \\
  &  & \scalebox{-1}[1]{$\ddots$} &  &  \\
  & 1 &  &  &  \\
1 &   &  &  &
\end{array}\right].
\end{gather*}

Later we will show that the coordinates \eqref{eq:iota(O)} of $\iota(O)$  satisfies a set of quartic equations.  In \S\S8--10, we will study these equations in more detail.

\section{Theta functions}

From now on, we assume that the base field is the field of complex numbers~$\C$.  We construct functions $X_{0},X_{1},\dots,X_{N-1}$ in \S4 using theta functions.  Then, we find quadratic relations among them from the relations among theta functions.

\begin{definition}\label{def:theta}
For a pair of real numbers $(p,q)$, we define the theta function $\theta_{(p,q)}(z,\tau)$ with characteristic $(p,q)$ by
\[
\theta_{(p,q)}(z,\tau):=\sum_{n\in\Z}
\mathbf{e}}%{\boldsymbol{e}\bigl({\tfrac{1}{2}(n+p)^2\tau + (n+p)(z+q)}\bigr),
\]
where $z\in\C$ and $\tau\in\mathfrak{H}} %\def\H{\mathcal{H}=\{\tau\in\C\mid \operatorname{Im}\tau>0\}$, and $\mathbf{e}}%{\boldsymbol{e}(x)=e^{2\pi i x}$.
\end{definition}

This conforms with the definition in Mumford \cite[Ch.~I. \S3]{Mumford:Tata-I}.
We have the following fundamental formulas.
\begin{proposition}\label{prop:theta-basic}
Suppose $p,q,r,s\in \R$, and $l,m\in\Z$. Then, we have
\begin{enumerate}
\setlength{\itemsep}{\medskipamount}
\item $\theta_{(p,q)}(z+s,\tau) = \theta_{(p,q+s)}(z,\tau)$.
\item $\theta_{(p,q)}(z+r\tau,\tau)=
\mathbf{e}}%{\boldsymbol{e}(-\frac{1}{2}r^{2}\tau-rz-rq)\theta_{(p+r,q)}(z,\tau)$.
\item $\theta_{(p+l,q+m)}(z,\tau)
= \mathbf{e}}%{\boldsymbol{e}(pm)\,\theta_{(p,q)}(z,\tau)$.
\end{enumerate}
\end{proposition}

Let $N$ be a positive integer.  Although we are interested mainly in the case where $N$ is even, we do not restrict ourselves to even $N$ throughout this section.

\begin{definition}\label{def:theta-N}
Let $N$ be a positive integer.  For an integer or a half-integer $k$, define
\begin{align*}
\theta^{(N)}_k(z,\tau)
 &:=\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}(Nz,N\tau)
\\
&\phantom{:}
=\sum_{n\in\Z}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}N\bigl(n-\tfrac{k}{N}+\tfrac1{2}\bigr)^2\tau
+N\bigl(n-\tfrac{k}{N}+\tfrac1{2}\bigr)\bigl(z+\tfrac12\bigr)}\Bigr).
\end{align*}
\end{definition}

It is easy to verify that $\theta^{(N)}_{k+N}(z,\tau)=
\theta^{(N)}_k(z,\tau)$, and thus $\theta^{(N)}_k(z,\tau)$ depends only on the class $k\bmod N$.

\subsection{Basic properties of $\theta^{(N)}_{k}(z,\tau)$ as a function of~$z$}

First, we fix a positive integer $N$ and a point $\tau\in \mathfrak{H}} %\def\H{\mathcal{H}$.
\begin{proposition}\label{prop:theta-basic2}
For any positive integer $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$, and any integer or half-integer $k\in \frac{1}{2}\Z$, the following relations hold.
\begin{enumerate}
\setlength{\itemsep}{\medskipamount}
\item $\theta^{(N)}_k(z+1,\tau) = (-1)^{N+2k}\,\theta^{(N)}_k(z,\tau)$.
\item $\theta^{(N)}_k(z+\tau,\tau)
= (-1)^{N}\mathbf{e}}%{\boldsymbol{e}\bigl(-\frac{N}{2}\tau - Nz\bigr)\,\theta^{(N)}_k(z,\tau)$.
\item $\theta^{(N)}_k\bigl(z+\tfrac{1}{N},\tau\bigr)
= -\mathbf{e}}%{\boldsymbol{e}\bigl(-\frac{k}{N}\bigr)\theta^{(N)}_k(z,\tau)$.
\item $\theta^{(N)}_k\bigl(z+\tfrac{\tau}{N},\tau\bigr)
= -\mathbf{e}}%{\boldsymbol{e}\bigl(-\frac{\tau}{2N} -z\bigr)\,\theta^{(N)}_{k-1}(z,\tau)$.
\item  $\theta^{(N)}_k\bigl(z+\tfrac{\tau}{2N},\tau\bigr)
= \mathbf{e}}%{\boldsymbol{e}\bigl(-\tfrac{\tau}{8N} -\tfrac{z}{2}-\tfrac{1}{4}\bigr)\,\theta^{(N)}_{k-\frac{1}{2}}(z,\tau)$.
\item $\theta^{(N)}_k(-z,\tau) = (-1)^{N+2k}\theta^{(N)}_{-k}(z,\tau)$.
\end{enumerate}
\end{proposition}

\begin{proof}
These formulas follow easily from the definition and Proposition~\ref{prop:theta-basic}.
See also \cite{Mumford:Tata-I} for details.
\end{proof}

From now on, we denote $\theta^{(N)}_k(z,\tau)$ simply by $\theta_k(z)$.

\begin{lemma}\label{lem:zeros}
Let $Z^{(N)}_{k}$ be the set of zeros of $\theta_{k}(z)$.
\begin{enumerate}
\item If $N$ is odd, then we have
$Z^{(N)}_{k}=\left\{\left.\frac{m}{N}+\frac{k}{N}\tau+n\tau\,\right|\, m,\,n\in\Z\right\}$.
\[
\includegraphics[scale=0.8]{zeros-n-odd.eps}
\]
\item If $N$ is even, then
$Z^{(N)}_{k}=\left\{\left.\frac{1}{2N}+\frac{m}{N}+\frac{k}{N}\tau+n\tau\,\right|\, m,\,n\in\Z\right\}$.
\[
\includegraphics[scale=0.8]{zeros-n-even.eps}
\]
\end{enumerate}
\end{lemma}

\begin{proof}
It follows easily from the fact that the set of  zeros of $\theta_{(p,q)}(z,\tau)$ is given by $\{(l+\frac{1}{2}-p)\tau+
(m+\frac{1}{2}-q)\mid l,m\in\Z\}$.
\end{proof}

\begin{lemma}\label{lem:lin-indep}
$\theta_{0}(z), \theta_{1}(z),\dots,\theta_{N-1}(z)$ are linearly independent over $\C$.
\end{lemma}

\begin{proof}
This is easily seen by looking at the series expansions of $\theta_i(z)$'s.
\end{proof}

\begin{definition}\label{def:S-T-in-C}
For $\tau\in\mathfrak{H}} %\def\H{\mathcal{H}$, let $\Lambda_{\tau}=\<1,\tau\>$ be the lattice in $\C$ spanned by $1$ and~$\tau$, and let $E_{\tau}=\C/\Lambda_{\tau}$ be the elliptic curve with modulus~$\tau$.  For $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$, define points $S,T\in E_{\tau}[N]$, and $ \tilde S,\tilde T \in E_{\tau}[2N]$ by
\[
\begin{array}{ll}
S=\frac{\tau}{N} \bmod \Lambda_{\tau},&T=\frac{1}{N} \bmod \Lambda_{\tau},
\\[\medskipamount]
\tilde S=\frac{\tau}{2N} \bmod \Lambda_{\tau}, \quad &\tilde T=\frac{1}{2N} \bmod \Lambda_{\tau}.
\end{array}
\]
\end{definition}

\begin{lemma}\label{lem:weil-pairing-in-C}
Let $\zeta_{N}=\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{1}{N})=e^{2\pi i/N}$ and $\tilde \zeta_{N}=\zeta_{2N}=\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{1}{2N})=e^{2\pi i/2N}$.  Then $(S,T)$ is a level $N$ structure with $e_{N}(S,T)=\zeta_{N}$ and $(\tilde S,\tilde T)$ is a level $2N$ structure above $(S,T)$ with $\tilde \zeta_{N}^{2}=\zeta_{N}$.
\end{lemma}

\begin{proof}
Define the function $f(z)$ on $\C$ by
\[
f(z)=
\frac{\theta_{1}(z)}
{\theta_{0}(z)}
\quad \text{if $N$ is odd,}
\quad
f(z)=\frac{\theta_{1}(z-\frac{1}{2N})}
{\theta_{0}(z-\frac{1}{2N})}
\quad\text{if $N$ is even.}
\]
Then, the divisor of $f$ is given by
\[
\operatorname{div} f = \sum_{j=0}^{N-1}
\left(\bigl\{\tfrac{\tau}{N}+\tfrac{j}{N}\bigr\}
- \bigl\{\tfrac{j}{N}\bigr\}\right).
\]
Thus, by Theorem~\ref{thm:weil-pairing} and Proposition~\ref{prop:theta-basic2} (3),  we have
\[
e_{N}(S,T)
= e_{N}\left(\tfrac{\tau}{N}\bmod \Lambda_{\tau},
	\tfrac{1}{N}\bmod \Lambda_{\tau}\right)
=\frac{\action{T}\!f(z)}{f(z)}=\frac{f\left(z-\frac{1}{N}\right)}{f(z)}
= \zeta_{N}.
\]
By the same token we have $e_{2N}(\tilde S,\tilde T)=\zeta_{2N}=\tilde\zeta_{N}$, which implies $(\tilde S,\tilde T)$ is above $(S,T)$.
\end{proof}

\subsection{Projective immersion via theta functions}

\begin{proposition}\label{prop:immersion}
Let $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$ and $\tau\in\mathfrak{H}} %\def\H{\mathcal{H}$, and let $\Lambda_{\tau}=\<1,\tau\>$ be the lattice in $\C$ spanned by $1$ and~$\tau$.
The map
\[
z \longmapsto   \bigl(\theta_{0}(z),\theta_{1}(z),
\dots,\theta_{N-1}(z)\bigr) \in \C^{N}.
\]
induces an immersion $\Theta$ of the elliptic curve $E_{\tau}=\C/\Lambda_{\tau}$ into the projective space $\P^{N-1}${\upshape:}
\[
\begin{array}{rcccc}
\Theta:&E_{\tau}=\hskip-\arraycolsep&\C/\Lambda_{\tau} & \longrightarrow & \P^{N-1}
\\
&&z & \longmapsto & \bigl(\theta_{0}(z):\theta_{1}(z):
\dots:\theta_{N-1}(z)\bigr).
\end{array}
\]
The image $E_{\Theta}$ of $\Theta$ is a curve of degree~$N$ defined as the intersection of $N(N-3)/2$ quadrics in $\P^{N-1}$.
\end{proposition}

\begin{proof}
Let $S,T,\tilde S$, and $\tilde T$ be as in Definition~\ref{def:S-T-in-C}.

If $N$ is odd, let $D}%{\mathfrak{d}_{T}=\sum_{k=0}^{N-1}\{kT\}=\sum_{k=0}^{N-1}\{\frac{k}{N}\}$. Then, for $k\in\Z/N\Z$, we have $\div \theta_{k}(z)/\theta_{0}(z)=\action{kS}D}%{\mathfrak{d}_{T}-D}%{\mathfrak{d}_{T}$, and it has the same divisor as $X_{k}$ in \S\ref{subsec:4.1}. This shows that $\Theta$ is an immersion to $\P^{N-1}$ and has the properties in the statement.

