# Friday March 6th

## Analytic Uniformization and Tate Curves

Gives us a way to use analysis to understand a variety.

Tate curves: let $(K, \abs{\wait}) = \RR, \CC$ or a complete nonarchimedean field.
Recall that if $K$ is $p\dash$adic, then $(K, +)$ has no nontrivial lattices (closed, discrete subgroups).
For $\CC$ we have an infinite-dimensional covering map 
$$
\operatorname{exp}: \CC/\ZZ \mapsvia{z\mapsto e^{2\pi i z}} (\CC\units, \cdot)
$$ 
which is an isomorphism of $\CC\dash$lie groups.

If $\tau \in \HH$ and $\Lambda_s = \generators{1, \tau}$, then taking the exponential yields 
\begin{align*}
C/\Lambda_S \mapsvia{\operatorname{exp}} \CC\units/\generators{q_\tau}
\end{align*}
where $q_\tau = e^{2\pi i q}$ and $0 < \abs{q_\tau} < 1$, which is a type of "additive uniformization".

Thus if $E/\CC$ is any elliptic curve, then there exists such a $q\in \CC$ such that 
$$
E(\CC) \cong C\units/\generators{q}
.$$

:::{.remark}
The map $q \mapsto j(q)$ is generally not injective.
Note that $j$ factors through the quotient


\begin{tikzcd}
j\HH \arrow[rr] \arrow[rrdd, dashed] &  & \CC                                    \\
                                     &  & \\
                                     &  & {\liesl(2, \ZZ)\ \HH} \arrow[uu, "\cong"]
\end{tikzcd}


In other words, for $(a, b; c, d) \in \liesl(2, \ZZ)$ we have $j(\tau) = j\qty{\frac{a\tau + b}{c\tau + d}}$.
The indeterminacy here is given by $e^{2\pi i \tau_1} = e^{2\pi i \tau_2}$ iff $j_2 = j_1 + b$ for some $b\in  \ZZ$ and $(1, b; 0, 1) \xi = \xi + b$.
:::

Last time we saw that $(K, \abs{\wait})$ is locally compact and $q\in K$ with $0< \abs q < 1$, then $\generators{q} = q^\ZZ$ is a full lattice in $K\units$.

> Proof last time: a one-liner using compactness of the unit ball.

Exercise
: If $\Lambda \in K\units$ is a full lattice (discrete, closed, compact quotient/cocompact), then $\Lambda = q^\ZZ$ for a unique $q$ with $0 < \abs q < 1$.
  The content: can't take $\abs{q} = 1$.
  Think about what happens if you take a $p\dash$adic field and start taking powers, see what it is dense in.

> Modding out by finite group doesn't lose compactness.

Exercise
: What are the full lattices in $(K\units)^g$?
  Higher dimensional abelian varieties are also uniformized by lattices in $\CC^g$.
  Claim: all isomorphic to $\ZZ^g$.

Exercise
: For $K = \RR$, $0 < \abs{q} < 1$, then 

\begin{align*}
R\units /\generators{q} \cong 
\begin{cases}
\RR/\ZZ \cross \ZZ/2\ZZ \quad q > 0 \\
\RR/\ZZ \quad q < 0
\end{cases}
.\end{align*}

> Primary source material for Tate Curves: Silverman Chapter 5

In fact, for all $E/\RR$ there exists a $0 < \abs q < 1$ such that $E(\RR) \cong E_q$ as $\RR\dash$lie groups, where $E_q \cong \RR\units / \generators{q}$.
So the Tate parameter $q$ is telling you whether or not 2-torsion exists.
Also relates sign of $q$ to sign of discriminant.
Can take $q$ series in the sense of modular forms, starting next week.

> Note: these $q$ series will still converge in $K$.

For our purposes, take $(K, \abs{\wait})$ a complete nonarchimedean field, $0< \abs{q} < 1$ in $K$, $E_q \definedas K\units/\generators{q}$ is a compact commutative 1-dimensional $K\dash$analytic lie group.

How all of this will end up working:

Theorem (Tate, Part A)
: For $(K, \abs{\wait})$ a complete nonarchimedean field and $q$ a Tate parameter, there exists an elliptic curve $E_q/K$ and a $K\dash$analytic group isomorphism
  $$
  \phi: K\units/\generators{q} \mapsvia{\cong} E_q(K)
  $$
  Moreover, $E_q$ has split multiplicative reduction.

