# Monday April 6th

We've been talking about quadratic twists of Tate curves.
Let $K$ be a field, $\ch K \neq 2$, $V/K$ a nice variety, $\iota: \aut(V)$ an involution, $L = K(\sqrt d)/K$, then $J(V, \iota, L) = V^d/K$ is a quadratic twist.
We have $V^d / K(\sqrt d) = V_{K(\sqrt d)}$, and $\Aut K(\sqrt d) / K = \theset{1, \sigma}$ is a two element set, so
\begin{align*}
V^d(K) \definedas \theset{ p \in V(K(\sqrt d)) \suchthat \sigma(p) = \iota(p) }
.\end{align*}

Example:
For $E/k$ an elliptic curve with $\iota = [-1]$, if $E: y^2 =x^3 + Ax + B$ yields $E^d: dy^2 = x^3 + Ax + B \sim y^2 = x^3 + d^2Ax + d^2 B$.
This works out similarly for hyperelliptic curves $C$: for $y^2 = f(x)$, we obtain $C^d:  dy^3 = f(x)$.
Note that there is not such a description in equations for abelian varieties.

Theorem (C- Stankewicz, 2018)
:   Let $K$ be a number field and $C/K$ a "nice" curve (conditions to be described) with $\iota$ an involution on $C$.
    Then

    a. For all but finitely many $d \in K\units / (K\units)^2$, $C^d(K) = \emptyset$.
    b. There exist infinitely many $d$ such that $C^d$ has $K_v\dash$rational points for all places $v$.
    c. If $K = \QQ$, then as $x\to\infty$ the number of $d$ with $\abs{d} \leq x$ such that $C^d$ violates the Hasse principle is $\gg x/\log x$

Punchline: under some conditions, you get a lot of Hasse principle violations.

Conditions:

1. $\theset{p\in C(K) \suchthat \iota(p) = p} = \emptyset$
2. $\theset{p \in C(\bar K) \suchthat \iota(p) = p} \neq \emptyset$.
3. There exists some $d\in K\units / \qty{K\units}^2$ such that $C^d$ has points everywhere locally.
4. $(C/\iota)(K)$ is finite.

This works for hyperelliptic curves, conditional on the abc conjecture.

For condition 4, we can just take the genus $>2$ and apply Falting's theorem.
We can apply this to Atkin-Laner (?) twists of elliptic curves.

> Note that there is a lot of work being done on quadratic twists.

For $A/K$ an abelian variety, take $\iota = [-1]$ and $K(\sqrt{d}) / K$ with $\ch K \neq 2$.
What is the quadratic twist $A^d/K$?
For $N\geq 3 \in \ZZ^+$ and $\dim A = g$, there is a representation $\rho_N: g_K \to \Gl(2g, \ZZ/n\ZZ) = \Aut(A[N]) =\Aut(\qty{\ZZ/N\ZZ}^{2g})$.
This is the "mod $N$ galois representation".

What is the galois representation on the quadratic twist $\rho^d_N: g_K \to \GL(2g, \ZZ/N\ZZ)$?
They both restrict to the same function on $g(K(\sqrt d))$.
The succinct answer is that $\phi^d_N = \phi_N \tensor \chi_d$, where $\chi_d: g_k \to \theset{\pm 1}$ where $\sigma \mapsto 1$ iff $\restrictionof{\sigma}{K(\sqrt d)} = 1$ and -1 otherwise.
What is this tensor?
We can think of $\theset{\pm 1}\in \qty{\ZZ/N\ZZ}\units = \Gl(1, \ZZ/N\ZZ)$.
In other words, for all $\sigma \in g_k$, we have $\rho_N^d = \chi_d(\sigma) \rho_N(\sigma) = \pm \rho_N(\sigma)$, where we identify $\pm 1$ with the scalar matrices $\pm I$ in $\Gl(2g, \ZZ/N\ZZ)$.
Thus it's a twist by a quadratic character.

Now let $G \subset A[n] \subset A$ where $G = \gens{p}$ is cyclic of order $N$ and $p\in A(K\sep)$.
Then $G$ is $g_k\dash$stable, so we can choose a basis $p = p_1, \cdots, p_{2g}$ of $A[n] / \ZZ/N\ZZ$.
We can then write
\begin{align*}
\phi_N(\sigma) =
\begin{bmatrix}
\chi_G(\sigma) & \ast \\
0 & \ast
\end{bmatrix}
\end{align*}

where $\sigma(p) = \chi_G(\sigma) p$, and thus
\begin{align*}
\begin{bmatrix}
\chi_d(\sigma) \cdot \chi_G(\sigma) & \ast \\
0 & \ast
\end{bmatrix}
.\end{align*}

We then have a diagram of inclusions

\begin{tikzcd}
G \ar[r, hook] \ar[d, hook] & A^d[N] \ar[r, hook] & A^d \ar[d, hook] \\
G \ar[r, hook]              & A[N] \ar[r, hook]   & A^d
\end{tikzcd}


What if we now want an actual $K\dash$rational point of order $N$?
Cyclic order $N$ $K\dash$rational subgroups of $A$ are equal to cyclic order $N$ $K\dash$rational subgroups of $A^d$, so

For $E/K$, we proved that it has a unique cyclic order $\ell$ subgroup corresponding to $\mu_\ell$.
So the same holds for every quadratic twist, and every rational point has to come from the original $\mu_\ell$ since cyclic subgroups aren't changing under quadratic twists.
Thus only the isogeny character can change.

So when does quadratic twisting gain or lose a rational point?
The image of the isogeny character must be order 1 or 2, corresponding to a quadratic extension, and the isogeny character is trivialized.
We'll eventually conclude that none of the quadratic twists will have a rational point of order $N$.