TITLE: Bound on Minimal Length of Vectors in Lattice and its Dual Lattice
QUESTION [6 upvotes]: Let $\Lambda$ be a lattice in $\mathbb{R}^n$ and $\Lambda^\ast$ its dual lattice. Let $d=\min_{v\in\Lambda} (v,v)$ and $d^\ast =\min_{v\in\Lambda^\ast} (v,v)$ be the minimal squared lengths of vectors in the two lattices. I'm interested in lattices in which both $d$ and $d^\ast$ are large.
For $n=24$, I know of the self-dual Leech lattice which has $d=d^\ast=4$. I'd like to find lattices with $d,d^\ast\ge 8$. Is this at all possible? Self-duality is not so important for me, so if I can lower $n$ at the expense of self-duality that's OK.

REPLY [3 votes]: The product $d\times d^*$ cannot be "so" large, as a consequence of the so-called  Transference theorems Particularly, Thm. 2.1. of the paper shows that  $d\times d^* \leq n^2$ (hence in your example, $n$ should be at least $8$). So the best you can do in terms of the product is $\Omega(n^2)$. The aforemetioned Conway-Thompson Theorem (see [1], p. 46) shows that indeed it is possible to achieve $\Omega(n^2)$.
In the case of $d, d^* \geq 8$, we have the lower bound $n \geq 8$ and the achievable upper bound $72$ (due to the unimodular table). Nevertheless, there is still a huge gap between both bounds. I am not aware of any result that shows that smaller $n$ are possible.
[1]. J. Milnor and D. Husemoller, ``Symmetric Bilinear Forms,'' Springer-Verlag, New York,
1973