# Monday February 17th

## Canonical Heights for Abelian Varieties

Theorem (Canonical Heights for Abelian Varieties)
:   If $A/K$ is an abelian variety, $D\in \div(A)$, ample, $\ell(D) \geq 2$, $[D]$ symmetric, then there exists a canonical height $$\hat h_{A, D}: A(K\sep) \to \RR$$  such that

    1. $\hat h_{A, D} = h_{A, D} + O(1)$
    2. $\hat h_{A, D}$ is a *quadratic form* (defined last time).

In particular, $\hat h_{A, D} \circ [n] = n^2 \hat h_{A, D}$.
We can define an associated bilinear form 

\begin{align*}
\inner{P}{Q} = \frac 1 2 \qty { \hat h_{A,D}(P+Q) - \hat h_{A, D}(P) - \hat h_{A, D}(Q) }
,\end{align*}

so consider the **linear extension**

\begin{align*}
\hat h_{A, D}: A(K\sep) \tensor \RR &\to \RR \\
\sum_{i=1}^n [P]\tensor \alpha_i &\mapsto \sum_{i=1}^n \alpha_I \hat h_{A, D}
.\end{align*}

Exercise
:   This yields a quadratic form on the domain.

*Question:*
What is $\ker(A(K\sep) \to A(K\sep) \tensor \RR)$?

*Answer:*
It contains the torsion, but turns out to be precisely this.
Note that $\ker(M \to M\tensor \QQ) = M[\tors]$, and we can tensor up in two steps:
$A(K\sep) \tensor_\ZZ \RR = (A(K\sep) \tensor_\ZZ \QQ) \tensor_\QQ \RR = A(K\sep)[\tors]$.

Theorem (Linear Extension is Positive Definite)
:   With the same setup as the previous theorem, the linear extension is *positive definite*, i.e. 
    $$
    q(x) \geq 0 ~\forall x \quad\text{and}\quad q(x) = 0 \iff x=0
    $$

Definition (Regulator)
:   If $P_1, \cdots, P_n \in A(K\sep) \tensor \RR$ are $\RR\dash$linearly independent, then we can define the **regulator** 
    $$\reg(P_1, \cdots, P_2) \definedas \det(\inner{P_i}{P_j}) > 0.$$

Note that this is the determinant of the Gram matrix, forms can be diagonalized into $\sum a_i x_i^2$, and positive-definite implies $a_i \geq 0$? 

If $\theset{P_i}$ is a $\ZZ\dash$basis for $A(K\sep) / A(K\sep) [\tors]$, then we define $\reg(A, K, D) \definedas \det(\inner{P_i}{P_j}) > 0$.

> This is well-defined for exactly the same reason that the discriminant of a number field is well-defined.

Proof
:   Mostly algebra, mostly omitted here.
    Uses the following lemma.

Lemma
:   Let $A \cong \ZZ^n$ and $q:A\to \RR$ be a quadratic form such that 

    1. For all $x\in A$, $q(x) \geq 0$ and $q(x) = 0 \iff x=0$,
    2. The Northcott Property holds, i.e. $\forall R\in \RR, \# \theset{x\in A \suchthat q(x) \leq R} < \infty$.

    Then $q\tensor \RR: A\tensor \RR \to \RR$ is still positive-definite.

Example
: See Cassels-Silverman.
  Let $K$ be a real quadratic field, e.g. $K = \ZZ[\sqrt 2] \subset \RR$.
  Define $q:A \to \RR, x\mapsto x^2$; this is positive-definite but for $q\tensor \RR$, there exist $x\neq 0$ such that $q(x) = 0$.

*Note:*
There exists a **canonical** canonical height which is used for the BSD conjecture.

### Constructing Canonical Heights for Abelian Varieties

**Step 1:**
For elliptic curves $E/K$, $D = 2[O] \iff D_x$ where $x: E\to \PP^1$ is taking the $x\dash$coordinate, note that if $y^2 = f(x)$, then $x([-P]) = x([P])$ making this an even function.
So taking the quotient $E \mapsvia{q} E/[-1]$ yields a 2-to-1 map, which factors through an isomorphism $E/[-1] \cong \PP^1$.
However, $2[-O] = [-1]^* D = D$, but we also have $2[O] = x^*([\infty])$.
Then $K(x) \subset K(E)$ contains all of the even functions, i.e. every even function is a function of the $x\dash$coordinate.
So take any rational function $f\in K(x) \setminus K$ yields a composition $F: E\mapsvia{x} \PP^1 \mapsvia{f} \PP^1$ where $\deg(F) = 2\deg(f)$.