If $N$ is even, let $D_{\tilde T}=\sum_{k=0}^{N-1}\{\tilde T+kT\}=\sum_{k=0}^{N-1}\{\frac{1}{2N}+\frac{k}{N}\}$.  Then, for $k\in\Z/N\Z$, we have $\div \theta_{k}(z)/\theta_{0}(z)=\action{kS}D}%{\mathfrak{d}_{\tilde T}-D}%{\mathfrak{d}_{\tilde T}$, and it has the same divisor as $X^{(\tilde S,\tilde T)}_{k}$ in Definition~\ref{def:X_k-even}.  Again, this shows that $\Theta$ has the required properties in the statement.
\end{proof}

\begin{theorem}\label{thm:theta-immersion}
Let $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$ and $\tau\in\mathfrak{H}} %\def\H{\mathcal{H}$ be as in Proposition~\ref{prop:immersion}.  Let $E_{\Theta}$ be the image of $\Theta$ in Proposition~\ref{prop:immersion}.  Then, $E_{\Theta}$ is an elliptic curve whose group operation is such that $\Theta$ is a homomorphism.  By an abuse of notatin, we denote $\Theta\bigl(\frac{\tau}{N}\bigr)$ and $Q=\Theta\bigl(\frac{1}{N}\bigr)$ also by $S$ and $T$.  They are points of order~$N$ in $E_{\Theta}$, and we have
\begin{align*}
&\bigl(\theta_{0}(z):\theta_{1}(z):
\dots:\theta_{N-1}(z)\bigr) + S
= \bigl(\theta_{N-1}(z):\theta_{0}(z):\theta_{1}(z):
\dots:\theta_{N-2}(z)\bigr),
\\
&\bigl(\theta_{0}(z):\theta_{1}(z):
\dots:\theta_{N-1}(z)\bigr) + T
= \bigl(\theta_{0}(z):\zeta_{N}^{-1}\theta_{1}(z):
\dots:\zeta_{N}^{-N+1}\theta_{N-1}(z)\bigr).
\end{align*}
In other words, the translations of $E_{\Theta}$ by $S$ and $T$ are given by the matrices $M_{S}$ and $M_{T}$ in \eqref{eq:matrices}, respectively.  Moreover, the map $[-1]$ on $E_{\Theta}$ is given by the matrix $M_{[-1]}$ in \eqref{eq:matrices}.  If $N$ is odd, the coordinate system of $E_{\Theta}\in \P^{N-1}$ given by $\Theta$ is exactly the one described in Propositions~\ref{prop:proj-coord-odd}.  If $N$ is even, it is exactly the one described in Propositions~\ref{prop:proj-coord-even} with the level $2N$ structure $(\tilde S,\tilde T)=\bigl((\Theta\bigl(\frac{\tau}{2N}\bigr),\Theta\bigl(\frac{1}{2N}\bigr)\big)$.
\end{theorem}

\begin{proof}
The formulas for the translations by $S$ and $T$, and the inversion $[-1]$ are consequences of Proposition~\ref{prop:theta-basic2} (3)(4)(5), respectively.  The latter part of the statement follows from Proposition~\ref{prop:immersion}.
\end{proof}

\subsection{Transformation formula for $\theta^{(N)}_{k}(z,\tau)$}

We now consider $\theta^{(N)}_{k}(z,\tau)$ as a function of $\tau$.  The matrix $\sltwo(a,b;c,d)\in SL_{2}(\Z)$ acts on the upper half plane by $\tau\mapsto \frac{a\tau+b}{c\tau+d}$.  We would like to know how this $SL_{2}(\Z)$ action affects the immersion $\Theta:E_{\tau}\to\P^{N-1}$.

The following transformation formula for the theta function is classical and fundamental.
\begin{lemma}[cf. Igusa \cite{Igusa:theta}]\label{lem:tr_formula}
Let $M=\sltwo(a,b;c,d)\in SL_{2}(\Z)$.  Then we have
\begin{equation}\label{eq:tr_formula}
\theta_{(p,q)}\Bigl(\frac{z}{c\tau+d},\frac{a\tau+b}{c\tau+d}\Bigr)
=\kappa(M)\mathbf{e}}%{\boldsymbol{e}(\phi_{M}(p,q))\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{cz^{2}}{2(cz+d)}\bigr)\sqrt{c\tau+d}\,
\theta_{(p',q')}(z,\tau),
\end{equation}
where $\kappa(M)$ is an eighth root of unity that does not depend on neither $\tau$ nor $(p,q)$,
\begin{align*}  \phi_{M}(p,q) &=  -\frac{1}{2}\bigl(abp^{2} + 2bcpq + cdq^{2}
- bd(ap+cq)\bigr), \\
(p',q')&=\bigl(ap+cq-\frac{1}{2}ac, bp+dq-\frac{1}{2}bd\bigr),
\end{align*}
and $\sqrt{c\tau+d}$ is the principal value.
\end{lemma}

\begin{proposition}\label{prop:tr_formula}
The following transformation formulas hold.
\begin{align}
&\theta_k^{(N)}(z,\tau+1)=c\, \mathbf{e}}%{\boldsymbol{e}\bigl(-\tfrac{k(N-k)}{2N}\bigr)\, \theta_k^{(N)}(z,\tau),
\tag 1
\\
&\theta_k^{(N)}\Bigl(\frac{z}{\tau},-\frac1{\tau}\Bigr)
=c'\,\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{z}{2}\bigr)\sqrt{\tfrac{\tau}{N}}\,
\sum_{j=0}^{N-1} \zeta_{N}^{-jk} \theta_j^{(N)}(z,\tau).
\tag 2
\end{align}
Here,  $c$ and $c'$  are constants independent of $k$, and $\zeta_N=\mathbf{e}}%{\boldsymbol{e}(\tfrac1N)$.
\end{proposition}

\begin{proof}
(1) \ Since $\theta_k^{(N)}(z,\tau+1)
=\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}(Nz,N\tau+N)$, we use the transformation formula \eqref{eq:tr_formula} with $M=\sltwo(1,N;0,1)$.  Then we have
\begin{align*}
\theta_k^{(N)}(z,\tau+1)&=\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}(Nz,N\tau+N)
\\
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{N}{2}(\tfrac{1}{4}-\tfrac{k^{2}}{N^{2}})\bigr)
\,\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2}-k)}(Nz,N\tau)
\\
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{N}{2}(\tfrac{1}{4}-\tfrac{k^{2}}{N^{2}}\bigr)
\mathbf{e}}%{\boldsymbol{e}\bigl(-(\tfrac{1}{2}-\tfrac{k}{N})k\bigr)
\,\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}(Nz,N\tau)
\\
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{N}{8}-\tfrac{k(N-k)}{2N}\bigr)
\,\theta_{k}^{(N)}(z,\tau)
\\
&=c\,\mathbf{e}}%{\boldsymbol{e}\bigl(-\tfrac{k(N-k)}{2N}\bigr)\, \theta_{k}^{(N)}(z,\tau).
\end{align*}
(2) \ Since $\theta_k^{(N)}(\frac{z}{\tau},-\frac{1}{\tau})
=\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}(\frac{z}{\tau/N},-\frac{1}{\tau/N})$, we let $\tau'=\tau/N$ and we use the transformation formula \eqref{eq:tr_formula} with $M=\sltwo(0,-1;1,0)$.  Then, we have $\phi_{M}(p,q)=\frac{N}{4}-\frac{k}{2}$ and $(p',q')=\left(\frac{N}{2},-\frac{1}{2}+\frac{k}{N}\right)$, and
\begin{align*}
\theta_k^{(N)}\Bigl(\frac{z}{\tau},\frac{-1}{\tau}\Bigr)
&=\theta_{(\frac{1}{2}-\frac{k}{N},\frac{N}{2})}
\Bigl(\frac{z}{\tau'},\frac{-1}{\tau'}\Bigr) \\
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{N}{4}-\tfrac{k}{2}\bigr)\mathbf{e}}%{\boldsymbol{e}(\tfrac{z}{2})\sqrt{\tau'}
\,\theta_{(\frac{N}{2},-\frac{1}{2}+\frac{k}{N})}(z,\tau').
\end{align*}
We compute the series of $\theta_{(\frac{N}{2},-\frac{1}{2}+\frac{k}{N})}(z,\tau')$ by splitting it into $N$ parts:
\begin{multline*}
\theta_{(\frac{N}{2},-\frac{1}{2}+\frac{k}{N})}(z,\tau')
\\
\begin{aligned}
&=\sum_{n\in\Z}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}\bigl(n+\tfrac{N}{2}\bigr)^2\tfrac{\tau}{N}
+\bigl(n+\tfrac{N}{2}\bigr)\bigl(z-\tfrac{1}{2}+\tfrac{k}{N}\bigr)}\Bigr)
\\
&=\sum_{j=0}^{N-1}\biggl(
\sum_{n+j\equiv 0\bmod N}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}\bigl(n+\tfrac{N}{2}\bigr)^2\tfrac{\tau}{N}
+\bigl(n+\tfrac{N}{2}\bigr)\bigl(z-\tfrac{1}{2}+\tfrac{k}{N}\bigr)}\Bigr)
\biggr).
\end{aligned}
\end{multline*}
Since $n+j\equiv 0\bmod N$ if and only if $n=Nm-j$ for some $m\in\Z$, we have
\begin{align*}
&\sum_{n+j\equiv 0\bmod N}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}\bigl(n+\tfrac{N}{2}\bigr)^2\tfrac{\tau}{N}
+\bigl(n+\tfrac{N}{2}\bigr)\bigl(z-\tfrac{1}{2}+\tfrac{k}{N}\bigr)}\Bigr)
\\
&\qquad=\sum_{m\in\Z}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}\bigl((Nm-j)+\tfrac{N}{2}\bigr)^2\tfrac{\tau}{N}
+\bigl((Nm-j)+\tfrac{N}{2}\bigr)\bigl(z-\tfrac{1}{2}+\tfrac{k}{N}\bigr)}\Bigr)
\\
&\qquad=\sum_{m\in\Z}\mathbf{e}}%{\boldsymbol{e}\Bigl(
\tfrac{1}{2}N\bigl(m-\tfrac{j}{N}+\tfrac{1}{2}\bigr)^2\tau
+N\bigl(m-\tfrac{j}{N}+\tfrac{1}{2}\bigr)\bigl(z+\tfrac{1}{2}\bigr)
\\
\noalign{\hfill$
+\bigl(Nm-j+\tfrac{N}{2}\bigr)(\tfrac{k}{N}-1\bigr)\Bigr)$
\quad\null}
&\qquad=\sum_{m\in\Z}\mathbf{e}}%{\boldsymbol{e}\Bigl(
{\tfrac{1}{2}N\bigl(n-\tfrac{j}{N}+\tfrac{1}{2}\bigr)^2\tau
+N\bigl(n-\tfrac{j}{N}+\tfrac{1}{2}\bigr)\bigl(z+\tfrac{1}{2}\bigr)}\Bigr)
\mathbf{e}}%{\boldsymbol{e}(-\tfrac{jk}{N}+\tfrac{k-N}{2})
\\
&\qquad =\mathbf{e}}%{\boldsymbol{e}(\tfrac{k-N}{2})\zeta_{N}^{-jk}\theta_{j}^{(N)}(z,\tau)
\end{align*}
Therefore,
\begin{align*}
\theta_k^{(N)}\Bigl(\frac{z}{\tau},\frac{-1}{\tau}\Bigr)
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{N}{4}-\tfrac{k}{2}\bigr)\mathbf{e}}%{\boldsymbol{e}(\tfrac{z}{2})\sqrt{\tfrac{\tau}{N}}
\,\theta_{(\frac{N}{2},-\frac{1}{2}+\frac{k}{N})}(z,\tau')
\\
&=\kappa(M)\mathbf{e}}%{\boldsymbol{e}\bigl(-\tfrac{N}{4}\bigr)\mathbf{e}}%{\boldsymbol{e}\bigl(\tfrac{z}{2}\bigr)\sqrt{\tfrac{\tau}{N}}\,
\sum_{j=0}^{N-1} \zeta_{N}^{-jk}\theta_{j}^{(N)}(z,\tau).
\qedhere
\end{align*}
\end{proof}

