Timeline for Lower bounds on the rank of a unimodular lattice, given the binlinear pairing of a subset of basis vectors
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2014 at 21:29 | comment | added | Alex Degtyarev | @D.S.Park: Just one last addition: sometimes, you have to add $8$ to get the signature right and satisfy the hypothesis of Nikulin's theorem (without computing the determinant for $\mathrm{length}=\mathrm{rank}$). On the other hand, there's no need to add $[-1]$: if $n$ is odd, the result is odd automatically, if $n$ is even, you take the quadratic extension $\langle(n+1)/n\rangle$ rather than $\langle1/n\rangle$, so the quadratic forms are not identified via an anti-isomorphism and the ambient lattice is still odd. Now, everything seems taken care of, and it's still $2k+8$ :) | |
| Dec 23, 2014 at 21:12 | comment | added | D. S. Park | Thank you very much! Your explanation is crystal clear now. :) | |
| Dec 23, 2014 at 21:11 | vote | accept | D. S. Park | ||
| Dec 23, 2014 at 18:19 | comment | added | Alex Degtyarev | Accidentally, the answer to your Question 5 is also $\le2k+8$, with the same proof :) | |
| Dec 23, 2014 at 18:16 | history | edited | Alex Degtyarev | CC BY-SA 3.0 |
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| Dec 23, 2014 at 17:59 | comment | added | Alex Degtyarev | I will try to put a short proof to the answer. | |
| Dec 23, 2014 at 15:35 | comment | added | D. S. Park | Thanks for the answer. I'm surprised at your claim that $b(n,k) \leq 2k + {\rm const}$---can you explain this in more detail? I expected $\alpha_n$ to be larger than $2$ for, say, $n=12$. Note that I've fixed the signature of the lattice to be $(1,T)$ and assumed that the lattice is odd. | |
| Dec 22, 2014 at 21:21 | history | answered | Alex Degtyarev | CC BY-SA 3.0 |