Timeline for Algebraic curves and torsion points of its Jacobian
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2020 at 11:12 | comment | added | Michael Stoll | (3) I think one expects that 50% of all Jacobians of odd degree hyperelliptic curves of fixed genus have rank 0 and 50% have rank 1. So, combining (2) and (3), you'll get at least 50% heuristically for curves with a rational Weierstrass point. But almost all of them will have trivial torsion. It might be possible to prove something for curves with two rational Weierstrass points (forcing the rational torsion subgroup to have even order); also in this case, the heuristic expectation is that 100% of all curves have no additional rational points and 50% have positive rank. | |
| Nov 8, 2020 at 11:09 | comment | added | Michael Stoll | (1) The density of curves with a rational point is expected to be zero and known to tend to zero as the genus grows (by work of Bhargava). (2) Restricting to curves with $a_{2g+2}=0$ to force the existence of a rational point, it is again expected that this is the only rational point outside a density zero set, and it is known that the density of curves with additional rational points tends to 0zero as the genus grows (by work of Poonen and myself; the definition of height is a bit different here, though). TBC... | |
| Nov 8, 2020 at 0:15 | comment | added | Stanley Yao Xiao | Thank you for the answer. Do you have some idea as to how dense (say, with hyperelliptic curves of genus $g$ given by $y^2 = a_{2g+2}x^{2g+2} + \cdots + a_1 x + a_0, a_i \in \mathbb{Z}$, ordered by the height $\max |a_i| \leq X$) the curves with a rational point and such that $C(\mathbb{Q})$ embeds into $\text{Jac}_C(\mathbb{Q}^{\text{tors}}$ under the Abel-Jacobi map, and such that the rank of the Jacobian is positive? | |
| Nov 8, 2020 at 0:12 | vote | accept | Stanley Yao Xiao | ||
| Nov 6, 2020 at 14:28 | history | answered | Michael Stoll | CC BY-SA 4.0 |