The action of $SL_{2}(\Z)$ on the upper half plane induces coordinate changes of $\P^{N-1}$ via the immersion
$\Theta$ by theta functions.  As a consequence we have a representation $\rho_{\theta}:SL_{2}(\Z)\to PGL_{N-1}(\C)$.

\begin{theorem}\label{thm:sl2z-action}
The images of $\rho_{\theta}$ of the generators $S=\sltwo(0,-1;1,0)$ and $T=\sltwo(1,1;0,1)$ of
$SL_{2}(\Z)$ are given by
\[
\rho_{\theta}\sltwo(0,-1;1,0) = \bigl(\zeta^{ij}\bigr)_{i,j=0}^{n-1}
\quad \text{and} \quad
\rho_{\theta}\sltwo(1,\ 1;0,1)
= \operatorname{Diag}\bigl(\zeta^{-i^{2}/2}\bigr)_{i=0}^{n-1}.
\]
The kernel of $\rho_{\theta}$ is equal to $\Gamma^{(N)}(2N)$.
\end{theorem}

\begin{proof}
Immediate from Proposition~\ref{prop:tr_formula}.
\end{proof}

Finally, by using Lemma~\ref{lem:tr_formula} (theta transformation formula), we can describe
the transformation properties of the ``Theta Null Werte"
\[ a_k^{(N)}(\tau):=\theta_k^{(N)}(0,\tau)=\theta_{(\frac12-\frac{k}N,\frac{N}2)}(0,N\tau)\quad (k\in\Z)\]
under the action of the group $\Gamma(N)$.

By Proposition~\ref{prop:theta-basic2}~(5), we  see
\[
a_{k}^{(N)}(\tau) = (-1)^N a_{-k}^{(N)}(\tau) = (-1)^Na_{N-k}^{(N)}(\tau).
\]
Thus, $a_0^{(M)}(\tau)=0$ if $N$ is odd, and we use $a_{k}^{(N)}(\tau)$ for $k=1,\ldots, (N-1)/2$
to embed the modular curve $\mathfrak{H}} %\def\H{\mathcal{H}/\Gamma(N)$
into $\P^{(N-3)/2}$ when $N$ is odd.
On the other hand, when $N$ is even,  we use $a_{k}^{(N)}(\tau)$ for $k=0,\ldots, N/2$
for an immersion of $\mathfrak{H}} %\def\H{\mathcal{H}/\Gamma^{(N)}(2N)$, {\it not} for $\mathfrak{H}} %\def\H{\mathcal{H}/\Gamma(N)$, as
the following Proposition suggests.  Recall that the principal congruence
subgroup $\Gamma(N)$ is generated by the subgroup $\Gamma^{(N)}(2N)$ and the elements
$\sltwo(1,N;0,1)$ and $\sltwo(1,0;N,1)$.  The difference between $N$ being odd or even
is also clear here.

\begin{proposition}  Let $N$ be a positive integer.  Write $a^{(N)}_{k}(\tau)=a_{k}$ for short.
\begin{enumerate}
\setlength{\itemsep}{\smallskipamount}
\item If $N$ is odd, then the ratio $(a_{1}:\dots:a_{\frac{N-1}{2}})$ is invariant under the action of $\Gamma(N)$.
\item If $N$ is even, then the ratio $(a_{0}:a_{1}:\dots:a_{\frac{N}{2}})$ is invariant under the action of
$\Gamma^{(N)}(2N)$.  Moreover, the following hold.
\smallskip
\begin{enumerate}
\setlength{\itemsep}{\smallskipamount}
\item
The action $\tau\mapsto \tsltwo(1,N;0,1)\tau=\tau + N$ induces the action
\[
(a_{0}:a_{1}:\dots:a_{k}:\dots:a_{\frac{N}{2}}) \mapsto
(a_{0}:-a_{1}:\dots:(-1)^{k}a_{k}:\dots:(-1)^{\frac{N}{2}}a_{\frac{N}{2}}).
\]
\item
The action $\tau\mapsto \tsltwo(1,0;N,1)\tau=\frac{\tau}{N\tau + 1}$ induces the action
\[
(a_{0}:a_{1}:\dots:a_{k}:\dots:a_{\frac{N}{2}}) \mapsto
(a_{\frac{N}{2}}:a_{\frac{N}{2}-1}:\dots:a_{\frac{N}{2}-k}:\dots:a_{0}).
\]
\end{enumerate}
\end{enumerate}
\end{proposition}

\begin{proof}
Straightforward calculation using Lemma~\ref{lem:tr_formula}.
\end{proof}

\section{Quadratic equations satisfied by theta functions}

Let $E$ be an elliptic curve $E$ over a filed $K\hookrightarrow\C$.  If we embed $E$ via the complete linear system $|N\{O_{E}\}|$ for some integer $N$, there exists a coordinate system $(X_{0}:X_{1}:\cdots:X_{N-1})\in\P^{N-1}$ described in Proposition~\ref{prop:proj-coord-odd}.  Then, by a suitable choice of $\tau\in\mathfrak{H}} %\def\H{\mathcal{H}$, we have a map $\C/\Lambda_{\tau}\to E\subset\P^{N-1}$ given by $X_{k}=\theta_{k}^{(N)}(z,\tau)$.  In order to describe $N(N-3)/2$ quadratic equations satisfied by $E$ explicitly, we would like to find relations satisfied among $\theta_{k}^{(N)}(z,\tau)$.  It turns out that the situation is quite different depending on the parity of $N$.

We use the method of Jacobi \cite{Jacobi:Theta}, which is completely elementary and algebraic
(no function theory is used).
\begin{definition}\label{def:Jacobi-basic}
Jacobi's basic theta functions $\vartheta_i(z)$ ($i=0,1,2,3$) are
defined by the following formulas:
\begin{alignat*}{2}
\vartheta_0(z)&=\theta_{(0,\frac12)}(z,\tau)
&&=\sum_{n\in\Z} \mathbf{e}}%{\boldsymbol{e}\bigl({\tfrac{1}{2}n^2\tau + n(z+\tfrac12)}\bigr),\\
\vartheta_1(z)&=\theta_{(\frac12,\frac12)}(z,\tau)
&&=\sum_{n\in\Z} \mathbf{e}}%{\boldsymbol{e}\bigl({\tfrac{1}{2}(n+\tfrac12)^2\tau + (n+\tfrac12)(z+\tfrac12)}\bigr),\\
\vartheta_2(z)&=\theta_{(\frac12,0)}(z,\tau)
&&=\sum_{n\in\Z} \mathbf{e}}%{\boldsymbol{e}\bigl({\tfrac{1}{2}(n+\tfrac12)^2\tau + (n+\tfrac12)z}\bigr),\\
\vartheta_3(z)&=\theta_{(0,0)}(z,\tau)
&&=\sum_{n\in\Z} \mathbf{e}}%{\boldsymbol{e}\bigl({\tfrac{1}{2}n^2\tau + nz}\bigr).
\end{alignat*}
\end{definition}

Note that our definition differs slightly from Jacobi's original notation by some rescaling and sign.  Although we should write $\vartheta_{i}(z,\tau)$ instead of $\vartheta_{i}(z)$, we omit $\tau$ for simplicity.

The function $\vartheta_1(z)$ is an odd function and the others are even functions:
\begin{equation}\label{eq:theta-odd-even}
\begin{array}{ll}
\vartheta_0(-z)=\vartheta_1(z), \ &
\vartheta_1(-z)=-\vartheta_1(z), \\
\vartheta_2(-z)=\vartheta_2(z), \ &
\vartheta_3(-z)=\vartheta_3(z),
\end{array}
\end{equation}
The following formulas are immediate from the definition.
\begin{equation}\label{jacobitheta}
\begin{aligned}
&\vartheta_0(z+\tfrac12)=\vartheta_3(z),\quad
\vartheta_1(z+\tfrac12)=-\vartheta_2(z),\\
&\vartheta_2(z+\tfrac12)=\vartheta_1(z),\quad
\vartheta_3(z+\tfrac12)=\vartheta_0(z).
\end{aligned}
\end{equation}
The starting point is the identity (A)-(1) in  Jacobi \cite[p.507]{Jacobi:Theta}.

For independent variables $w,x,y,z$, define variables $w',x',y',z'$ by
\begin{equation}\label{eq:xyzw}
\left\{\begin{aligned}
w' &=\tfrac{1}{2}(w+x+y+z), \\
x' &=\tfrac{1}{2}(w+x-y-z), \\
y' &=\tfrac{1}{2}(w-x+y-z), \\
z' &=\tfrac{1}{2}(w-x-y+z),
\end{aligned}\right.
\quad\text{or}\quad
\setlength{\arraycolsep}{2pt}
\left(\begin{array}{c}
w' \\ x' \\ y' \\ z'
\end{array}\right)
=
\frac{1}{2}
\left(\begin{array}{rrrr}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{array}\right)
\left(\begin{array}{c}
w \\ x \\ y \\ z
\end{array}\right).
\end{equation}
Denote by $A$ the $4\times 4$ matrix above.  In other words, define $A$ such that ${}^{t}(w',x',y',z')=A\,{}^{t}(w,x,y,z)$.  Then, we have
$A\in O(4)$, $A^{2}=I$, and $\det A=-1$.
This shows that the transformation given by \eqref{eq:xyzw} induces an involution on the set
\[ \{(w,x,y,z)\in \Z^4\,\mid\,w\equiv x\equiv y\equiv z\!\pmod2\} \]
and preserving the norm $w^2+x^2+y^2+z^2$.

