Timeline for 2-dimensional sublattices with all vectors having very big square (in absolute value)
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2015 at 14:42 | vote | accept | Misha Verbitsky | ||
| Sep 12, 2015 at 18:33 | comment | added | Noam D. Elkies | Not sure about a canonical reference but there's a lot of material (and references) about indefinite lattices and "gluing" in Conway and Sloane's "SPLAG" = Sphere Packings, Lattices and Groups. | |
| Sep 12, 2015 at 15:04 | comment | added | Misha Verbitsky | Many thanks for the idea, I tried to realize it (determinant 1 is OK, but we want it to be equivalent over ${\mathbb Q}$ with a unimodular lattice). I guess this is just my ignorance, could you point me to appropriate reference? | |
| Sep 10, 2015 at 0:08 | comment | added | Noam D. Elkies | One standard approach is to "glue" $\Lambda_0$ to an appropriate rank-1 lattice $M$, obtaining a determinant-1 lattice of rank 3 that contains $\Lambda_0 \oplus M$ with finite index, and is thus isomorphic with $\Lambda$ and contains a copy of $\Lambda_0$. This can be done for non-unimodular $\Lambda$ too as long as $\Lambda$ is the only lattice in its genus (which is common in the indefinite case). | |
| Sep 9, 2015 at 6:56 | comment | added | Misha Verbitsky | I still don't quite understand how it can be embedded to a given lattice (non-unimodular). For unimodular it's fine, except that I still cannot find a proper reference for an embedding result even in this generality (Morrison quotes Nikulin, but this result is not easy to find in Nikulin's paper). | |
| Sep 9, 2015 at 2:14 | comment | added | Noam D. Elkies | ...(up to finite index and variations such as changing $M^4+1$ to $M^4-1$). | |
| Sep 9, 2015 at 2:13 | comment | added | Noam D. Elkies | Thanks. There are two questions here: constructing $\Lambda_0$ and proving it embeds into $\Lambda$. (Not $\Lambda$ into $\Lambda_0$, right?) The former I gave a sketch of that can be completed with little difficulty. The latter, well it should certainly be OK for all but the smallest ranks (if $\Lambda$ has some really bad primes it might accommodate only a sublattice of $\Lambda_0$, but that's good enough. If $\Lambda$ has rank only $3$ it gets trickier but even there you can usually do anything that's not forbidden by a local obstruction, so some $M$ should work [cont'd] | |
| Sep 9, 2015 at 1:47 | history | bounty ended | David E Speyer | ||
| Sep 9, 2015 at 1:47 | comment | added | David E Speyer | I'm not sure I understand whether this is an actual proof or just a sketch? But it definitely seems like the sketch which comes closest to working, and I am supposed to give out the bounty, so I'll give it to you. If this is a actual proof, I'd love to understand how to fill in the gaps. (Why does this $\Lambda$ necessarily embed into $\Lambda_0$, for example?) | |
| Sep 9, 2015 at 0:41 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |