\documentclass[leqno, 12]{article}
       \usepackage{amsfonts,amsmath}
\usepackage[all]{xy}


\newcommand{\Z}{\mathbb Z}

\newcommand{\C}{\mathbb C}

\newcommand{\Q}{\mathbb Q}

\newcommand{\R}{\mathbb R}

\newcommand{\Sh}{\makebox[0pt][l]{\hspace{.005in}\rule{.151in}
{.003in}}{\rm III}}

\newcommand {\Hom}{{\rm Hom}}

\newcommand {\Gal}{{\rm Gal}}

\newcommand {\Spec}{{\rm Spec}}

\newcommand {\Ind}{{\rm Ind}}

\newcommand {\Res}{{\rm Res}}

\newcommand {\nov}{\not\:\mid}

\newcommand {\st}{:\!\!\mid}

\newcommand {\qed} {{}\newline\nopagebreak \noindent $\square $}

\begin{document}

\begin{center} 
\huge Notes on Kato-Siegel functions
\end{center}

\bigskip

\noindent {\bf A special function.}

First consider the elliptic curve $E_\tau$ over $\C$ corresponding to the lattice 
$\Z +\tau\Z $, where $\mathrm {Im} (\tau )>0$.

Consider the function
$$\Theta (u,\tau )=q^{\frac 1{12}}(t^{\frac 12}-t^{-\frac 12})
\prod _{n>0}(1-q^nt)(1-q^nt^{-1}),$$
where $q=e^{2\pi i\tau }$ and $t=e^{2\pi iu}$.
For any integer $D$ prime to 6, construct the function
$$\theta _D^{E_\tau /\C }:=(-1)^{\frac {D-1}2}\Theta (u,\tau )^{D^2}\Theta (Du,\tau )^{-1}.$$
Then  $\theta _D^{E_\tau /\C }$ enjoys the following properties.

\medskip

\noindent {\bf (i)}\qquad $\theta _D^{E_\tau /\C }$ has divisor 
$D^2(e)-\ker [\times D]$.

\noindent {\bf (ii)}\qquad For any isogeny 
$\alpha :E\rightarrow E'$ of degree prime to $D$ between two such curves,
$\alpha _*\theta _D^{E/\C }=\theta _D^{E'/\C }$.

\noindent {\bf (iv)}\,\,
$\bullet $\quad $\theta _{-D}=\theta _D$\quad (This is obvious.)
\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 

\smallskip

\noindent (The reason for the numbering will be become apparent in a moment.)

\bigskip

\noindent {\bf Aim.}

We wish to exhibit the algebraic nature of this phenomenon, 
and show that it generalizes to elliptic
curves over arbitrary schemes, behaving well in families. Let $f:E\rightarrow S$ be an
elliptic curve over an arbitrary scheme
$S$. Let 
$\boldsymbol\omega _{E/S}$ be the invertible sheaf 
$$f_*\Omega ^1_{E/S}=e^*\Omega ^1_{E/S},$$ which, since 
$\Omega ^1_{E/S}$ is free along the fibres of $f$, is
$$=x^*\Omega ^1_{E/S}$$ for any section $x\in E(S)$.

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

\noindent {\bf (ii)}\qquad The assignment is compatible with isogeny. Precisely, for any isogeny 
\newline $\alpha :E\rightarrow E'$ of degree prime to $D$ between elliptic curves over $S$, we have
$\alpha _*\theta _D^{E/S}=\theta _D^{E'/S}$.

\noindent {\bf (iii)}\qquad The assignment is compatible with base change. Precisely,
for any base change
$$\xymatrix{E \ar_f[d] & E \times _SS'\ar[d]\ar[l]_g \\ S & S'\ar[l]}$$
we have $g^*(\theta _D^{E/S })=\theta _D^{E \times _SS'/S'}$.

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

\bigskip

\noindent {\bf Note} that (i) and (ii) determine $\theta _D^{E/S}$ uniquely.
(By (i) any other possible assignment is 
$u\theta _D^{E/S}$, for some 
$u\in \mathcal O^*(S)$.
Applying (ii) with $\alpha =[\times 2]$ and $[\times 3]$ (2 and 3 are prime to $D$) gives
\begin{align*}
&u\theta _D^{E/S}\\
=&\alpha _*(u\theta _D^{E/S})\\
=&\alpha _*(u)\alpha _*\theta _D^{E/S}\\
=&\alpha _*(u)\theta _D^{E/S}\\
=&\mbox{ both }u^4\theta _D^{E/S} \mbox{ and }u^9\theta _D^{E/S},
\end{align*}
whence $u^4=u=u^9, \Rightarrow u=1$.

\bigskip

\noindent {\bf Example.} Suppose
$S=\Spec \,k$, $k$ algebraically closed, 
$\alpha :E\rightarrow E'$ separable. Then (ii) becomes
$$\underset {\alpha (x)=y}{\prod _{x\in E(k)}}
\theta _D^{E/k}(x)=\theta _D^{E'/k}(y)\qquad\forall y\in E'(k),$$ 
which is the familiar distribution relation.

\bigskip

\noindent {\bf Proof of main result.}  First note that if $S=\Spec\,k$, then
$\ker [\times D]-D^2(e)$ is principal. (Because $D$ is odd, the sum of 
$\ker [\times D]$ on the elliptic curve is 0.) 

