# Friday February 14th

## Canonical Heights

Last time: the Néron-Tate Canonical Height.
Simple if you buy the Weil Height Machine!
Recall the statement of the theorem:

Theorem (Existence of Canonical Height Functions)
:   Let $K$ be a global field, $V/K$ a nice variety, $D\in \div(V)$.
    Suppose there exists $\phi:V \to V$ such that $D$ is an "eigendivisor", i.e. $\phi^* D \sim \alpha D$ for some $\alpha \in \QQ$.

    Then there exists a unique function $\hat h_{V, \phi, D}: V(K\sep) \to \RR$ such that

    - (CH1) $\hat h_{V, \phi, D} = h_{V, D}  O(1)$
    - (CH2) $\hat{h_{V, \phi, D}} \circ \phi = \alpha \hat{h_{V, \phi, D}}$, which is constructed in the following way:
        For all $p\in K\sep$, $\hat h_{V, \phi, D}(p) = \lim_n \frac{1}{\alpha^n} h_{V, D}(\phi^n(p))$.


Two important examples:

Example 
:   Morphism from $\PP \to \PP$, then $\pic \PP = \ZZ$ so everything's an eigendivisor.
    So just take a degree $\geq 2$ map and a hyperplane divisor.

Example (Key example)
:   $A/K$ an abelian variety, $D$ ample, $\ell(D) \geq 2$ (so at least 2 sections of the line bundle), and symmetric in the sense that $[-1]^* D \sim D$.
    Then for all $n\in \ZZ$, $[n]^* D \sim n^2 D$, so take $\phi$ to be multiplication by $n$ for $\abs{n} \geq 2$.

    In this case $\hat h_{A, D} = \hat h_{A, [n], D}$.
    It agrees with the usual Weil height, and moreover $\hat h_{A, D}([n]p) = n^2 \hat h_{A, D}(p)$.
    Note that this is at least one property we'd want from a quadratic form.

The geometry associated to the height function on other types of varieties can become more interesting, e.g. $K_3$ surfaces and Calabi-Yau.

Next up:
Showing $\hat h_{A, D}: A(K\sep) \to \RR$ is a "quadratic form", but what is a quadratic form on an abelian group?

## Quadratic Functions

Let $A, B$ be abelian groups written additively.
Then a map $f: A\to B$ is *quadratic* if the associated function 

\begin{align*}
B_f: A\cross A &\to B \\
(x, y) &\mapsto f(x+y) - f(x) - f(y) + f(0)
\end{align*}
is bilinear.

A function is *homogeneous quadratic* iff $f$ is quadratic and $f(0)= 0$.
A *quadratic form* is homogeneous, quadratic, and symmetric in the sense that $f(-x) = f(x)$ for all $x\in A$.

> Note: multiplication on $\RR$ by 2 is an isomorphism, i.e. $\RR$ is *uniquely 2-divisible*. 
> If $B$ doesn't have this property, things can get slightly more complicated.

Lemma
:   $f: A\to B$ is quadratic iff for all $x, y, z\in A$ we have

    \begin{align*}
    f(x + y + z) - f(x + y) - f(x + z) - f(y+z) + f(x) + f(y) + f(z) - f(0) = 0
    .\end{align*}

In other words, knowing what it does on values and pairs determines what it does on triples.

Proof
:   Straightforward computation.

Lemma
:   Let $f:A\to B$ such that

    1. $f$ is a quadratic form.
    2. $f$ satisfies the *parallelogram law* (also: quadraticity) $f(x + y) + f(x-y) = 2f(x) + 2f(y)$.

    Then 1 implies 2, and if the 2-torsion $B[2]$ is trivial, then 2 implies 1.

Lemma
:   Suppose $f$ is a quadratic function, then $f$ is a quadratic form iff $\forall n\in \ZZ,~\forall x\in A$, $f(nx) = n^2f(x)$.

Note that $n=0$ implies homogeneity, and $n=1$ implies symmetric.
Proceed by strong induction.