\begin{proposition}\label{prop:Jacobi-A-1}
Let $w,x,y,z$ be independent variables, and let $w',x',y',z'$ be the variables defined by \eqref{eq:xyzw}.
Then, the following identity holds:
\begin{equation}\label{eq:Jacobi4}
\begin{aligned}
\vartheta_{3}(w)\vartheta_{3}(x)&\vartheta_{3}(y)\vartheta_{3}(z)
+\vartheta_2(w)\vartheta_2(x)\vartheta_2(y)\vartheta_2(z)
\\
&= \vartheta_{3}(w')\vartheta_{3}(x')\vartheta_{3}(y')\vartheta_{3}(z')
+\vartheta_2(w')\vartheta_2(x')\vartheta_2(y')\vartheta_2(z'),
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
Jacobi's identity \eqref{eq:Jacobi4} follows from the properties of \eqref{eq:xyzw}. The reader is encouraged to consult the beautiful, original article of Jacobi \cite{Jacobi:Theta}.
\end{proof}

\begin{theorem}\label{th:three-term}
Let $w$, $x$, $y$, $z$ be independent variables.  As in Proposition~\ref{prop:Jacobi-A-1}, define $w'$, $x'$, $y'$, $z'$ by \eqref{eq:xyzw}.  Furthermore, define variables $w'',x'',y'',z''$ by
\begin{equation}\label{eq:xyzw3}
{}^{t}\!(w'', x'', y'', z'') = A\, {}^{t}\!(w,x,y,-z).
\end{equation}
Then, the following three-term identity holds:
\begin{multline}\label{eq:threeterm}
\vartheta_1(w)\vartheta_1(x)\vartheta_1(y)\vartheta_1(z)
\\
+\vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
\\
-\vartheta_0(w'')\vartheta_0(x'')\vartheta_0(y'')\vartheta_0(z'')=0.
\end{multline}
\end{theorem}

\begin{proof}
Replacing $w$ by $w+1$ in \eqref{eq:Jacobi4} and using \eqref{jacobitheta}, we obtain
\begin{multline}\label{eq:Jacobi2}
\vartheta_{3}(w)\vartheta_{3}(x)\vartheta_{3}(y)\vartheta_{3}(z)
-\vartheta_2(w)\vartheta_2(x)\vartheta_2(y)\vartheta_2(z)
\\
= \vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
+\vartheta_1(w')\vartheta_1(x')\vartheta_1(y')\vartheta_1(z').
\end{multline}
Adding and subtracting \eqref{eq:Jacobi4} and \eqref{eq:Jacobi2}, we obtain
\begin{multline*}
2\vartheta_3(w)\vartheta_3(x)\vartheta_3(y)\vartheta_3(z)\\
=\vartheta_{3}(w')\vartheta_{3}(x')\vartheta_{3}(y')\vartheta_{3}(z')
+\vartheta_2(w')\vartheta_2(x')\vartheta_2(y')\vartheta_2(z')\\
+\vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
+\vartheta_1(w')\vartheta_1(x')\vartheta_1(y')\vartheta_1(z'),
\end{multline*}
and
\begin{multline*}
2\vartheta_2(w)\vartheta_2(x)\vartheta_2(y)\vartheta_2(z)\\
=\vartheta_{3}(w')\vartheta_{3}(x')\vartheta_{3}(y')\vartheta_{3}(z')
+\vartheta_2(w')\vartheta_2(x')\vartheta_2(y')\vartheta_2(z')\\
-\vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
-\vartheta_1(w')\vartheta_1(x')\vartheta_1(y')\vartheta_1(z').
\end{multline*}
Replace $w,x,y,z$ by $w+\tfrac12,x+\tfrac12,y+\tfrac12,z+\tfrac12$ in these two identities.  Then $w'$ becomes $w'+1$, and $x',y',z'$ unchanged.  By \eqref{jacobitheta} we obtain
\begin{multline}\label{eq:th0}
2\vartheta_0(w)\vartheta_0(x)\vartheta_0(y)\vartheta_0(z)\\
=\vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
-\vartheta_1(w')\vartheta_1(x')\vartheta_1(y')\vartheta_1(z')\\
-\vartheta_2(w')\vartheta_2(x')\vartheta_2(y')\vartheta_2(z')
+\vartheta_{3}(w')\vartheta_{3}(x')\vartheta_{3}(y')\vartheta_{3}(z'),
\end{multline}
and
\begin{multline}\label{eq:th1}
2\vartheta_1(w)\vartheta_1(x)\vartheta_1(y)\vartheta_1(z)\\
=-\vartheta_{0}(w')\vartheta_{0}(x')\vartheta_{0}(y')\vartheta_{0}(z')
+\vartheta_1(w')\vartheta_1(x')\vartheta_1(y')\vartheta_1(z')\\
-\vartheta_2(w')\vartheta_2(x')\vartheta_2(y')\vartheta_2(z')
+\vartheta_{3}(w')\vartheta_{3}(x')\vartheta_{3}(y')\vartheta_{3}(z').
\end{multline}

Since the relation between $w,x,y,z$ and $w',x',y',z'$ are symmetric, we have
\begin{multline}\label{eq:th0-inv}
2\vartheta_0(w')\vartheta_0(x')\vartheta_0(y')\vartheta_0(z')\\
=\vartheta_{0}(w)\vartheta_{0}(x)\vartheta_{0}(y)\vartheta_{0}(z)
-\vartheta_1(w)\vartheta_1(x)\vartheta_1(y)\vartheta_1(z)\\
-\vartheta_2(w)\vartheta_2(x)\vartheta_2(y)\vartheta_2(z)
+\vartheta_{3}(w)\vartheta_{3}(x)\vartheta_{3}(y)\vartheta_{3}(z),
\end{multline}
By the definition of the transformation \eqref{eq:xyzw3}, the identity \eqref{eq:th0-inv} can be translated to
\begin{multline}\label{eq:th0-inv-2}
2\vartheta_0(w'')\vartheta_0(x'')\vartheta_0(y'')\vartheta_0(z'')\\
=\vartheta_{0}(w)\vartheta_{0}(x)\vartheta_{0}(y)\vartheta_{0}(-z)
-\vartheta_1(w)\vartheta_1(x)\vartheta_1(y)\vartheta_1(-z)\\
-\vartheta_2(w)\vartheta_2(x)\vartheta_2(y)\vartheta_2(-z)
+\vartheta_{3}(w)\vartheta_{3}(x)\vartheta_{3}(y)\vartheta_{3}(-z).
\end{multline}
Now, calculating \eqref{eq:th1} $+$ \eqref{eq:th0-inv} $-$ \eqref{eq:th0-inv-2}, and using \eqref{eq:theta-odd-even}, we obtain the relation \eqref{eq:threeterm}.
\end{proof}

\subsection{Odd case}
Assume $N$ is odd.

From our definition of $\theta^{(N)}_{k}(z,\tau)$ and Proposition~\ref{prop:theta-basic}, we have the relations
\[
\theta^{(N)}_{0}(z,\tau)=  (-1)^{\frac{N-1}{2}}\vartheta_{1}(Nz,N\tau)\quad\text{and}\quad
\theta^{(N)}_{\frac{N}{2}}(z,\tau)= \vartheta_{0}(Nz,N\tau).
\]
Dropping the superscript ``$(N)$'', \eqref{eq:threeterm} translates to
\begin{multline}\label{eq:finalthreeterm-odd}
\theta_0(w)\theta_0(x)\theta_0(y)\theta_0(z)
\\
+\theta_{\frac{N}2}(w')\theta_{\frac{N}2}(x')\theta_{\frac{N}2}(y')\theta_{\frac{N}2}(z')
-\theta_{\frac{N}2}(w'')\theta_{\frac{N}2}(x'')\theta_{\frac{N}2}(y'')\theta_{\frac{N}2}(z'')=0.
\end{multline}

We use this equation to obtain our quadratic equations.

\begin{theorem}\label{th:eq-odd}
Suppose $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$ is odd.  Let $E_{\Theta}$ be the image of $\Theta$ in Proposition~\ref{prop:immersion}.  Let $V=H^{0}(\P^{N-1},\mathcal{I}_{E_{\Theta}}(2))$, and $V_{k}=V\cap \<X_{i}X_{j}\mid i+j\equiv k \mod N\>$, where $X_{i}$ $(i=0,\dots,N-1)$ are the coordinate functions of $\P^{N-1}$ in Proposition~\ref{prop:proj-coord-odd}. Let $a_{i}=\theta^{(N)}_{i}(0,\tau)$.  Then, the quadratic forms
\begin{equation}\label{eq:deg2-N-odd}
a_{j+1}a_{N-j}X_{0}^{2}
-a_{\frac{N-1}{2}-j}a_{\frac{N+1}{2}+j}
X_{\frac{N+1}{2}}X_{\frac{N-1}{2}}
+a_{\frac{N-1}{2}}a_{\frac{N+1}{2}}X_{\frac{N-1}{2}-j}
X_{\frac{N+1}{2}+j}
\end{equation}
with $j=1,\ldots,\tfrac{N-3}2$ forms a basis of the vector space $V_{0}$.\end{theorem}

\begin{proof} First, we show that if we replace $X_{i}$ by $\theta^{(N)}_{i}(z,\tau)$, the quadratic forms in \eqref{eq:deg2-N-odd} vanishes.  In \eqref{eq:threeterm},  let
\[
(w,x,y,z) =\bigl(z,z,\tfrac{j\tau}{N},-\tfrac{(j+1)\tau}{N}\bigr).
\]
Then, we have
\begin{align*}
(w',x',y',z') &= \bigl(z-\tfrac{\tau}{2N},z+\tfrac{\tau}{2N},\tfrac{(2j+1)\tau}{2N},-\tfrac{(2j+1)\tau}{2N}\bigr),
\\
(w'',x'',y'',z'') &=\bigl(z+\tfrac{(2j+1)\tau}{2N},z-\tfrac{(2j+1)\tau}{2N},-\tfrac{\tau}{2N},\tfrac{\tau}{2N}\bigr).
\end{align*}
Then, use Proposition~\ref{prop:theta-basic2}~(4) to see that \eqref{eq:deg2-N-odd} is satisfied by $X_{i}=\theta^{(N)}_{i}(z,\tau)$ and $a_{i}=\theta^{(N)}_{i}(0,\tau)$.  Recall that $a_k$ depends
only on $k \bmod N$, and we have $a_0=0$ and $a_k=-a_{N-k}$. Thus, we may restrict the range of $j$ to $1\le j\le (N-3)/2$.  ($j=(N-1)/2$ gives the trivial relation.)

Next we show that these $(N-3)/2$ relations are independent.  Indeed, the coefficients of the terms $X_{\frac{N-1}{2}-j}X_{\frac{N+1}{2}+j}$ for $1\le j\le (N-3)/2$ are all equal to $a_{\frac{N-1}{2}}a_{\frac{N+1}{2}}$, which is nonzero, and the terms $X_{0}^{2}, X_{\frac{N+1}{2}}X_{\frac{N-1}{2}}$ are different from $X_{\frac{N-1}{2}-j}X_{\frac{N+1}{2}+j}$ for $1\le j\le (N-3)/2$.  Therefore those $(N-3)/2$ quadratic forms are independent.
\end{proof}

We briefly describe below the classical examples $N=5$ and 7.

\begin{example} When $N=5$, \eqref{eq:deg2-N-odd} yields $(5-3)/2=1$ quadratic form, which is
\[
a_1a_2X_0^2-a_1^2X_{2}X_{3}+a_2^2X_{1}X_{4}.
\]
$V=H^{0}(\P^{N-1},\mathcal{I}_{E_{\Theta}}(2))$ is generated by this form and its various permutations by $M_{Q}$ in Proposition~\ref{prop:proj-coord-odd}. If we define $\phi(\tau)$ by
\[
\phi(\tau)=-\frac{a_1(\tau)}{a_2(\tau)}
=q^{\frac{1}{5}} - q^{\frac{6}{5}} + q^{\frac{11}{5}} - q^{\frac{21}{5}} + q^{\frac{26}{5}} - q^{\frac{31}{5}} + \cdots\quad
(q=\mathbf{e}}%{\boldsymbol{e}(\tau)=e^{2\pi i \tau}),
\]
we have the equations
\[ X_i^2+\phi(\tau)X_{i+2}X_{i-2}-\frac1{\phi(\tau)}X_{i+1}X_{i-1}=0
\quad (i=0,\ldots,4).
\]
This set of equations is the well-known Bianchi normal form \cite{Bianchi}.
\end{example}

\begin{example}  When $N=7$, quadratic forms \eqref{eq:deg2-N-odd} give two equations
\begin{align*}
&a_1a_2X_0^2-a_2^2X_{3}X_{4}+a_3^2 X_{2}X_{5}=0,\\
&a_2a_3X_0^2-a_1^2X_{3}X_{4}+a_3^2 X_{1}X_{6}=0.
\end{align*}
By the specialization $z=0$ in $X_i$ ($X_i\mapsto a_i$), we obtain the unique relation
\[a_1^3a_2=a_2^3a_3+a_1a_3^3. \]
This is the renowned Klein's quartic, which is a model of the modular curve $\mathfrak{H}} %\def\H{\mathcal{H}/\Gamma(7)$ of genus 3.
\end{example}

