# Friday February 28th

## Clark-Xarles Paper

Continuing Clark-Xarles paper on torsion points of abelian varieties $A/k$, where $[k: \QQ_p] = ef$ and $q = p^f$.
If $A/k$ has potentially good reduction, the prime to $p$ torsion injects into the torsion of the residue field, which gave an estimate 
$$
\# A(k)[\tors]' \text{(prime to $p$)}
\leq 
\floor{ \qty{1 + \sqrt q  }^2  }^ g
= 
\floor{ q + 2\sqrt q + 1  }^ g
$$

Can also get a bound on $\# A(k)[p^\infty]$ in terms of $p, g, e$, and if $p \gg e$, the bound is precisely 1.

We want to generalize this: let $R$ be a valuation ring, then there exists a finite type smooth commutative (not necessarily proper) $R\dash$group scheme $A/R$ with generic fiber $A/k$, the Néron model, such that $A(R) = A(K)$ and 
$$
0 \to A'(R) \to A(R) = A(k) \mapsvia{r} A(\FF_q) \to 0
.$$
Note that $A'(R)$ here comes from a formal group law.

For $A/\FF_q$ a smooth commutative algebraic group, and $A/k$ has good reduction $\iff A/\FF_q$ is an abelian variety.

We obtain 
$$
0 \to A^0/\FF_q \to A/\FF_q \to \Phi \to 0
,$$ 
where the first entry is the component group and the third is a finite etale $\FF_q\dash$group scheme.
Identify $\Phi$ with $\Phi(\bar \FF_q)$ as a $g_{\FF_q}\dash$module, then $\Phi(\FF_q) = \Phi^{g_{\FF_q}}$ coming from a Galois action.

Note that $A^0/\FF_q$ is a connected commutative smooth algebraic group over the perfect field $\FF_q$, so we can apply some classification theory.

## The Chevalley Decomposition

Let $k$ be any perfect field and $G/k$ be a commutative connected algebraic group, then there exists a unique SES of such groups 
$$
0 \to T\oplus U\to G \to B \to 0
.$$
Plugging in $\bar k$ points would yield a SES of Galois modules.
Plugging in $k$ points may not work, because a surjection of elliptic curves (isogeny) is not necessarily surjective on $k\dash$points.
Say $\dim G= g, \dim B = \beta$ be the abelian ranks.

Since $B/k$ is an abelian variety, $T/k$ is a linear torus and $T/\bar k = \GG_m^\mu$, so define $\dim T = \mu$ the "toric rank".
Then $U/k$ is a commutative unipotent *linear algebraic group*, so it embeds in some $\GL(n, k)$ with eigenvalues only 1.
Note that $\GG_a = \theset{ [1, x; 0, 1] \suchthat x\in k}$, and anything here will be a repeated extension of the additive group.
Let $\alpha = \dim U$ be the additive rank.

If $A/\FF_q$ is an abelian variety of dimension $\beta$, we have the bound $\# A(\FF_q) \leq \floor{ q + 2\sqrt q + 1  }^\beta$, if $T_{\FF_q}$ is a torus of dimension $u$, then $\# T(\FF_q) \leq \qty{q+1}^\mu$.
Note that if $\dim T = 1$, then $T$ is isomorphic to either $\GG_m$ or $$N = \ker( \GG_m/\FF_{q^2} \to \GG_m/\FF_q  )$$ (Weyl restriction) over $\FF_q$.
Moreover, $N(\FF_q) = \ker(\FF_{q^2}\units \to \FF_q\units)$ which is cyclic of order $q+1$.

Note that if we have a nodal cubic, and the group splits to $\GG_m$ if the slopes of the two tangents at the node do (??) lie in $\FF_q$, and lands in the second case if the slopes are Galois conjugate.

If $U/\FF_q$ is a commutative unipotent algebraic group of dimension $\alpha$, then $\# U(\FF_q) = q^[\alpha]$, the number of points of $\GG_m^\alpha$ (?).

We'll find a filtration 
$$
0 = G^3 \subset G^2 \subset G^1 \subset G^0 = A(k)
.$$
Take $G^1$ to be the points reducing into $A^0(\FF_q)$, i.e. $$\ker( A(k) \mapsvia{r} A(\FF_q) \mapsvia{?} \Phi(\FF_q) ).$$
Then $G^0 / G^1 \injects \Phi(\FF_q)$.
Take $$G^2 = \ker(A(k) \mapsvia{r} A(\FF_q)),$$ then $G^1/ G^2 \injects A^0(\FF_q)$.
Note that $G^2$ comes from the formal group law.

Now let $T^i = G^i \intersect A(k)[\tors]$, then 
$$
\# A(k)[\tors]' \leq \# \Phi(\FF_Q) \# A^0(\FF_q),
,$$ 
and since we know everything about $A^0(\FF_q)$, we have

\begin{align*}
\#A(k)[\tors]' \leq \# \Phi(\FF_q) \floor{ q + 2\sqrt q + 1  }^\beta \qty{q + 1}^\mu
.\end{align*}

Note that $q^\alpha$ is $p\dash$torsion, so we don't need the unipotent piece here.

Theorem (Lestra-Oort)
:   Suppose $\mu = 0$, so no toric part.
    Then $\# \Phi' \leq 2^{2\alpha} \leq 2^{2g}$.

Thus if $\mu = 0$, we have

\begin{align*}
\#A(k)[\tors]' \leq \# \Phi(\FF_q) ~\floor{ q + 2\sqrt q + 1  }^\beta ~\qty{q + 1}^\mu ~2^{2\alpha}
.\end{align*}

Noting that $\alpha + \mu + \alpha = g$, this can give a better bound than in the case of good reduction.

Definition (Anisotropic Torii)
:   Let $T/k$ be a torus, then $T$ is *anisotropic* iff there does not exist an embedding $\GG_m \injects T$ over $k$.

Definition (Anisotropic Reduction)
: $A/k$ has *anisotropic reduction* (or AR) iff the torus part is anisotropic.

Theorem (Clark-Xarles)
:   If $A/k$ has anisotropic reduction, then
  
    \begin{align*}
    \#A(k)[\tors]' 
    &\leq \# \Phi(\FF_q) \floor{ q + 2\sqrt q + 1  }^\beta \qty{q + 1}^\mu 2^{\alpha} \\
    &\leq \floor{ q + 2\sqrt q + 1 }^g
    .\end{align*}

This generalizes the potentially good reduction case, since p.g.r. implies anisotropic.

Next up: need to understand $p\dash$adic uniformization of abelian varieties, and particularly for abelian varieties.