Timeline for Which proportion of elliptic curves over Q is proved to satisfy BSD conjecture?
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11 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2018 at 16:58 | comment | added | Wojowu | @StanleyYaoXiao That answer in its entirety addresses an old paper of Morita. A completely new one was posted in March (and updated a week ago) on arXiv. | |
| Jun 18, 2018 at 18:48 | comment | added | Just-A-Poster-Here | The whole ordeal of "at least X% satisfy BSD" is to my opinion a rather kitschy way of stating the Bhargava-school results in the first place (though for publicity reasons I can understand it). Though for that matter, I can't think that upping 40% to 41% for zeta-zeros is numerically that interesting either, yet the novelty in methods of course has some regard. | |
| Jun 18, 2018 at 10:08 | comment | added | Stanley Yao Xiao | @SylvainJULIEN there are many reasons why there isn't as much interest in the proportion of curves satisfying BSD as say proportions of zeroes of $L$-functions on the half-line. One that I've heard is that the height used by Bhargava and Shankar to get that the average rank is bounded is not considered canonical. If one succeeds in counting elliptic curves by discriminant (and also Selmer elements), then I think people might be more interested | |
| Jun 18, 2018 at 10:05 | comment | added | Stanley Yao Xiao | @JoeSilverman it appears there is an error in Morita's argument, as identified by j.c. in the first comment (mathoverflow.net/questions/244459/congruent-number-problem/…). There is an answer that claims that "it comes down to an elementary but unfixable blunder in p-adic Hodge theory". | |
| Jun 17, 2018 at 21:35 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Added link
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| Jun 17, 2018 at 20:43 | comment | added | Joe Silverman | If correct, this would be a huge breakthrough. Note that even the statement "100% of $E/\mathbb Q$ satisfy BSD" would likely involve first proving that 100% of the curves have rank at most 1. But Morita claims to prove BSD for all $E/\mathbb Q$ having CM, which would include large numbers of curves with rank $\ge2$. I am very curious where he's finding the points of infinite order in the case that $L$ vanishes to order $r\ge2$. It's known that the Heegner point construction fails in that case. | |
| Jun 17, 2018 at 20:28 | comment | added | Sylvain JULIEN | Thank you very much Stanley. This is quite strange as tiny improvements on the proportion of critical zeros of the Riemann zeta function lead to several publications. | |
| Jun 17, 2018 at 20:24 | comment | added | Stanley Yao Xiao | Various refinements since the Bhargava-Skinner-Zhang paper means that it is now possible to improve the percentage... at a talk in 2016 Bhargava mentioned that (as of September 2016) it should be possible to push the percentage above the 80s. Since this is quite tedious and somewhat unrewarding work, it appears that it has not been written down yet. | |
| Jun 17, 2018 at 20:18 | comment | added | Chris Wuthrich | Very few curves over $\mathbb{Q}$ have complex multiplication. This would not change the % anyway. | |
| Jun 17, 2018 at 20:17 | comment | added | j.c. | The work of Morita has been discussed here before mathoverflow.net/a/270820 mathoverflow.net/a/244478 | |
| Jun 17, 2018 at 20:06 | history | asked | Sylvain JULIEN | CC BY-SA 4.0 |