\subsection{Even case}

Next suppose $N$ is even.
Again we start with Jacobi's \eqref{eq:Jacobi4}, which in terms of our theta's is written as
\begin{equation}\label{eq:start-even}
\begin{aligned}
\theta_{\frac{N}{2}}(w)\theta_{\frac{N}{2}}(x)&\theta_{\frac{N}{2}}(y)\theta_{\frac{N}{2}}(z)
+\theta_{0}(w)\theta_{0}(x)\theta_{0}(y)\theta_{0}(z)
\\
&= \theta_{\frac{N}{2}}(w')\theta_{\frac{N}{2}}(x')\theta_{\frac{N}{2}}(y')\theta_{\frac{N}{2}}(z')
+\theta_{0}(w')\theta_{0}(x')\theta_{0}(y')\theta_{0}(z').
\end{aligned}
\end{equation}
We can deduce our quadratic relations directly from this with the following specializations.
For integers $i$ and $j$, consider the four substitutions

\begin{align*}
 (w,x,y,z)=
&\left(z,-\tfrac{j\tau}{N}, -\tfrac{j\tau}{N}, z\right),\quad
\left(-\tfrac{2j\tau}{N},0, z, z\right),\\
&\left(z-\tfrac{\tau}{N},-\tfrac{j\tau}{N}, -\tfrac{(j+1)\tau}{N}, z\right),\quad
\left(-\tfrac{(2j+1)\tau}{N},0,z, z-\tfrac{\tau}{N}\right),
\end{align*}
which, respectively, yield
\begin{multline*}
 (w',x',y',z')=
\left(z-\tfrac{j\tau}{N},0,0, z+\tfrac{j\tau}{N}\right),\
\left(z-\tfrac{j\tau}{N},-z-\tfrac{j\tau}{N}, -\tfrac{2j\tau}{N},-\tfrac{2j\tau}{N}\right),\\
\left(z-\tfrac{(j+1)\tau}{N},0, -\tfrac{\tau}{N}, z+\tfrac{j\tau}{N}\right),\
\left(z-\tfrac{(j+1)\tau}{N},-z-\tfrac{j\tau}{N}, -\tfrac{j\tau}{N},-\tfrac{(j+1)\tau}{N},\right).
\end{multline*}
Applying these to~\eqref{eq:start-even} and using Proposition~\ref{prop:theta-basic2},
we obtain the following set of equations.
\renewcommand{\arraystretch}{1.4}
\setlength{\arraycolsep}{3pt}
\begin{gather}
\left\{
\begin{array}{rcrcrcr}
a_{j}^{2}\,X_{0}^{2} & + & a_{\frac{N}{2}+j}^{2}\,X_{\frac{N}{2}}^{2} & = &
a_{0}^{2}\,X_{j}X_{N-j} & + & a_{\frac{N}{2}}^{2}\,X_{\frac{N}{2}+j}X_{\frac{N}{2}-j},
\\
a_{0}a_{2j}\,X_{0}^{2} & + & a_{\frac{N}{2}}a_{\frac{N}{2}+2j}\,X_{\frac{N}{2}}^{2} & = &
a_{j}^{2}\,X_{j}X_{N-j} & + & a_{\frac{N}{2}+j}^{2}\,X_{\frac{N}{2}+j}X_{\frac{N}{2}-j},
\end{array}  \label{eq:515}
\right.
\\[\smallskipamount]
\def\+{\hskip 4pt+\hskip 4pt}
\left\{
\begin{array}{rcl}
a_{j}a_{j+1}\,X_{0}X_{1} & + & a_{\frac{N}{2}+j}a_{\frac{N}{2}+j+1}\,X_{\frac{N}{2}}X_{\frac{N}{2}+1}
\\
& = & a_{0}a_{1}\  X_{j+1}X_{N-j}  \+  a_{\frac{N}{2}}a_{\frac{N}{2}+1} X_{\frac{N}{2}+j+1}X_{\frac{N}{2}-j},
\\
a_{0}a_{2j+1}\,X_{0}X_{1} & + & a_{\frac{N}{2}}a_{\frac{N}{2}+2j+1}\,X_{\frac{N}{2}}X_{\frac{N}{2}+1}
\\
& = & a_{j}a_{j+1}\,X_{j+1}X_{N-j} \+  a_{\frac{N}{2}+j}a_{\frac{N}{2}+j+1}\,X_{\frac{N}{2}+j+1}X_{\frac{N}{2}-j}.
\end{array}  \label{eq:516}
\right.
\end{gather}
Here, as before, the indices are considered modulo $N$.  Equations \eqref{eq:515} (resp. \eqref{eq:516})
belongs to the space $V_{0}$ (resp. $V_{1}$) (see \S2.3).
Because of the relation $a_j=a_{N-j}$,
it is easy to see that we may restrict ourselves to the case $0\le j\le \frac{N}2$.
Furthermore, $j=0\text{ and }\frac{N}2$ give trivial relations, and  \eqref{eq:516} becomes also trivial for $j=\frac{N}2-1$.  And finally, the second equation of \eqref{eq:515} (resp. \eqref{eq:516})
is unchanged if we replace $j$ by $\frac{N}2-j$ (resp. $\frac{N}2-j-1$).  Out of these equations, first equations of \eqref{eq:515} and \eqref{eq:516}
give $N-3$ independent equation.

\begin{proposition}\label{prop:deg2-N-even}
The equations
\begin{equation}\label{eq:V2i}
a_{j}^{2}\,X_{0}^{2}  +  a_{\frac{N}{2}+j}^{2}\,X_{\frac{N}{2}}^{2}  =
a_{0}^{2}\,X_{j}X_{N-j}  +  a_{\frac{N}{2}}^{2}\,X_{\frac{N}{2}+j}X_{\frac{N}{2}-j}
\end{equation}
with $j=1,\ldots,\frac{N}{2}-1$ form a basis of $V_{0}$, and the equations
\begin{equation}\label{eq:V2ip1}
\begin{aligned}
a_{j}a_{j+1}\,X_{0}X_{1}  &+  a_{\frac{N}{2}+j}a_{\frac{N}{2}+j+1}\,X_{\frac{N}{2}}X_{\frac{N}{2}+1} \\
& =  a_{0}a_{1}\  X_{j+1}X_{N-j}  +  a_{\frac{N}{2}}a_{\frac{N}{2}+1} X_{\frac{N}{2}+j+1}X_{\frac{N}{2}-j}
\end{aligned}
\end{equation}
with $j=1,\ldots,\frac{N}{2}-2$ form a basis of $V_{1}$.
\end{proposition}

\begin{proof}  Looking at the indices of $X$, only possible
dependency \eqref{eq:V2i} occurs between equations with $j=j_0$ and $j=\frac{N}2-j_0$. But then the
determinant  $a_0^4-a_{\frac{N}2}^4$ of the two by two matrix of coefficients of $X_{j_0}X_{-j_0}$
and $X_{\frac{N}2+j_0}X_{\frac{N}2-j_0}$ is non-zero, as is seen by looking at its Fourier
series. The same argument applies for equations in \eqref{eq:V2ip1}.
\end{proof}

\begin{remark}
As will be seen in the examples in the following sections, the whole set of equations
\eqref{eq:515} and \eqref{eq:516} are needed to obtain relations among $a_i$'s.
\end{remark}

\subsection{Other equations for $N$ even}

If $N$ is even, our $\theta_0(z)=\theta^{(N)}_{0}(z,\tau)$  and $\theta_{\frac{N}2}(z)=\theta^{(N)}_{\frac{N}2}(z,\tau)$
are related to the Jacobi theta functions in the following way:
\[
\theta_{0}(z)=(-1)^{\frac{N}{2}}\vartheta_2(Nz,N\tau)\quad
\text{and}\quad
\theta_{\frac{N}{2}}(z)=\vartheta_3(Nz,N\tau).
\]
Since we have
\[
(-1)^{\frac{N}{2}}\theta_{0}\left(z+\tfrac{1}{2N}\right)
=\vartheta_{1}(Nz,N\tau), \quad
\theta_{\frac{N}{2}}\left(z+\tfrac{1}{2N}\right)
=\vartheta_{0}(Nz,N\tau),
\]
the relation \eqref{eq:threeterm} translates to
\begin{multline}\label{eq:finalthreeterm-even}
\theta_{0}\left(w+\tfrac{1}{2N}\right)\theta_0\left(x+\tfrac{1}{2N}\right)\theta_0\left(y+\tfrac{1}{2N}\right)\theta_0\left(z+\tfrac{1}{2N}\right)\\
-\theta_{\frac{N}2}\left(w'+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(x'
+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(y'+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(z'+\tfrac{1}{2N}\right)\\
+\theta_{\frac{N}2}\left(w''+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(x''+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(y''+\tfrac{1}{2N}\right)\theta_{\frac{N}2}\left(z''+\tfrac{1}{2N}\right)=0.
\end{multline}

\begin{theorem}\label{th:eq-even-1}
Suppose $N\in \mathbf {N}} \def\Z{\mathbf {Z}} \def\P{\mathbf {P}$ is even.  Let $E_{\Theta}$ be the image of $\Theta$ in Proposition~\ref{prop:immersion}.  Let $V=H^{0}(\P^{N-1},\mathcal{I}_{E_{\Theta}}(2))$, and $V_{k}=V\cap \<X_{i}X_{j}\mid i+j\equiv k \mod N\>$, where $X_{i}$ $(i=0,\dots,N-1)$ are the coordinate functions of $\P^{N-1}$ in Definition~\ref{def:X_k-even}. Let
\[
s_{i}=\theta^{(N)}_{i}(\tfrac{1}{2N},\tau).
\]  Then, the quadratic forms
\begin{equation}\label{eq:deg2-N-even-0}
s_{j}s_{N-j}X_{0}^{2}
+s_{\frac{N}{2}-j}s_{\frac{N}{2}+j}
X_{\frac{N}{2}}^{2}
-s_{\frac{N}{2}}^{2}X_{\frac{N}{2}-j}
X_{\frac{N}{2}+j}
\end{equation}
with $j=1,\ldots,\tfrac{N}2-1$ forms a basis of the vector space $V_{0}$,
and
\begin{equation}\label{eq:deg2-N-even-1}
s_{j+1}s_{N-j}X_{0}X_{1}
+s_{\frac{N}{2}-j}s_{\frac{N}{2}+j+1}
X_{\frac{N}{2}}X_{\frac{N}{2}+1}
-s_{\frac{N}{2}}s_{\frac{N}{2}+1}X_{\frac{N}{2}-j}
X_{\frac{N}{2}+j+1}
\end{equation}
with $j=1,\ldots,\tfrac{N}2-2$ forms a basis of the vector space $V_{1}$
\end{theorem}

\begin{proof}
If we let
\[(w,x,y,z) = \bigl(z-\tfrac{1}{2N},
z-\tfrac{1}{2N},\tfrac{j\tau}{N},-\tfrac{j\tau}{N}\bigr),
\]
then, we have
\begin{align*}
(w',x',y',z') &= \bigl(z-\tfrac{1}{2N},
z-\tfrac{1}{2N},\tfrac{j\tau}{N},-\tfrac{j\tau}{N}\bigr)
\\
(w'',x'',y'',z'') &=\bigl(z+\tfrac{j\tau}{N}-\tfrac{1}{2N},
z-\tfrac{j\tau}{N}-\tfrac{1}{2N},0,0\bigr)
\end{align*}
Substituting these in \eqref{eq:finalthreeterm-even},
and using Proposition~\ref{prop:theta-basic2}, we obtain \eqref{eq:deg2-N-even-0}.

