# Wednesday March 4th

## Analytic Uniformization

:::{.remark}
Let $M/\CC$ be a connected manifold, then it has a universal cover $\tilde M$ and there is a map $\pi: \tilde M \to M$ such that $M \cong \tilde M / \pi_1(M)$.
For free, there is a unique way to make $\pi$ holomorphic, and $\tilde M$ is simply connected.

When is this useful?

1. When $\tilde M$ is simpler than $M$, or
2. $\pi$ has infinite degree $\iff \pi_1(M)$ is infinite.

Note that in AG, infinite degree covers don't occur.
:::

:::{.theorem title="Uniformization for Riemann surfaces"}
If $\dim M = 1$, then there are three possibilities up to biholomorphism:

1. $\CP^1$, positive curvature
2. $\CC$, 0 curvature
3. $\HH$ (by open mapping?), negative curvature
:::

:::{.remark}
Note that (2) and (3) are contractible, and (1) is compact and reasonable to understand, and admit metrics of constant curvature.
What do these spaces uniformize?

1. Only itself: can only have finite degree covering map (by compactness, cover can't be infinite degree)
    (Note that the only quotient here would be $\RP_2$, but these reverse orientation.
    
2. For $M$ compact, $M = \CC/\Lambda$ (i.e. torii, genus 1)

3. Curves of genus $g > 2$.

Note that $\Aut_\CC(\HH) = \PSL(2, \RR)$, and every curve $X$ arises as the quotient by a discrete subgroup $\Gamma$ (Fuchsian), i.e. $M = \Gamma \hq \HH$.
:::

:::{.slogan}
Compact manifolds $\approx$ projective varieties.
:::

:::{.example title="?"}
$K_3$ surface: compact and simply connected, so uniformizes itself.
Difficult to understand as a manifold, the 2nd betti number is 22.
:::

:::{.remark}
Note that in case 2, $\CC$ is a Lie group.
The best case is when $\tilde M$ is a Lie group, in which case $\pi_1(M) \injects \tilde M$ is a discrete subgroup.
:::

:::{.definition title="?"}
If $G$ is a locally compact topological group, then a **full lattice in $G$** is a closed, discrete subgroup $\Lambda \subset G$ such that $G/\Lambda$ is compact.

If $G = \CC^g$, then a **full lattice** $\Lambda$ is a closed discrete subgroup isomorphic to $\ZZ^{2g}$.
:::

:::{.remark}
Note that every genus $g$ abelian variety is uniformized by $\CC^g$.
:::

:::{.question}
What is an analog when $K$ is a $p\dash$adic field (finite extension of $\QQ_p$?
I.e. can we view $E(K)$ as $G/\Lambda$ where $G$ is a noncompact $K\dash$analytic Lie group and $\Lambda$ is a full lattice in some useful way?
:::

:::{.answer}
First try: take $G = (K, +) = \GG_a$.
This fails because $(K, +)$ has no non-trivial closed discrete groups. 
Why? Since $p^n \to 0\in K$ and $x\neq 0\in \Lambda$, then $p^n x \to 0$, so 0 is not an isolated point and the group is not discrete.
:::

:::{.remark}
There is a discrete subgroup $\FF_q[\frac 1 t] \subset \FF_q((t$, since the valuation doesn't decrease in powers of $t\inv$.
  Note that $\FF_q[t] \subset \FF_q((\frac 1 t))$ which is analogous to $\RR \from \ZZ$.
  
**Idea (Tate):**
Use the multiplicative group $\GG_m = K^\times$.
There is a surjective group morphism $\CC \to \CC\units$ where $z \mapsto e^{2\pi i z}$ with kernel $\ZZ$, thus there is a complex Lie group isomorphism
$$
(\CC, +) / (\ZZ, +) \mapsvia{\cong} (\CC\dual, \wait)
$$
The RHS is isomorphic to $S^1 \cross \RR^{>0}$, which is a cylinder.

Note that $\CC/\Lambda \cong \CC/\alpha \Lambda$ only depends on the homothety class of $\Lambda$ where $\alpha a\in \CC\units$.
WLOG, we can choose a generator to be 1, i.e. $\Lambda = \ZZ\cdot 1 \oplus \ZZ\cdot\tau$ where $\tau$ can be chosen such that $\tau \in \HH$.

We define $q = \im(\tau) \definedas e^{2\pi i \tau}$, and thus the exponential map gives an isomorphism
$$
\exp: E_s = \CC / \Lambda_\tau \mapsvia{\cong} \CC\units / \generators{q}
$$

In fact, every elliptic curve admits such a **multiplicative uniformization**.
:::

:::{.question}
How do we view the $N\dash$torsion 
\[
(\CC\units/\generators{q})[N] \cong \mu_N \times \generators{q^{\frac 1 N}}/\generators{q}
,\]
which has a basis $i_N = e^{2\pi i / N} \cross q^{\frac 1 N}$.
Note that since $q\in \HH$, we have $0 < \abs{q} < 1$.
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:::{.proposition title="?"}
Let $(K, \norm{\wait})$ be a locally compact field, $q\in K$, and $0 < \abs{q} < 1$.
Then $\generators{q}$ is closed, discrete, isomorphic to $\ZZ$, and thus is a full lattice in $K\units$ and $K\units/\generators{q}$ is compact.
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:::{.proof title="?"}
Let $x\in K\units$, then there exists an $n$ in $\ZZ$ such that $\abs{q} \leq \abs{xq^n} \leq 1$.
So take $\pi: K\units \to K\units/\generators{q}$ and restrict to the unit disc in $K$, which is compact since $K$ is locally compact.
Then $K\units/\generators{q}$ is the image of the closed unit disc in $K$, i.e. the image of a compact set, which is thus compact.
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:::{.exercise title="?"}
Take $K = \RR$ with $0 < \abs{q} < 1$.
What is $\RR\units/ \generators{q}$?
:::