Timeline for 2-dimensional sublattices with all vectors having very big square (in absolute value)
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2015 at 6:50 | comment | added | Misha Verbitsky | strike the previous comment, my proof was based on false assumption that any lattice is commeasurable with one which is proportional to unimodular | |
| Sep 8, 2015 at 12:40 | comment | added | Misha Verbitsky | Yes, and this seems to give an answer indeed. I will post it in a couple of days, once I am entirely sure there are no errors | |
| Sep 2, 2015 at 18:43 | comment | added | Noam D. Elkies | But then it feels like the $N \cdot H$ construction or something much like it does work even when $\Lambda$ is not unimodular. | |
| Sep 2, 2015 at 18:10 | comment | added | Misha Verbitsky | thanks! anyway, rank $\geq 6$ or $\geq 7$ is a usual assumption in these kind of applications | |
| Sep 1, 2015 at 23:16 | comment | added | Noam D. Elkies | In any case this approach must fail in rank $n=3$ because then there are only finitely many $N$ for which a lattice of the form $N\cdot H$ can embed primitively into $\Lambda$. | |
| Sep 1, 2015 at 22:33 | history | answered | Misha Verbitsky | CC BY-SA 3.0 |