Lemma
:   Let $f: A\to B$ be a quadratic function.
    Define the even and odd parts of $f$ in the following way:

    \begin{align*}
    q: x &\mapsto \qty{ f(x) - f(0) } + \qty{ f(-x) - f(0)  } \\
    l: x &\mapsto \qty{ f(x) - f(0) } - \qty{ f(-x) - f(0)  } \\
    .\end{align*}

    Then 

    a. $q$ is a quadratic form, $l$ is a group morphism, and $2f = q + l + 2f(0)$.

    b. If $B$ is uniquely 2-divisible, then dividing by 2 yields $f = \frac 1 2 q + \frac 1 2 l + f(0)$ where $q$ is a quadratic form, $l$ is a linear morphism, and $f(0)$ is constant.
      This decomposition is unique.

Exercise
:   Show that if $f: A\to B$ a quadratic function,

    a. $f(A[\tors]) \subseteq B[\tors]$.

    b. For $n$ odd, $f(A[n]) \subset B[n]$.

    c. The map $f: \ZZ/2\ZZ \to \ZZ/4\ZZ$ where $0\mapsto 0, 1\mapsto 1$ is a quadratic form.
        Thus the $n$ odd condition above is necessary.

This comes up in Galois cohomology of abelian varieties? See Pete's paper.
It takes an $n$ torsion to an element that is no worse than $2n$ torsion.

From now on, take $B = \RR$.
Note that if $B=\RR$, height functions kill all torsion.

Define a pairing

\begin{align*}
\inner{\wait}{\wait}: A\cross A &\to B \\
\inner{x}{y} &= \frac 1 2 \qty{ f(x+y) - f(x) - f(y)  }
.\end{align*}

If $f$ is a quadratic form, $\inner{x}{x} = f(x)$, so this reduces to the usual bilinear form associated to a quadratic form.

Exercise
:   a. Show that for all $x\in A, y\in A[\tors]$, we have $f(x+y) = f(x)$.
      Thus $f$ factors through a morphism $\hat f: A/A[\tors] \to \RR$ through the torsion-free quotient.

      > This is stronger than just killing torsion, since $f$ isn't necessarily linear.

    b. If $f$ is a quadratic form, $\hat f$ is as well.

Theorem (Canonical Height Descent)
:   Let $h: A\to \RR$ be a quadratic form and suppose $\exists n\geq 2$ such that 

    1. $\cok{A \mapsvia{\times 2} A} = A/nA < \infty$,
    
    2. The Northcott property holds, i.e. $\theset{x\in A \suchthat h(x) \leq R} < \infty$ for each $R\in \RR$,

    Letting $y_1, \cdots, y_r$ be coset representatives for $nA \subset A$ and $C_0 \definedas \max h(y_i)$, then $A$ is generated by the finite set $\theset{x\in A \suchthat h(x) \leq C_0}$.

Proof:
: Omitted.

Theorem (The Néron-Tate Height is a Quadratic Form)
:   Let $K$ be a global field, $A/K$ an abelian variety, $D$ ample and $\ell(D) \geq 2$ with $[D]$ symmetric.
    Then there exists a unique function $\hat h_{A, D}: A(K\sep) \to \RR$ such that

    1. $\hat h_{A, D} = h_{A, D} + O(1)$

    2. $\hat h_{A, D}$ is a *quadratic form* (stronger conclusion), and in particular $\hat h_{A, D}([n]p) = n^2 \hat h_{A, D}(p)$.

Proof
:   We have 1, we'll show 2 by establishing the parallelogram law and use a previous lemma.

    > Note $\phi$ is suppressed here, since by uniqueness, multiplication by $n$ always yields the same function. So we'll take multiplication by 2.

    For all $P, Q \in A(K\sep)$, we can compute

    \begin{align*}
    \hat h_{A, D}(P + Q) + \hat h_{A, D}(P-Q) 
    &=
    \lim 4^{-n} \qty{ h_{A,D}([2^n]P + [2^n]Q) + h_{A, D}\qty{ [2^n]P - [2^n]Q  }  } \\
    &=
    \lim 4^{-n} \qty{ 2 h_{A,D}([2^n]P) + 2 h_{A, D}([2^n]Q) + O(1) } \\
    &=
    2 \hat h_{A, D}(P) + 2\hat h_{A, D}(Q)
    .\end{align*}