Suppose that a hyperelliptic curve $C$ of genus $g \geq 4$ is given by the equation

$$\displaystyle C: y^2 = a_0 x^{2g+2} + a_1 x^{2g+1} + \cdots + a_{2g+2} = f(x).$$

The Jacobian variety $J(C)$ of the curve $C$ can be thought of as the group of linear equivalence classes of degree $0$ divisors on $C(\overline{\mathbb{Q}})$. It is a finitely generated abelian group, by the Mordell-Weil theorem, with rank $r \geq 0$.

Let us denote a quadratic twist $C_d$ of $C$ as a hyperelliptic curve given by the equation

$$\displaystyle C_d: dy^2 = a_0 x^{2g+2} + \cdots + a_{2g+2}.$$

$C$ and $C_d$ are isomorphic over the quadratic extension $\mathbb{Q}(\sqrt{d})$, and isomorphic over $\mathbb{Q}$ if and only if $d$ is a square.

It is known from the following paper:

S. Petersen, *The rank of hyperelliptic Jacobians in families of
quadratic twists*, Journal de Th´eorie des Nombres
de Bordeaux, **18** (2006), 653-676

that all elliptic curves defined over $\mathbb{Q}$ admit infinitely quadratic twists of rank at least one. If one further imposes that the $j$-invariant of the curve is not $0$ nor $1728$, then one can show that there are infinitely many quadratic twists with rank at least $2$.

What about the higher genus situation? Does there exist a hyperelliptic curve $C$ (of possibly large genus) such that every quadratic twist $C_d$ of $C$ has rank at most $g+1$? Is this expectation sharp for every genus $g$?

at least2, but then you ask about hyperelliptic curves of rankat most$g+1$. Do you meanat most, orat least? $\endgroup$ – Joe Silverman Aug 23 '15 at 17:34