If we let
\[
(w,x,y,z) = \bigl(z-\tfrac{\tau}{N}-\tfrac{1}{2N},
z-\tfrac{1}{2N},\tfrac{j\tau}{N},-\tfrac{(j+1)\tau}{N}\bigr),
\]
then we have
\begin{align*}
(w',x',y',z') &= \bigl(z-\tfrac{\tau}{N}-\tfrac{1}{2N},
z-\tfrac{1}{2N},\tfrac{j\tau}{N},-\tfrac{(j+1)\tau}{N}\bigr)
\\
(w'',x'',y'',z'') &= \bigl(z-\tfrac{j\tau}{N}-\tfrac{1}{2N},
z-\tfrac{(j+1)\tau}{N}-\tfrac{1}{2N},-\tfrac{\tau}{N},0\bigr).
\end{align*}
Substituting these in \eqref{eq:finalthreeterm-even},
and using Proposition~\ref{prop:theta-basic2}, we obtain \eqref{eq:deg2-N-even-1}.
\end{proof}

\section{Level 4}\label{sec:level-4}

Consider the case $N=4$.  When $i$ and $j$ run from $0$ to $2$ in the formulas in
Proposition~\ref{prop:deg2-N-even} the only nontrivial relations we obtain are
{\renewcommand{\arraystretch}{1.4}
\setlength{\arraycolsep}{3pt}
\begin{gather*}
V_{0}:\left\{\begin{array}{*{2}rc}
a_{1}^{2}(X_{0}^{2} + X_{2}^{2}) &\ =\ &
(a_{0}^{2}+a_{2}^{2})X_{1}X_{3},
\\
a_{0}a_{2}(X_{0}^{2} + X_{2}^{2}) &\ =\ &
2a_{1}^{2}X_{1}X_{3},
\end{array}\right.
\\
V_{2}:\left\{\begin{array}{*{2}rc}
a_{1}^{2}(X_{1}^{2} + X_{3}^{2}) &\ =\  &
(a_{0}^{2}+a_{2}^{2})X_{0}X_{2},
\\
a_{0}a_{2}(X_{1}^{2} + X_{3}^{2}) &\ =\ &
2a_{1}^{2}X_{0}X_{2}.
\end{array}\right.
\end{gather*}
}%
Note that $V_{1}=V_{3}=0$ in this case.
If we let $z=0$, then $X_{i}$ becomes $a_{i}$.  Considering the fact $a_{1}=a_{3}$, we obtain a unique nontrivial relation from the above equations
\begin{equation}\label{eq:X4_8}
 a_{0}a_{2}(a_{0}^{2} + a_{2}^{2}) = 2a_{1}^{4}.
\end{equation}
This is a nonsingular curve in $\P^{2}$, which is a genus~$3$ curve.  On the other hand, the congruence subgroup $\Gamma^{(4)}(8)$ is the subgroup of $\Gamma(1)$ of index~$96$ and genus~$3$.  Thus, the curve~\eqref{eq:X4_8}
is nothing but $X^{(4)}(8)$.

Using the relation \eqref{eq:X4_8} we see that the two relations in $V_{0}$ and the two relations in $V_{1}$ are equivalent, respectively.  Thus, the following two equations determine the universal elliptic curve $\mathscr{E}^{(4)}(8)$ over $X^{(4)}(8)$:
\begin{equation}\label{eq:E4_8}
\renewcommand{\arraystretch}{1.4}
\setlength{\arraycolsep}{2pt}
\mathscr{E}^{(4)}(8) : \
\left\{
\begin{array}{*3c}
a_{0}a_{2}(X_{0}^{2} + X_{2}^{2}) &= & 2a_{1}^{2}\, X_{1}X_{3},
\\
2a_{1}^{2}\, X_{0}X_{2} &= & a_{0}a_{2}(X_{1}^{2} + X_{3}^{2}).
\end{array}\right.
\end{equation}

\begin{proposition}\label{prop:E4_8}
The curve $\mathscr{E}^{(4)}(8)$ is isomorphic over $K$ to
\begin{equation}\label{eq:Weier_E4_8}
Y^{2}=X\bigl(X-(a_{0}-a_{2})^4\bigr)\bigl(X-(a_{0}+a_{2})^4\bigr).
\end{equation}
The points on $X^{(4)}(8)$ at which the fiber of $\mathscr{E}^{(4)}(8)$ becomes a N\'eron polygon of four sides are
\[
(a_{0}:a_{1}:a_{2})
= (0:0:1), (1:0:0), (\pm\zeta_{8}^{2}:0:1), (1:\zeta_{8}^{k}:1) \ (k=0,\dots,7),
\]
where $\zeta_{8}$ is a primitive $8$th root of unity.
\end{proposition}

\begin{proof}
From two quadrics \eqref{eq:E4_8}, eliminate $X_{1}$ to obtain a quartic
equation.  Then, using the rational point $(X_{0}:X_{1}:X_{2}:X_{3})=(a_{0}:a_{1}:a_{2}:a_{1})$, we obtain a Weierstrass equation.  After some simplification, we obtain \eqref{eq:Weier_E4_8} with the change of coordinates
\begin{equation}\label{eq:chang_E4_8}
\left\{\begin{aligned}
X &=
(a_{0}^{2}-a_{2}^{2})^2\frac{
(a_{1}^{2} X_{0} X_{2}+a_{0}a_{2} X_{1}X_{3}) }
{(a_{1}^{2}X_{0} X_{2}-a_{0}a_{2} X_{1} X_{3}) },
\\
Y &=
4a_{1}^{2}(a_{0}^{2}-a_{2}^{2})^2\frac{
(X_{1}+X_{3})(X_{0}+X_{2})(a_{0}a_{2}X_{0}X_{2}-a_{1}^{2} X_{1} X_{3})}
{(X_{1}-X_{3})(X_{0}-X_{2})(a_{1}^{2} X_{0} X_{2}-a_{0}a_{2} X_{1} X_{3})}.
\end{aligned}\right.
\end{equation}

The locus of degenerate fibers are where the right hand side of \eqref{eq:Weier_E4_8} has a multiple root.
\end{proof}

The action of $SL_{2}(\Z)/\Gamma^{(4)}(8)$ on $\mathscr{E}^{(4)}(8)$ and $X^{(4)}(8)$ are given as follows
\begin{alignat*}{2}
\rho\tsltwo(0,-1;1,0)
&=\left[\begin{array}{rcrc}
1 & 1 & 1 & 1 \\
1 &\zeta_{8}^{2} & -1 & -\zeta_{8}^{2} \\
1 & -1 & 1 & -1 \\
1 &-\zeta_{8}^{2} & -1 & \zeta_{8}^{2}
\end{array}\right],
\quad &
\rho\tsltwo(1,1;0,1)
&=\left[\begin{array}{rcrc}
1 & 0 & 0 & 0 \\
0 &\zeta_{8}^{3} & 0 &0 \\
0 & 0 & -1 & 0 \\
0 & 0 &  0 & \zeta_{8}^{3}
\end{array}\right],
\\
\bar\rho\tsltwo(0,-1;1,0)
&=\left[\begin{array}{rcr}
1 & 2 & 1 \\
1 & 0 & -1\\
1 & -2 & 1 \\
\end{array}\right],
\quad &
\bar\rho\tsltwo(1,1;0,1)
&=\left[\begin{array}{rcrc}
1 & 0 & 0\\
0 &\zeta_{8}^{3} & 0 \\
0 & 0 & -1 \\
\end{array}\right].
\end{alignat*}
In particular, we have
\begin{align*}
\rho\tsltwo(1,4;0,1)&:
(X_{0}:X_{1}:X_{2}:X_{3}) \mapsto
(X_{0}:-X_{1}:X_{2}:-X_{3}),
\\
\rho\tsltwo(1,0;4,1)&:
(X_{0}:X_{1}:X_{2}:X_{3}) \mapsto
(X_{2}:X_{3}:X_{0}:X_{1}),
\\
\bar\rho\tsltwo(1,4;0,1)&:
(a_{0}:a_{1}:a_{2}) \mapsto
(a_{0}:-a_{1}:a_{2}),
\\
\bar\rho\tsltwo(1,0;4,1)&:
(a_{0}:a_{1}:a_{2}) \mapsto
(a_{2}:a_{1}:a_{0}).
\end{align*}
Let $G=\Gamma(4)/\Gamma^{(4)}(8)=\<\tsltwo(1,4;0,1)
\Gamma^{(4)}(8),\tsltwo(1,0;4,1)
\Gamma^{(4)}(8)\>\simeq \Z/2\Z\times \Z/2\Z$.  In order to obtain a model of $\mathscr{E}(4)\to X(4)$,
we take the quotient of $\mathscr{E}^{(4)}(8)\to X^{(4)}(8)$ by the action of~$G$.
To do so, we first dehomogenize the equations by letting
\[
x_{0}=\frac{X_{0}}{X_{3}}, \quad
x_{1}=\frac{X_{1}}{X_{3}}, \quad
x_{2}=\frac{X_{2}}{X_{3}}, \quad
\alpha_{0}=\frac{a_{0}}{a_{1}}, \quad
\alpha_{2}=\frac{a_{2}}{a_{1}},
\]
and consider the function field $K(\alpha_{0},\alpha_{2},x_{0},x_{1},x_{2})$.  Then, these variables satisfy the following relations:
\begin{gather}\label{eq:X4_8-a}
\alpha_{0}\alpha_{2}(\alpha_{0}^{2} + \alpha_{2}^{2}) = 2, \\
\label{eq:E4_8-a}
\left\{
\begin{aligned}
& \alpha_{0}\alpha_{2}(x_{0}^{2} + x_{2}^{2}) = 2 x_{1},\\
& 2 x_{0}x_{2} =  \alpha_{0}\alpha_{2}(x_{1}^{2} + 1).
\end{aligned}\right.
\end{gather}
Let $\sigma_{1}=\tsltwo(1,4;0,1)\Gamma^{(4)}(8)$ and $\sigma_{2}=\tsltwo(1,0;4,1))\Gamma^{(4)}(8)$ be the generators of $G$.  Then, they act as follows:
\begin{align*}
\sigma_{1}&:
(\alpha_{0},\alpha_{2},x_{0},x_{1},x_{2}) \mapsto
(-\alpha_{0},-\alpha_{2},-x_{0},x_{1},-x_{2}),
\\
\sigma_{2}&:
(\alpha_{0},\alpha_{2},x_{0},x_{1},x_{2}) \mapsto
\left(\alpha_{2},\alpha_{0},\frac{x_{2}}{x_{1}},\frac{1}{x_{1}},\frac{x_{0}}{x_{1}}\right).
\end{align*}
It is easy to see that $K(\alpha_{0},\alpha_{2})^{G}=K(\alpha_{0}\alpha_{2},\alpha_{0}^{2}+\alpha_{2}^{2})$.  Considering \eqref{eq:X4_8-a}, we see that $K(\alpha_{0},\alpha_{2})^{G}=K(\lambda)$ with $\lambda=\alpha_{0}\alpha_{2}$.
This shows that $X(4)\simeq \P^{1}_{\lambda}$.  Also, we have
\begin{align*}
\lambda(\tau) &= \frac{a_{0}(\tau)a_{2}(\tau)}{a_{1}(\tau)^{2}} =
2\left(\frac{\eta(\tau)\eta(4\tau)^2}{\eta(2\tau)^3}\right)^2\\
&=2q^{\frac14} -4q^{\frac54}+10q^{\frac94}-20q^{\frac{13}4}+36q^{\frac{17}4}+\cdots
\quad (q=e^{2\pi i \tau}),
\end{align*}
where $\eta(\tau)$ is the Dedekind eta function.
Further calculations show that the fixed field $K(\alpha_{0},\alpha_{2},x_{0},x_{1},x_{2})^{G}$ is generated by
\[
\lambda=\alpha_{0}\alpha_{2}, \quad
\xi_{0}=\frac{x_{0}x_{2}}{x_{1}}, \quad
\xi_{1}=x_{1}+\frac{1}{x_{1}}, \quad
\xi_{2}=\frac{(x_{1}+1)(x_{0}+x_{2})}{(x_{1}-1)(x_{0}-x_{2})}.
\]
Now, \eqref{eq:E4_8-a} can be written in terms of $\lambda$, $\xi_{0}$ and $\xi_{1}$:
\[
X=\frac{4(1-\lambda^4)}{\lambda^{2}}
\frac{(\xi_{0}+\lambda)}{(\xi_{0}-\lambda)}, \quad
Y=\frac{16(1-\lambda^4)}{\lambda^{2}}
\frac{\xi_{2}(1-\lambda \xi_{0})}
{(\xi_{0}-\lambda)}.
\]
If we let $X'=\lambda^{2}X/4$ and $Y'=\lambda^{2}Y/8$,
then the relation above becomes
\[
\mathscr{E}(4): Y'^{2} = X'\bigl(X'-(\lambda^2-1)^{2}\bigr)
\bigl(X'-(\lambda^2+1)^{2}\bigr).
\]
The action of $PSL_{2}(\Z/4\Z)$ is given by
\begin{align*}
\left[\begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix}\right]:
(\lambda,X',Y')&\mapsto
\left(\frac{-\lambda+1}{\lambda+1},
-\frac{4(X'+(\lambda^2+1)^2)}{(\lambda+1)^4},
-\frac{8\zeta_{8}^{2}Y'}{(\lambda+1)^6}\right),\\
\left[\begin{smallmatrix} 1 & 1 \\ 0 & 1\end{smallmatrix}\right]:
(\lambda,X',Y')&\mapsto ( -\zeta_{8}^{2}\lambda, X',Y').
\end{align*}

