# Monday February 24th

## Classification of Locally $K\dash$Analytic Lie Groups

Let $K$ be a locally compact field with discrete valuation, $R$ a valuation ring with $\mfm = (\pi)$ its maximal ideal where $R/\mfm \cong \FF_q = \FF_{p^a}$.
There are two cases:

- If $\ch K = 0$, then $\QQ_p \subset K$ and $d = [L: \QQ_p]$
- If $\ch K = p > 0$, then $K \cong \FF_q((t))$.

Fact (Lie Theoretic)
:   Let $G$ be a compact commutative $g\dash$dimensional $K\dash$analytic lie group (i.e. locally looks like $k^n$ with transition harts given by convergent power series) which is 2nd countable.

    a. There exists a filtration by open subgroups $G = G^0 \supset G^1 \supset \cdots \supset G^n \supset \cdots$ such that
      0. $G^0/G^i$ is finite and discrete for all $i$, 
      1. $G^i \cong (\mfm^i)^g$, with addition given by a $g\dash$dimensional formal group law,
      2. $\intersect G^i = (0)$, so the filtration is exhaustive,
      3. $G/G^{i+1}$ is $p\dash$torsion,
      4. $G^1[\tors] = G^1[p^\infty]$
    b. If $\ch(K) = 0$, then there exists an open subgroup $U$ of $G$ such that $U \cong (R^g, +)$ as $K\dash$analytic Lie groups.

> Analog in Lie theory: Lie groups with isomorphic Lie algebras yield isomorphic universal covers.
> Can then recover the formal group from the Lie algebra.
> Wildly false in characteristic $p$, since we lose information about the height of the formal group.

Theorem (C-Lacy)
:   a. If $\ch(K) = 0$ (i.e. in a $p\dash$adic field), then $G \cong (R, +)^g \oplus G[\tors]$ as topological groups, where $G[\tors]$ is finite, which is in turn isomorphic to $\ZZ_p^{dg} \oplus G[\tors]$.
    
    b. If $\ch(K) = p$, then there exists a countable set $I$ such that $G \cong \prod_{i\in I} \ZZ_p \oplus G[\tors]$ as topological groups, where each of the groups on the RHS are closed subgroups.
      Moreover, $G[\tors] < \infty \iff G[p] < \infty$, and when these conditions hold, $I$ is infinite.

Lemma 
:   If $H$ is a a commutatve torsionfree pro-$p$ group, then $H \cong \prod_{i\in I} \ZZ_p$.
    If $H$ is 2nd countable, then $I$ is countable.

:::{.proof title="of lemma, sketch"}
We'll take *Pontryagin duals*.
Recall that if $G$ is an locally compact abelian (LCA) group, then $G\dual \definedas \hom(G, \RR/\ZZ)$ is an LCA group.
Note that the dual of a profinite group (inverse limit) is an ind-finite group (direct limit), which are discrete torsion groups.

$H\dual$ is a discrete $p\dash$primary torsion group.
Example: $\ZZ_p\dual = \QQ_p/\ZZ_p$, which flips the following exact sequence:


\begin{tikzcd}
0 \arrow[rr]   &  & \ZZ_p \arrow[rr] &  & \QQ_p \arrow[rr] &  & \QQ_p/\ZZ_p \arrow[rr] \arrow[llll, bend left=49] &  & 0
\end{tikzcd}


Then $H\dual = [p] H\dual$, and thus $H\dual$ is divisible.
We then apply the structure theory of divisible group to get a direct sum, then applying duality again yields a direct product, which proves the lemma.
:::

Proof (of Theorem)
:   Assume that $G[\tors]$ is finite.
    We have a SES of commutative profinite groups:

    \begin{align*}
    0 \to G[\tors] \to G \to G/G[\tors] \to 0
    ,\end{align*}

    and taking Pontryagin duals yields
    
    \begin{align*}
    0 \to (G/G[\tors])\dual \to G\dual \to G[\tors]\dual
    .\end{align*}

    Then $G/G[\tors]$ is torsionfree and has a finite index pro-$p$ subgroup,
    so $G/G[\tors]$ is itself a pro-$p$ group.
    By the lemma, $G/G[\tors] \cong \prod_{i\in I} \ZZ_p$, so 
    $$
    \qty{G/G[\tors]}\dual \cong \bigoplus_{i\in I} \QQ_p/\ZZ_p
    .$$
    But this is divisible, and hence injective since we're over a PID, so the dual sequence above splits.
    So the original sequence splits.

    We thus have an isomorphism of topological groups 
    $$
    G \cong G/G[\tors] \oplus G[\tors] \cong \prod_{i\in I} \ZZ_p \oplus G[\tors]
    ,$$ 
    where $G[\tors]$ was assumed finite.

    Suppose $\ch(K) = 0$.
    We have two open subgroups of $G$, $\prod_{i\in I} \ZZ_P \leq G$ (open since its complement is finite) and $(R, +)^g \cong \ZZ_p^{dg}$ by Serre.
    It follows that $\abs{I} = dg$.

    Suppose instead that $\ch(K) > 0$.
    The claim is that $[G: pG]$ is infinite and thus $\abs{I}$ is infinite.
    This is because the cokernel of multiplication by $p$ on $\prod \ZZ_p \oplus G[\tors]$ is infinite iff $I$ is infinite, so it suffices to check the size of this cokernel.
    
    Consider the formal group law in characteristic $p$ given by 

    \begin{align*}
    [p] \in R[[X_1^p, \cdots, X_g^p]]^g
    .\end{align*}

    It suffices to restrict to $G^1 = (t \FF_q[[t]]^g, \text{fgl} )$.
    Then $pG_1 \subseteq t \FF_q[[t]]^g \intersect \FF_q[[t^p]]^g$.
    But $[\FF_q[[t]] : \FF_q[[t^p]] ]$ is infinite, so $[G: pG_1]$ is infinite, so $[G: pG]$ is infinite and thus $I$ is infinite.

If we know the torsion is finite, can we find bounds on their size?
We'll need to revisit Néron models (as covered in the abelian varieties course), and introduce Tate curves.