1
$\begingroup$

I am looking for a certain notion of sparseness of lattices.

I want to find a vector in $\mathbb{Z}^N$ that the minimal possible inner product with all the vectors of a given lattice. Or at least, I would like to know what this number is.

In other words, for a positive-definite lattice $\Gamma$ or rank $N$, what is $$ \min_{\vec{v} \in \mathbb{Z}^N} \max_{\vec{u} \in \Gamma} \frac{\vec{u} \cdot \vec{v}}{u v} $$

If $\Gamma$ is a hypercubic lattice, then I can always choose $\vec{u}$ to be $\vec{v}$, which means this quantity is just 1. Can I choose $\Gamma$ such that it is provably less than 1?

What is the value of this quantity for the smallest unimodular even lattice ($E_8$)?

I would expect that unimodular lattices with no roots (vectors of norm 1 or 2) are sparse in the above sense; the intuition being that they miss many of the short vectors in $\mathbb{Z}^N$.

Edit: normalization

$\endgroup$
5
  • 1
    $\begingroup$ I am having trouble parsing the definition of the quantity in question. Why wouldn't the maximum over $u$ be infinite? $\endgroup$ Apr 7, 2014 at 2:39
  • $\begingroup$ Also, how is $\mathbb{Z}^N$ related to $\Gamma$? Is $\Gamma = \mathbb{Z}^N$? $\endgroup$
    – Simon Rose
    Apr 7, 2014 at 2:47
  • $\begingroup$ Indeed, I forgot to divide by the length of $u$ itself. Fixed now. The only relationship between $\Gamma$ and $\mathbb{Z}^N$ is that they have the same rank. The point of my question is to understand what the above quantity can be depending on $\Gamma$. $\endgroup$ Apr 7, 2014 at 3:59
  • $\begingroup$ For any choice of $\vec{v}$, there will be $\vec{u} \in \Gamma$ with arbitrarily small angular separation from the line $\mathbb{R}\vec{v}$. This means your quantity is always 1. Do you want to restrict $\vec{v}$ to be in some ball? $\endgroup$
    – S. Carnahan
    Apr 7, 2014 at 8:31
  • $\begingroup$ Thank you Scott. I see now that inner product can be made arbitrarily close to 1. $\endgroup$ Apr 7, 2014 at 18:37

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.