Theorem (Tate, Part B)
:   Moreover, for all finite extensions $L/K$, note that we can extend the norm to the algebraic closure uniquely (by Number Theory II), and the following diagram commutes
  
    
    \begin{tikzcd}
    K\units/\generators{q} \arrow[dd, hook] \arrow[rr] &  & E_q(K) \arrow[dd, hook] \\
                                                      &  &                         \\
    L\units/\generators{q} \arrow[rr]                  &  & E_q(L)                 
    \end{tikzcd}
    

Taking the $\directlim$ over finite $L/K$, 
$$
\phi: \bar K\units /\generators{q} \mapsvia{\cong} E_q(\bar k)
.$$
Moreover, $\phi$ is equivariant for $\aut(\bar K/K)$, i.e. for all 

\begin{align*}
\forall \sigma \in \aut(\bar k/k),~ x\in \bar K\units/\generators{q}, \quad\quad \phi(\sigma x) = \sigma(\phi x)
.\end{align*}

For $N \in \ZZ^+$ with $\ch K \notdivides N$, we have 
$$
E_q[N] \cong (K\sep)\units / \generators{q} [N]
$$ 
as $g_K\dash$modules.
Note that $\mu_N$ is a subgroup of the RHS, which is fixed under the galois action.

Punch line: every elliptic curve comes with a unique order $N$ subgroup which is galois-invariant.
Thus we reduce the question of computing the galois module structure of torsion points to a qual-level algebra problem of computing galois extensions.

Theorem (Tate, Part C)
: Suppose $K$ is discretely valued, $\abs{x} = C^{-v(x)}$, $v(K^x) = \ZZ$. 
  Then $E_q$ has **split** multiplicative reduction (by A), and 
  $$
  v(j(E_q)) = v\qty{\frac 1 q} = -v(q) \in \ZZ^+
  .$$
  This implies that the component group of the Néron special fiber is *constant* and isomorphic to $\ZZ/v(q)\ZZ$.

> See Kodaira-Néron classification

Major difference: only uniformizing curves with split multiplicative reduction, as opposed to all curves in $\RR, \CC$ case.

Theorem (Tate, Part D)
: For all $j\in K$ with $\abs{j} > 1$, there exists a unique $q\in K$ such that $0 \abs{q} < 1$ and $j(E_q) = j$.
  So the Tate parameter is uniquely recovered.


For every $q$, we build a curve $E_q$, and get a galois structure on its torsion points.
They must have split multiplicative reduction, and every such curve is isomorphic to $E_q$ for a unique $q$.
Pretty strong result!

Theorem (Tate, Part E)
: Assume $\ch K \neq 2$.
  If $E/K$ is any elliptic curve with $\abs{j(E)} > 1$ (so not integral), then there exists a unique $q$ such that $E/K$ is a *quadratic twist* of $E_q/ K$.
  For $K(\sqrt \alpha) / K$, we have 
  $$
  E/K(\sqrt \alpha) \cong E_q / K(\sqrt \alpha)
  .$$
  Moreover,
  
  \begin{align*}
  E/K \text{ has } 
  \begin{cases}
  \text{split mult. reduction} & K(\sqrt \alpha) = K \quad \sim \GG_m\\
  \text{non-split mult. reduction} & K(\sqrt \alpha)/K \text{ is unramified } \quad\sim T\text{ a nonsplit torus} \\
  \text{additive reduction} & K(\sqrt \alpha) / K \text{ is ramified }
  \end{cases}
  .\end{align*}

Remark: the torsion can be arbitrarily large.

Example
: Take $K = \QQ_p$ and $q = p^n$ so $\abs{p^n} < 1$.
  Then 
  $$
  E_q(\QQ_p) \cong \QQ_p\units / \generators{p^n}
  .$$
  What is the torsion subgroup?
  
  Note that roots of unity in $\QQ_p$ inject into the quotient, and $\mu(\QQ_p) \cong \FF_p\units \cong \ZZ/(p-1)$ (by Number Theory II).
  The class of $p$ is order $n$, and thus also torsion, so 
  $$
  T \cong \ZZ/(p) \times \ZZ/(p-1)
  ,$$ 
  where the containment is obvious and staring at this calmly will show that this in fact the entire torsion subgroup.

In general, for $(K, \abs{\wait})$ discretely valued, it's easy to give an upper bounds.
We have 
$$
E_q(K)[N] \injects \mu_N(K) \times Z
,$$ 
where $Z$ is cyclic of order $v(q)$.
Equality occurs if (not iff) $q$ is a perfect $v(q)$th power.

Taking $E_q(\bar K)[N]$ is generated by $\zeta_n$, a primitive $n$th root of unity, and any $n$th root of $q$, $q^{\frac 1 n}$.
Thus this is literally a Kummer extension, and this is no big deal to work out.

The culprit: the component group can be arbitrarily large!