# Friday April 10th

Let $(K, \abs{\wait})$ be complete and $q\in K\units$ be the Tate parameter with $\abs{q} < 1$.
The *Tate curve* is given in equations by
\begin{align*}
E_q: y^ + xy = x^3 + a_1(q) x + a_2(q)
.\end{align*}

For all $k\geq 1$, we define $S_k(q) = \sum_{n=1}^\infty {n^k q^n \over 1 - q^n}$ for which convergence in $K$ can easily be checked since in the non-Archimedean settings, series converge iff the terms go to zero.

Fact:
\begin{align*}
a_4(q) &= -S_3(q) \\
a_6(q) &= {-5 S_3(q) + 7 S_5(q) \over 12}
.\end{align*}

This is some multiplicative analog of exponentiating the Weierstrass elliptic equations in $\CC\units$.
The discriminant is given as in the theory of modular forms as
\begin{align*}
\Delta (E_q) = q \prod_{n=1}^\infty \qty{1 - q^n}^{?} \neq 0
,\end{align*}

so $E_q$ is an elliptic curve.

The $j\dash$invariant is given by
\begin{align*}
j(E_q) &= {1 \over q} + 744 +  196844q + \cdots \\
& = \frac 1 q \sum_{n=1}^\infty c(n) q^n,\quad c(n) \in \ZZ
.\end{align*}

> Connection to irreducible representations of the Monster group.

Note that $\abs{j(E_q)} = {1 \over \abs q} > 1$ and (claim) the norms of $S_k, a_4, a_q$ are less than 1.

After reducing mod $n$, we obtain $\ts{x\in K \st \abs x < 1}$ and
\begin{align*}
\tilde E_q:  y^2 + xy = x^3
\end{align*}
which becomes singular at $(0, 0)$.

Taking the lowest degree part yields $0 = y^3 + xy = y(y+x)$, meaning we have split multiplicative reduction.
We now want the Tate parameterization.

For $u \in K\units \sm q^\ZZ$, define
\begin{align*}
X(u, q) &\definedas \sum_{n\in \ZZ} {q^n u \over \qty{1 - q^n u}^2} - 2S_1(q)
Y(u, q) &\definedas \sum_{n\in \ZZ} {\qty{q^n u}^2 \over \qty{1 - q^n u}^2} - 2S_1(q)
.\end{align*}

> Note the similar formulas involving $g_2, g_3$ in the case of modular forms.

We have a uniformizer map
\begin{align*}
\phi: K\units &\to E_q(K) \\
u &\mapsto
\begin{cases}
(X(u, q), Y(u, q)) & u\in q^\ZZ \\
0 & u\in q^\ZZ
\end{cases}
.\end{align*}

> Usual group symmetry trick, just add up over orbit of the group.

This immediately yields a map $K\units / \gens{q^\ZZ} \to E_q(K)$, and the work goes into showing that $\phi$ is a $K\dash$analytic group isomorphism.
The hard part here is showing surjectivity, see Silverman II, Chapter 5 (4 page proof).

Since $(K, v)$ is complete and discretely valued (with multiplicative reduction), we have $v(j(E_q)) = -v(q)$.
See Kodaira-Neron.
Moreover, the identity component of the generic fiber is cyclic of order $v(q)$.

Question: is every curve with an integral $j\dash$invariant a Tate curve?

Lemma
:   Let $\alpha \in K$ with $\abs \alpha > 1$.
    Then there exists a unique $q \in K(\alpha)$ (a finite extension) such that $\abs q < 1$ and $j(E_q) = \alpha$.

Thus the function $j$ which maps $q \mapsto j(E_q)$ is bijective, so you can recover the Tate parameter $q$ from either the elliptic curve or the $j\dash$invariant, which is not true over $K = \CC$.

Proof
:   Define $f(q) \definedas {1 \over j(q)} = {1 \over 1 + ?q + ? q^2 + \cdots}$.
    By inverting formal power series, this is equal to $q - 744 q^2 + 35665 q^3 + \cdots \in \ZZ[[q]]$.
    To get a functional inverse, for any ring $R$, any  $f\in R[[q]]$ such that $g \circ f = f\ \circ g = \id$ iff $a_1(f)$ is a unit.
    Then there is a $\beta \in E, \abs \beta < 1$ such that $\abs{g(\beta)} = \abs{\beta}$.
    Defininng $q \definedas g(1/\alpha) \in K(\alpha)$, we have $0 < \abs q = \abs{1 \over \alpha} < 1$, and inverting yields ${1 \over J(q)} = f(q) = f(g(1/\alpha)) = 1\alpha$, so $\alpha = j(q)$.
    This shows existence, so the $j$ function is surjective.

    \

    For uniqueness, suppose $j(q) = j(q')$ with $\abs{q}, \abs{q'} < 1$.
    Then $f(q) = f(q')$ and so $0 = \abs{f(q) - f(q')} = \abs{q-q'} \abs{1 - 744(q+q') + 35665 \qty{q^2 + qq' + (q')^2} + \cdots}$.
    The second term has norm 1, so this is equal to $\abs{q - q'}$.
    So the $j$ function is injective.

Note that this proof doesn't go through for $\CC$, and the injectivity is false.
Over $\CC$, the function
\begin{align*}
j: \HH/ \SL(2, \ZZ) &\mapsvia{\cong} \CC
j(q) = j(q') &\iff q' = {aq + b \over cq + d},
\begin{bmatrix}
a & b \
c & d
\end{bmatrix}
\in \SL(2, \ZZ)
\end{align*}

where $q = e^{2\pi i j}, q' = e^{2\pi i j'}$, we have $j(q) = j(q') \notimplies q = q'$.
E.g. for curves over $\CC$, the $\tau$ parameter can be associated to multiple curves.

Next time: non-Archimedean uniformization of higher dimensional varieties, particularly analytic torii.
Any compact connected $\CC\dash$analytic Lie group is a torus.
Note that the moduli space of elliptic curves of dimension ? is greater than the corresponding moduli space of abelian varieties, so additional conditions are needed to ensure they are torii.
The abelian varieties that can be uniformized are exactly those with split multiplicative reduction.