\medskip

\noindent $\bullet$ \; To give a rule 
$\theta _D$ satisfying (i) and (iii) is equivalent to giving an isomorphism of line bundles
$$\mathcal O_E(\ker [\times D])\overset ~\rightarrow\mathcal O_E(D^2e)$$ compatible with base 
change. Taking any fibre of $E/S$, we have the situation $S=\Spec\, k$ above; hence the line 
bundles are indeed isomorphic when restricted to any fibre. Then, our task is equivalent to
finding, for each $E/S$, a trivialization of 
$$e^*\mathcal O_E(\ker [\times D])\otimes e^*\mathcal O_E(D^2e)^\vee ,$$
which is 
\begin{align*}
=&e^*\mathcal O_E(\ker [\times D])\otimes e^*\mathcal O_E(-D^2e)\\
=&e^*[\times D]^*\mathcal O_E(e)\otimes e^*\mathcal O_E(-D^2e)\\
=&e^*\mathcal O_E(e)^{\otimes (1-D^2)}\\
=&\boldsymbol\omega _{E/S}^{\otimes (D^2-1)},
\end{align*}
compatible with base change.

Now we know the set of nowhere-vanishing sections of 
$\boldsymbol\omega ^{\otimes 12d}$ over the moduli stack -- that is, the collections, of a section
of $\boldsymbol\omega ^{\otimes 12d}_{E/S}$ for each $E/S$,
compatible with base change and isogeny -- is 
$\{\pm\Delta ^d\}$, for any $d\in\Z $, where $\Delta $ is the discriminant.
Then (since $(D,6)=1\Rightarrow D\equiv 1\mod 12$)
we have 2 non-vanishing sections of 
$\boldsymbol\omega ^{\otimes (D^2-1)}$, that is,
$\pm\Delta (E/S)^{(D^2-1)/12}$. Let 
$\pm\phi^{E/S}_D$ be the corresponding functions on $E-\ker [\times D]$.
So both $E/S\mapsto\pm\phi _D^{E/S}$ satisfy (i) and (ii).

Change base so that $\alpha $ factors as a product of isogenies of prime degree. 
Thus to verify (ii) we may assume $\mathrm {deg}\,\alpha =p$ prime.

The quotient $g_p(E/S,\alpha ):=\alpha _*\phi ^{E/S}(\phi ^{E'/S})^{-1}\in\mathcal O^*(S)$
is compatible with base change.
The modular stack $\mathcal M_{\Gamma _0(N)}$ classifies pairs
$(E/S,\alpha )$ where 
$\alpha :E\rightarrow E'$ is a cyclic isogeny of degree $N$.
So $g_p$ is a modular unit $\in\Gamma (\mathcal M_{\Gamma _0(p)},\mathcal O^*)$, 
and so 
$g_p(E/S,\alpha )=\pm 1\,\,\,\forall (E/S,\alpha )$. And the sign depends only on $p$.

\medskip

\noindent $\bullet$\; To determine the sign evaluate $g_p(E/\Bbb F_p, \mathrm {Fr}_E)$. Now 
$\mathrm {Fr}_{E*}: \kappa (E)^*\rightarrow\kappa (E)^*$ is the norm map. So for $p$ odd, 
$g_p(E,\mathrm {Fr}_E)=1$. In particular this does not depend on our choice of 
$\pm\phi^{E/S}_p$.
For $p=2$, though, replacing one by the other replaces $g_2$ by $-g_2$. Therefore exactly one
of $\pm\phi ^{E/S}$ will make $\theta _2$ satisfy (ii).

\medskip

\noindent $\bullet$\; We check (iv).
$\theta _{-D}$ also satisfies (i) and (ii), which uniquely determine 
$\theta _D$. So $\theta _{-D}=\theta _D$. The constant 1 satisfies (i) and (ii) for $D=1$; hence 
$\theta _1=1$.

\smallskip

$[\times M]_*\theta _{MC}$ has divisor $M^2(C^2(e)-\ker [\times C])$ and is 
compatible with base change, so 
$[\times M]_*\theta_{MC}=\epsilon\theta _C^{M^2}$, 
$\epsilon =\pm 1$. Now (ii) gives 
\begin{align*}
\epsilon\theta _C^{M^2}=&[\times M]_*\theta_{MC}\\ 
=&[\times M]_*[\times 2]_*\theta _{MC}\\
=&[\times 2]_*[\times M]_*\theta _{MC}\\
=&[\times 2]_*(\epsilon\theta
 _C^{M^2})\\
=&\epsilon ^4\theta _C^{M^2}\\
=&\theta _C^{M^2}
\end{align*}
and so $\epsilon =1$.

Using $D=M$ and $C=1$ produces $[\times D]_*\theta _D=\theta _1^{D^2}=1$.

\smallskip

Now $\theta _C\circ [\times M]$ and 
$\theta _{MC}/\theta _M^{C^2}$
both have divisor
$$C^2\ker [\times M]-\ker [\times MC];$$
hence their ratio is a unit compatible with base change. So (i) and (ii) give the result as before.

\medskip

\noindent $\bullet$\; 
A final matter is to check that condition (v) holds -- that is, that our $\theta _D^{E/S}$ is
indeed a generalization of the analytic $\theta _D^{E_\tau /\C}$ given at first. This found by
calculating that 
$F(u,\tau ):=\Theta (u,\tau )^{D^2}\Theta (Du,\tau )^{-1}$ is a function on 
$E_\tau$ ($\approx\C /(\Z +\tau\Z )$) with divisor
$D^2(e)-\ker [\times D]$, and which is $\mbox{\textsc{sl}}_2(\Z )$-invariant. Then $F(u,\tau )$ is
a constant multiple of $\theta _D^{E/S}$ for $E=E_\tau $, the constant being independent of $\tau$.

Consideringt a curve $E_\tau $ defined over $\R $ with two real connected components (for example
$Y^2=X^3-X$, where $\tau =i$) we calculate that
$$[\times 2 ]_*F(u,\tau )=(-1)^{\frac {D-1}2}F(u,\tau ).$$\qed







\end{document}