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Timeline for Lattices containing $A_n$ and $D_n$

Current License: CC BY-SA 4.0

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Feb 6, 2022 at 15:24 answer added David E Speyer timeline score: 1
Feb 6, 2022 at 14:37 comment added David E Speyer Note that, even with rescaling, the dimension must be of the form $k^2-1$. Let $A$ and $D$ be the Gram matrices of the two lattices. Let $S$ be the rational change of basis matrix which turns the simple $A$-basis into the simple $D$-basis. Then $SAS^T = D$ so $\det(A) \det(S)^2 = \det(D)$. We have $\det(A) = n+1$ and $\det(D) = 4$, so we deduce that $\tfrac{n+1}{4}$ is a rational square. Given this, it would be nice to see whether or not $15$ dimensions is achievable.
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Oct 8, 2021 at 23:06 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Sep 8, 2021 at 23:25 comment added Marco Golla @LSpice: integral overlattices of an integral lattice $L$ correspond to isotropic (? I think this is the correct word) subgroups of the dual group of $L^*/L$.
Sep 8, 2021 at 22:41 history edited YCor
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Sep 8, 2021 at 22:14 comment added LSpice @MarcoGolla, dual group of what?
Sep 8, 2021 at 22:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Aug 9, 2021 at 20:27 answer added FermaX timeline score: 0
Aug 3, 2021 at 21:55 history edited Daniel Sebald CC BY-SA 4.0
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Aug 3, 2021 at 21:52 comment added Daniel Sebald Rescaling is fine.
Aug 3, 2021 at 20:21 comment added Marco Golla If you don't admit rescalings (i.e. if you only consider integral lattices), then overlattices of $A_n$ and $D_n$ are classified by subgroups of the dual group. In particular $D_n$ has at most three overlattices (one or two of which are diagonal).
Aug 3, 2021 at 20:14 comment added Roland Bacher There is a problem with the Leech lattice: it contains no roots. I think you accept rescalings.
Aug 3, 2021 at 20:11 comment added Roland Bacher Such lattices must be unimodular (of determinant 1) and of dimension $k^2-1$ (since $A_n$ has determinant $n+1$).
Aug 3, 2021 at 20:06 history asked Daniel Sebald CC BY-SA 4.0