\section{Level 6}\label{sec:level-6}

In this section we study the case $N=6$ in detail.
In this case, the nontrivial relations obtained from Proposition~\ref{prop:deg2-N-even} are as follows
\allowdisplaybreaks
{
\renewcommand{\arraystretch}{1.4}
\setlength{\arraycolsep}{2pt}
\begin{equation}\label{eq:level-6}
\begin{gathered}
V_{0}: \left\{
\begin{array}{rcrcrcr}
a_{1}^{2}X_{0}^{2} & + & a_{2}^{2}X_{3}^{2}
& = & a_{0}^{2}X_{1}X_{5} & + &a_{3}^{2}X_{2}X_{4}, \\
a_{2}^{2}X_{0}^{2} & + & a_{1}^{2}X_{3}^{2}
& = & a_{3}^{2}X_{1}X_{5} & + & a_{0}^{2}X_{2}X_{4}, \\
a_{0}a_{2}X_{0}^{2} & + & a_{1}a_{3}X_{3}^{2}
& = & a_{1}^{2}X_{1}X_{5} & + &a_{2}^{2}X_{2}X_{4},
\end{array}\right.
\\[\smallskipamount]
V_{1}:\left\{
\begin{array}{crc}
a_{1}a_{2} (X_{0}X_{1} + X_{3}X_{4})
& = & (a_{0}a_{1}+a_{2}a_{3} )X_{2}X_{5},  \\
a_{0}a_{3} (X_{0}X_{1} + X_{3}X_{4})
& = &2 a_{1}a_{2}X_{2}X_{5}.
\end{array}\right.
\end{gathered}
\end{equation}
}
By letting $z=0$ (i.e. $X_{i}=a_{i}$), we obtain quartic relations among
$a_i$'s.  Noting $a_5=a_1$ and $a_4=a_2$, we obtain only two relations:
\begin{equation}\label{eq:level-6-homogeneous}
\left\{
\begin{gathered}
a_{1}^{4}+a_{2}^{4}=a_{0}^{3}a_{2}+a_{1}a_{3}^{3} \\
a_{0}a_{3} \left( a_{0}a_{1}+a_{2}a_{3} \right)
=2\,a_{1}^{2}a_{2}^{2}
\end{gathered}
\right.
\end{equation}
Using these relations we see that the equations \eqref{eq:level-6} are not linearly independent, and we have $\dim V_{0}=2$, $\dim V_{1}=1$.

The curve in $\P^{3}$ defined by \eqref{eq:level-6-homogeneous} turns out to be reducible.  Computations using Groebner basis reveal that there are five irreducible components; four of them are lines
\begin{gather*}
a_{1}=a_{2}=0, \quad
a_{0}-a_{2}=a_{1}-a_{3}=0, \\
a_{0}-\omega a_{2}=\omega a_{1}-a_{3}=0, \quad
a_{0}-\omega^{2} a_{2}=\omega^{2} a_{1}-a_{3}= 0,
\end{gather*}
where $\omega$ is third root of unity, and the other component is an irreducible curve of genus~$13$.  On the other hand, the congruence subgroup
$\Gamma^{(6)}(12)$ is the subgroup of $\Gamma(1)$ of index~$288$ and genus~$13$.  Thus, the last
irreducible component is the modular curve $X^{(6)}(12)$ associated with the group $\Gamma^{(6)}(12)$.

\begin{gather*}
\bar\rho\tsltwo(0,-1;1,0)
=\left[\begin{array}{rcrc}
1 & 2 & 2 & 1 \\
1 & 1 & -1 & -1\\
1 & -1 & -1 & 1\\
1 & -2 & 2 & -1
\end{array}\right],
\quad
\bar\rho\tsltwo(1,1;0,1)
=\left[\begin{array}{*4c}
1 & 0 & 0 & 0\\
0 &\zeta_{12}^{5} & 0 & 0\\
0 & 0 &\zeta_{12}^{8} & 0 \\
0 & 0 & 0 &\zeta_{12}^{9}
\end{array}\right].
\end{gather*}
In particular, we have
\begin{align*}
\bar\rho\tsltwo(1,6;0,1)&:
(a_{0}:a_{1}:a_{2}:a_{3}) \mapsto
(a_{0}:-a_{1}:a_{2}:-a_{3}),
\\
\bar\rho\tsltwo(1,0;6,1)&:
(a_{0}:a_{1}:a_{2}:a_{3}) \mapsto
(a_{3}:a_{2}:a_{1}:a_{0}).
\end{align*}

Let us find a model of $X(6)$ by taking the quotient of $X^{(6)}(12)$ by
$G=\Gamma(6)/\Gamma^{(6)}(12)\simeq \Z/2\Z\times \Z/2\Z$.
Dehomogenize the coordinates by setting
\[
\alpha_{1}=\frac{a_{1}}{a_{0}}, \quad \alpha_{2}=\frac{a_{2}}{a_{0}}, \quad
\alpha_{3}=\frac{a_{3}}{a_{0}}.
\]
Now the equation \eqref{eq:level-6-homogeneous} becomes
\begin{equation}\label{eq:inhomogeneous}\left\{
\begin{gathered}
\alpha_{1}^{4}+\alpha_{2}^{4}=\alpha_{2}+\alpha_{1}\alpha_{3}^{3}, \\
\alpha_{3}(\alpha_{1}+\alpha_{2}\alpha_{3})=2\alpha_{1}^{2}\alpha_{2}^{2}.
\end{gathered}
\right.
\end{equation}
Let $\sigma_{1}$ and $\sigma_{2}$ denote the automorphisms induced on the function field $k(\alpha_{1},\alpha_{2},\alpha_{3})$ by $\bar\rho\tsltwo(1,6;0,1)$ and $\bar\rho\tsltwo(1,0;6,1)$, respectively.  We have
\[
\sigma_{1} : (\alpha_{1},\alpha_{2},\alpha_{3}) \mapsto (-\alpha_{1},\alpha_{2},-\alpha_{3}), \quad
\sigma_{2} : (\alpha_{1},\alpha_{2},\alpha_{3}) \mapsto
\Bigl(\frac{\alpha_{2}}{\alpha_{3}},\frac{\alpha_{1}}{\alpha_{3}},\frac{1}{\alpha_{3}}\Bigr).
\]
The fixed subfields by $\sigma_{1}$ and $\sigma_{2}$ are
\begin{align*}
&k(\alpha_{1},\alpha_{2},\alpha_{3})^{\sigma_{1}}
=k\Bigl(\alpha_{1}\alpha_{3},\frac{\alpha_{1}}{\alpha_{3}},\alpha_{2}\Bigr), \\
&k(\alpha_{1},\alpha_{2},\alpha_{3})^{\sigma_{2}}
=\Bigl(\alpha_{1}+\frac{\alpha_{2}}{\alpha_{3}},\alpha_{2}+\frac{\alpha_{1}}{\alpha_{3}},\alpha_{3}+\frac{1}{\alpha_{3}}\Bigr).
\end{align*}
Finally, define
\[
b_{1}=\alpha_{1}\alpha_{3}+\frac{\alpha_{2}}{\alpha_{3}^{2}}, \quad
b_{2}=\alpha_{2}+\frac{\alpha_{1}}{\alpha_{3}}, \quad
b_{3}=\alpha_{3}^{2}+\frac{1}{\alpha_{3}^{2}}.
\]
Clearly, $b_{1}$, $b_{2}$, and $b_{3}$ are fixed by $G=\<\sigma_{1},\sigma_{2}\>$.  It is easy to show that the fixed field by the group $G$ is given by
\[
k(\alpha_{1},\alpha_{2},\alpha_{3})^{G} = k(b_{1},b_{2},b_{3}).
\]
The system of equations \eqref{eq:inhomogeneous} becomes
\[
\left\{\begin{gathered}
b_{3}(b_{1}^2+b_{2}^2-b_{1}b_{2}b_{3})^2+b_{2}(2b_{1}-b_{2}b_{3})(b_{1}^2-b_{2}^2)(b_{3}^2-4)=b_{1}(b_{3}^{2}-4)^2
\\
b_{2}(b_{3}^{2}-4)^2 = 2(b_{1}^2+b_{2}^2-b_{1}b_{2}b_{3})^2
\end{gathered}\right.
\]
Using calculations based on Groebner basis, we find that if we let
\[
X = \frac{-2(b_{1}^2+b_{2}^2-b_{1}b_{2}b_{3})}{b_{3}^2-4},
\quad
Y = \frac{b_{2}(2b_{1}-b_{2}b_{3})}{b_{1}^2-b_{2}^2},
\]
they satisfy
\[
Y^{2} = X^{3} + 1,
\]
which is well-known model of the modular curve~$X(6)$.
Furthermore, we have
\[
b_{1} = \frac{X^2(Y^2-3)}{4Y},\quad
b_{2} = \frac{1}{2}X^2,\quad
b_{3} = \frac{Y^4-6X^2-3}{4Y},
\]
and thus $k(b_{1},b_{2},b_{3})=k(X,Y)$.

