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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 $g$?

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  • $\begingroup$ (1) The way you've phrased the problem is confusing, since you call the Jacobian the points over $\overline{\mathbb Q}$, but then you talk about the Mordell-Weil group of $J(C)$ without specifying a field. Presumably you mean $J(C)(\mathbb Q)$. (2) The theorem you quote says that there are elliptic curves of rank at least 2, but then you ask about hyperelliptic curves of rank at most $g+1$. Do you mean at most, or at least? $\endgroup$
    – Joe Silverman
    Commented Aug 23, 2015 at 17:34
  • $\begingroup$ @JoeSilverman: I do want an upper bound, perhaps with finitely many exceptions. And yes I do mean the rank of the Jacobian over the rationals. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 23, 2015 at 18:35
  • $\begingroup$ For elliptic curves there is a significant difference of opinion as to whether the rank of the twists $E_D(\mathbb Q)$ should be bounded or unbounded. Of course, there are also now some heuristic arguments suggesting that the rank $E(\mathbb Q)$ is bounded as one ranges over all elliptic curves. But are you suggesting, based on Petersen's article, that there exists an $E/\mathbb Q$ such that rank$(E_D(\mathbb Q))\le2$ for all $D$ (or at least for all but finitely many $D$ modulo squares)? That sounds bold. $\endgroup$
    – Joe Silverman
    Commented Aug 23, 2015 at 19:34
  • $\begingroup$ @JoeSilverman: I've done some more literature review on the subject in the last few weeks, and it seems that the heuristic that best supports that the rank of the Jacobian of hyperelliptic curves should usually be small compared to the genus is the Bhargava-Gross result on the average size of the $2$-Selmer group. The average is independent of genus, and hence as $g$ goes up, it is harder and harder for there to be Jacobians with rank exceeding $g+1$. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 31, 2015 at 22:43

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