# Bound on Minimal Length of Vectors in Lattice and its Dual Lattice

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.

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So $d=d_*=8$ is possible for $n=72$ and an integral self-dual lattice (thanks Yoav!). Can one find a lattice with an arbitrary large $\min(d,d_*)$? Can one find a lattice with $d=d_*\ge 8$ and $n<72$ abandoning integrality and/or self-duality? –  Slava Rychkov Feb 24 '14 at 6:59
The Conway-Thompson theorem asserts the existence of unimodular lattices of arbitrarily large minimal norm. Indeed, the maximal minimal norm for unimodular lattices of rank $r$ can be bounded below by a function that is asymptotic to $\frac{r}{2\pi e}$. –  S. Carnahan Feb 24 '14 at 10:18

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

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Thanks, this is very helpful. It seems to me that the $\lambda$'s in the transference theorem you cite are rather the square roots of my $d$'s, so that the theorem implies $d \times d^*\le n^2$. So it seems that this still allows $d,d^*\ge 8$ and $n$ as low as 8. –  Slava Rychkov Feb 24 '14 at 17:59
Indeed by Minkowski there are unimodular lattices with $d_\min = d^*_\min \gg n$, so $n^2$ is the right asymptotic growth. –  Noam D. Elkies Feb 24 '14 at 18:14
@NoamD.Elkies: is this Minkowki result the same as Conway-Thompson? Can you please give a reference? –  Slava Rychkov Feb 24 '14 at 18:56
I don't know what's "Conway-Thompson" but it does give the same $r/(2\pi e)$ asymptotic. Sorry I don't have a reference at my fingertips, but Minkowski = the mass formula saying that the average of theta functions $\theta_L$, weighted by $\left|{\rm Aut}(L)\right|^{-1}$, is an Eisenstein series. Yet another approach is to calculate that for each $\delta>0$ an average lattice $L$ has minimal norm at least $r/(2\pi e) - \delta r$ with probability $1 - \epsilon$ with $\epsilon \to 0$ as $r \to \infty$, so the same is true for both $L$ and $L^*$ with probability at least $1 - 2\epsilon$. –  Noam D. Elkies Feb 24 '14 at 20:16
@NoamD.Elkies This probability method is interesting. How do you define an average lattice? In whatever definition, I presume $L^*$ and $L$ are not independent, so your final step may be not allowed. –  Slava Rychkov Feb 24 '14 at 21:03

The table of unimodular lattices suggests this is possible for n = 72.

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Do you know what $N$ and $M$ stand for in the table? –  Atsushi Kanazawa Feb 25 '14 at 8:07
No, I was wondering the same. –  Yoav Kallus Feb 25 '14 at 9:21