On the curve $Y^{2}=X^{3}+1$, the actions of $\bar\rho\sltwo(0,1;-1,0)$ and $\bar\rho\sltwo(1,1;0,1)$ are given by
\begin{align*}
\bar\rho\tsltwo(0,1;-1,0)
&:(X,Y) \mapsto (2,-3) - (X,Y),
\\
\bar\rho\tsltwo(1,1;0,1)
&:(X,Y) \mapsto [-\omega](X,Y)=(\omega X,-Y),
\end{align*}
where the operation ``$-$'' in the first map is the group operation of $Y^{2}=X^{3}+1$, and the map $[\omega]$ is the complex multiplication of $Y^{2}=X^{3}+1$.
In terms of $a_{i}(\tau)=\theta_{i}^{(N)}(0,\tau)$, $X$ and $Y$ are expressed as follows:
\[
X = \frac{2a_{1}(\tau)a_{2}(\tau)}{a_{0}(\tau)a_{3}(\tau)}, \quad
Y = \frac{a_{0}(\tau)^{2}a_{1}(\tau)^{2}
-a_{2}(\tau)^{2}a_{3}(\tau)^{2}}
{a_{0}(\tau)^{2}a_{2}(\tau)^{2}
-a_{1}(\tau)^{2}a_{3}(\tau)^{2}}.
\]
Incidentally, $X$ and $Y$ (which are modular functions on $\Gamma(6)$) can also be written by using the
Dedekind eta function as
\[ X=\frac{\eta(2\tau)\eta(3\tau)^3}{\eta(\tau)\eta(6\tau)^3}=q^{-\frac13} + q^{\frac{2}{3}} + q^{\frac{5}{3}}
- q^{\frac{8}{3}} - q^{\frac{11}{3}} + q^{\frac{17}{3}} + 2\, q^{\frac{20}{3}} -\cdots
\]
and
\[
Y=\frac{\eta(2\tau)^4\eta(3\tau)^2}{\eta(\tau)^2\eta(6\tau)^4}=q^{-\frac12} + 2 q^{\frac{1}{2}} + q^{\frac{3}{2}}
- 2 q^{\frac{7}{2}} - 2 q^{\frac{9}{2}} +  2 q^{\frac{11}{2}} + 4 q^{\frac{13}{2}} + \cdots,
\]
where $q=e^{2\pi i \tau}$. The function $Y$ has a simpler expression in terms of $a_i(\tau)=\theta_i^{(6)}(0,\tau)$ as
\[ Y=\frac{a_0(\frac{\tau}{3})a_3(\frac{\tau}{3})}{a_0(\tau)a_3(\tau)}. \]

At the end of this section, we remark the connection to the ``Hesse cubic'', that is the elliptic normal
curve of degree 3.  Put
\begin{align*}  3\mu&=X^2-\frac{2}{X} =\frac{Y^2-3}{X}\\
&=q^{-\frac23}+5q^{\frac43}-7q^{\frac{10}3}+3q^{\frac{16}3}+15q^{\frac{22}3}-32q^{\frac{28}3}
+9q^{\frac{34}3}+\cdots.
\end{align*}
We can check that $\mu(\tau/2)$ is a modular function for the group $\Gamma(3)$,
and we have the relation
\[ X_0^3+X_2^3+X_4^3=3\mu X_0 X_2 X_4. \]
Actually, our theta functions $\theta_k^{(3)}(z,\tau)\ (k=0,1,2)$ in the case of $N=3$
can be obtained from those for $N=6$ ($k=0,2,4$) by changing the variables $z\to z/2-1/4, \,\tau\to\tau/2$.
We refer the reader to \cite{KKNT} for the derivation of the Hesse cubic in the same line of the current paper.

\section{Level 8}\label{sec:level-8}

In this section we study the case $N=8$ in detail.

For $N=8$, the quadratic equations obtained in \S3 are as follows.
\begin{gather*}
\setlength{\arraycolsep}{2pt}
V_{0}: \left\{
\begin{array}{*9c}
a_{1}^{2}X_{0}^{2} &+& a_{3}^{2}X_{4}^{2}
&=& a_{0}^{2}X_{1}X_{7}& & &+& a_{4}^{2}X_{3}X_{5}, \\
a_{2}^{2}X_{0}^{2} &+& a_{2}^{2}X_{4}^{2}
&=& & & (a_{0}^{2}+a_{4}^{2})X_{2}X_{6}, &
\\
a_{3}^{2}X_{0}^{2} &+& a_{1}^{2}X_{4}^{2}
&=& a_{4}^{2}X_{1}X_{7} & & &+& a_{0}^{2}X_{3}X_{5},
\\
a_{0}a_{2}X_{0}^{2} &+& a_{2}a_{4}X_{4}^{2}
&=& a_{1}^{2}X_{1}X_{7} & & &+& a_{3}^{2}X_{3}X_{5},
\\
a_{0}a_{4} X_{0}^{2} &+& a_{0}a_{4}X_{4}^{2}
&=& &  &2a_{2}^{2}X_{2}X_{6},
\end{array}\right.
\\
V_{1}: \left\{
\begin{array}{*9c}
a_{1}a_{2}X_{0}X_{1} &-& a_{0}a_{1}X_{2}X_{7}
&-& a_{3}a_{4}X_{3}X_{6} &+& a_{2}a_{3}X_{4}X_{5} =0,
\\
a_{2}a_{3}X_{0}X_{1} &-& a_{3}a_{4}X_{2}X_{7}
&-& a_{0}a_{1}X_{3}X_{6} &+& a_{1}a_{2}X_{4}X_{5} =0,
\\
a_{0}a_{3}X_{0}X_{1} &-& a_{1}a_{2}X_{2}X_{7}
&-& a_{2}a_{3}X_{3}X_{6} &+& a_{1}a_{4}X_{4}X_{5}=0.
\end{array}\right.
\end{gather*}
By letting $z=j\tau/N$, $j=0,1,\dots,$ in these formula, we obtain the following relations among $a_{i}$:
\begin{equation}\label{eq:level-8-homogeneous}
\left\{
\begin{aligned}
&a_{0}a_{4}(a_{0}^{2}+a_{4}^{2}) = 2a_{2}^{4}
\\
&a_{0}a_{4}(a_{1}^{2}+a_{3}^{2}) = 2a_{1}a_{3}a_{2}^{2}
\\
&a_{0}a_{2}a_{4}(a_{0}+a_{4})=2a_{1}^{2}a_{3}^{2}
\\
&a_{2}^{3}(a_{0}+a_{4}) = a_{1}a_{3}(a_{1}^{2}+a_{3}^{2})
\\
&a_{1}a_{3}(a_{0}^{2}+a_{4}^{2}) = a_{2}^{2}(a_{1}^{2}+a_{3}^{2})
\\
&a_{2}(a_{0}^{3} + a_{4}^{3}) = a_{1}^{4} + a_{3}^{4}
\end{aligned}
\right.
\end{equation}
The curve in $\P^{4}$ defined by \eqref{eq:level-8-homogeneous} turns out to be an irreducible, and this is a model of $X^{(8)}(16)$.

Dehomogenize \eqref{eq:level-8-homogeneous} by letting
\[
\alpha_{0}=\frac{a_{0}}{a_{2}}, \quad \alpha_{1}=\frac{a_{1}}{a_{2}}, \quad
\alpha_{3}=\frac{a_{3}}{a_{2}}, \quad \alpha_{4}=\frac{a_{4}}{a_{2}}.
\]
It turns out that, in the function field $k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})$, the following three equations are enough to generate the above six equations
\begin{equation}\label{eq:level8-rel}
\alpha_{0}\alpha_{4}(\alpha_{0}^{2}+\alpha_{4}^{2}) = 2,\quad
\alpha_{0}+\alpha_{4} = \alpha_{1}\alpha_{3}(\alpha_{1}^{2}+\alpha_{3}^{2}),\quad
\alpha_{1}\alpha_{3}(\alpha_{0}^{2}+\alpha_{4}^{2}) = \alpha_{1}^{2}+\alpha_{3}^{2}.
\end{equation}

Let $\sigma_{1}$ and $\sigma_{2}$ denote the automorphisms induced on the function field $k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})$ by $\bar\rho\tsltwo(1,8;0,1)$ and $\bar\rho\tsltwo(1,0;8,1)$, respectively.  We have
\begin{align*}
& \sigma_{1} : (\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4}) \longmapsto
(\alpha_{0},-\alpha_{1},-\alpha_{3},\alpha_{4}), \\
&\sigma_{2} : (\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4}) \longmapsto
(\alpha_{4},\alpha_{3},\alpha_{1},\alpha_{0}).
\end{align*}
It is easy to see that $\alpha$ and $\beta$ commute, and thus the group of automorphism $\<\alpha,\beta\>$ induced by $\alpha$ and $\beta$ is isomorphic to $(\Z/2\Z)^{2}$.  The fixed subfields by $\alpha$ and $\beta$ are
\begin{align*}
k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})^{\sigma_{1}}
&=k\Bigl(\alpha_{0},\alpha_{1}\alpha_{3},\frac{\alpha_{1}}{\alpha_{3}},\alpha_{4}\Bigr), \\
k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})^{\sigma_{2}}
&=k\left(\alpha_{0}+\alpha_{4},\alpha_{1}+\alpha_{3}, (\alpha_{1}-\alpha_{3})(\alpha_{0}-\alpha_{4}),\frac{\alpha_{1}-\alpha_{3}}{\alpha_{0}-\alpha_{4}}\right).
\end{align*}
Finally, define
\[
b_{0}=\alpha_{0}+\alpha_{4},\quad
b_{1}=\alpha_{1}\alpha_{3},\quad
b_{3}=\frac{\alpha_{1}}{\alpha_{3}}+\frac{\alpha_{3}}{\alpha_{1}},\quad
b_{4}=\frac{\alpha_{1}\alpha_{3}(\alpha_{0}-\alpha_{4})}{(\alpha_{1}+\alpha_{3})(\alpha_{1}-\alpha_{3})}.
\]
Then, $b_{1}$, $b_{2}$, and $b_{3}$ are fixed by $G=\<\sigma_{1},\sigma_{2}\>$.  It is easy to show that the fixed field by the group $G$ is given by

\[
k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})^{G}
=k(b_{0},b_{1},b_{3},b_{4}).
\]
Further calculations show that the relations \eqref{eq:level8-rel} translates to the relations
\[
b_{0}=b_{1}b_{3}, \quad b_{3}=\frac{1}{b_{4}^{2}}, \quad
b_{1}^{4}=4b_{4}^{6}+b_{4}^{2}.
\]
This shows that
\[
k(\alpha_{0},\alpha_{1},\alpha_{3},\alpha_{4})^{G}
=k(b_{1},b_{4}), \quad \text{with } b_{1}^{4}=4b_{4}^{6}+b_{4}^{2}
\]
It is easy to see that $b_{1}^{4}=4b_{4}^{6}+b_{4}^{2}$ defines a curve of genus~$5$.  Thus, an affine model of $X(8)$ is given by
\[
X(8): b_{4}^{2}(1+4b_{4}^{2})=b_{1}^{4}.
\]
Recall that an affine equation of $X^{(4)}(8)$ is given by $\alpha_{2}(1 + \alpha_{2}^{2}) = 2\alpha_{1}^{4}$ (cf. \eqref{eq:X4_8}).
Thus, there is a two-to-one map $X(8)\to X^{(4)}(8)$ given by $
(\alpha_{1},\alpha_{2})=(b_{1},2b_{4}^{2})$.

The modular functions $b_1$ and $b_4$ (on $\Gamma(8)$) can be written in terms of the Dedekind eta function as
\[
b_1=\frac{\eta(2\tau)^4\eta(8\tau)^2}{\eta(\tau)\eta(4\tau)^5}=q^{\frac{1}{8}} + q^{\frac{9}{8}} - 2 q^{\frac{17}{8}} - q^{\frac{25}{8}}
+ 4 q^{\frac{33}{8}} + 2 q^{\frac{41}{8}} -  7 q^{\frac{49}{8}}  -\cdots
\]
and
\[
b_4=-\frac{\eta(2\tau)\eta(8\tau)^2}{\eta(4\tau)^3}=q^{\frac{1}{4}} - q^{\frac{9}{4}} + 2 q^{\frac{17}{4}} - 3 q^{\frac{25}{4}} + 4 q^{\frac{33}{4}} -
 6 q^{\frac{41}{4}} + 9 q^{\frac{49}{4}}- \cdots.
\]
They also have neat expressions in terms of our theta series as
\[
b_1=\frac{a_2(\frac{\tau}{2})a_2(2\tau)}{a_2(\tau)^2},\quad b_4=\frac{a_2(2\tau)}{a_2(\tau)}.
\]

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