Define $D_f = F^*([\infty])$, which is a pullback of a symmetric function (since $x$ is only a function of one coordinate), so $[-1]^* D_f = D_f$, so $D_f$ is a symmetric divisor.
Moreover, $\deg D_f = 2\deg f$.
Take $\hat h_{E, f} = \hat h_{E, D_f}$, since by Riemann-Roch the number of sections is at least 2.
How does this depend on $f$? 
Does the regulator depend on $f$?

> There is a dependence, but it's understandable enough to be removed at the end.

We can apply the Weil Height Machine, and a comparison shows $h_{E, D_f} = \qty{\deg f} h_{E, D_x} + O(1)$.
Therefore $\hat h_{E, D_f} = \qty{\deg f} \hat h_E, D_x$ on the nose.
By a computation, $\reg(E, D_f) = \deg(f)^{n} \reg(E, D_x)$ where $n = \rank(E(K))$.
So there is a dependence, but nothing too scary.

Silverman defines the canonical height in the curve case as

\begin{align*}
\hat h_E \definedas \frac{\hat h_{E, D_f}}{2\deg f}
.\end{align*}

**Step 2:**
For abelian varieties, define a height pairing $\hat h: A(K\sep) \cross A\dual(K\sep) \to \RR$, where the dual variety is defined such that the functor of points $A\dual \iff \pic^0(A)$, where $A\dual(K) = \pic^0(A)$.

There is a (well-defined up to linear equivalence) *Poincare divisor* $\mcp \in \div(A\cross A\dual)$ which is ample and symmetric.
We can then define $\hat h_{A\cross A\dual, \mcp}$ and take the associated bilinear form

\begin{align*}
\inner{\wait}{\wait}_{A\cross A\dual, \mcp}:\qty{A(K) \cross A(K\sep)}^2 \to \RR
.\end{align*}

So choose bases $P_1, \cdots, P_R$ for the Mordell-Weil group mod torsion $A(K)/\tors$, and by isogeny, a basis $Q_1, \cdots, Q_R$ of $A\dual(K)/\tors$.
Then define $\reg(A, K) \definedas \abs{\det{\inner{P_i}{Q_j}}} > 0$.

> Note that swapping signs in the basis changes the determinant by $\pm 1$, hence the absolute value.
> This is the regulator that appears in the BSD conjecture.

Recall Mumford's formula:
If $D\in \div(A)$, then

\begin{align*}
[n]^* D \sim \left[ \frac{n^2 + n}{2}\right] D + \left[ \frac{n^2-n}{2}\right] [-1]^* D
.\end{align*}

If $D$ is symmetric, then $[n]^* D \sim n^2 D$, i.e. the quadratic eigendivisor condition.
In this case $\hat h_{A, D}$ is a quadratic *form*.

If $D$ is skew-symmetric, i.e. $[-1]^* D = -D \iff D \in \pic^0(A)$, then $[n]^* D = nD$ and we obtain a *linear* eigendivisor condition.

> Note: $D\in \pic^0(A)$ means algebraically equivalent to zero.

If $D\in \pic^0(A)$, then $\hat h_{A, D}: A(K\sep)\to \RR$ is a linear form.
By the Weil Height Machine, up to $O(1)$ we have $\hat h_{A, D_1 + D_2} = \hat h_{A, D_1} + \hat h_{A, D_2}$.
So we do have a bilinear pairing

\begin{align*}
\inner{\wait}{\wait}: A(K\sep)^2 &\to \RR \\
(P, Q\iff D\in \div(A) \text{ after extending bases  }) &\mapsto \hat h_{A, D}(P) \in \RR
.\end{align*}

> Note: we could equivalently define it on $A(L)$ for every finite separable extension $L/K$.

This explains the regulator appearing in BSD.
This concludes the discussion of height functions and the Mordell-Weil group.