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\begin{document}
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We want to assign to each $E/S$, and $D$ prime to 6, a section
$$\theta _D^{E/S}$$
of $\mathcal O^*(E-\ker [\times D])$ such that

\medskip

\noindent {\bf (i)}\; $\theta _D^{E/S }$ (as a rational function) has divisor 
$D^2(e)-\ker [\times D]$.

\noindent {\bf (ii)}\; The assignment is compatible with isogeny of degree prime to $D$. 

\noindent {\bf (iii)}\; The assignment is compatible with base change. 

\noindent {\bf (iv)}\,\,
$\bullet $\quad $\theta _{-D}=\theta _D$\quad 
\newline \mbox { }\,\qquad $\bullet $\quad $\theta _1=1$
\newline\mbox { }\,\qquad $\bullet $\quad $[\times M]_*\theta _{MC}=\theta _C^{M^2}
\in\mathcal O^*(E-\ker [\times C])$ (In particular, $[\times D]_*\theta _D=1$.)
\newline\mbox { }\,\qquad $\bullet $\quad $\theta _C\circ [\times M]=\theta _{MC}/\theta _M^{C^2}
\in\mathcal O^*(E-\ker [\times MC])$ \quad 

\noindent {\bf (v)}\; In the case where $E$ is an elliptic curve
over $\C$ corresponding to the lattice 
$\Z +\tau\Z $, $\mathrm {Im} (\tau )>0$, 
$$\theta _D^{E/\C }=(-1)^{\frac {D-1}2}\Theta (u,\tau )^{D^2}\Theta (Du,\tau )^{-1}.$$


\newpage

\noindent $\bullet$ \; (i) and (ii) determine $\theta _D^{E/S}$ uniquely.

\bigskip

\noindent $\bullet$ \; To give a rule 
$\theta _D$ satisfying (i) and (iii) is equivalent to giving a trivialization of 
$$e^*\mathcal O_E(\ker [\times D])\otimes e^*\mathcal O_E(D^2e)^\vee $$
which is compatible with base change.
We do this using the discriminant, finding $\theta _D^{E/S}$
up to a sign.

\bigskip

\noindent $\bullet$ \; We find that exactly one of these two satisfy condition (ii).

\bigskip

\noindent $\bullet$ \; We check (iv). The technique is to apply the properties (i) and (ii) to both
sides of the each equation.

\bigskip

\noindent $\bullet$ \; Property (v) is checked using an explicit calculation with the isogeny
$[\times 2]$.


\end{document}