Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This a shortened version (motivation from telecommunications stripped away) of a question I asked in MO in late May (no answers). I am mostly checking, if somebody has seen this or a related question studied during the history. Of course, all answers, suggestions and comments are welcome.

A detailed description of the question: Select two lattices $\Lambda_1$ and $\Lambda_2$ (here a lattice=additive free abelian group without accumulation points) of maximal rank two in the real plane. We normalize the lattices in such a way that their fundamental regions have area $=1$. Let $B(r)$ be a disk of radius $r$ centered at the origin. Let $N_1,N_2>0$ be two parameters. Select one non-zero vector $\vec{v}_1$ from $B(N_1)\cap \Lambda_1$ and another $\vec{v}_2$ from $B(N_2)\cap \Lambda_2$. Consider the area of the parallelogram with sides $\vec{v}_i,i=1,2.$ Let $D(N_1,N_2)$ be the minimum area of such parallelograms. We want to make the function $D(N_1,N_2)$ as large as possible by selecting the lattices $\Lambda_1$ and $\Lambda_2$ in a clever way. The simplest form of my question is: what can we say about the asymptotic behavior of $D(N_1,N_2)$ as the parameters $N_1,N_2\to\infty$ (either separately or jointly)? As additional questions I would like to ask: Does this ring a bell? What kind of methods have been used in studying this problem? What methods would you use?

What we know about the problem: The pigeon hole principle tells us that there is a constant $K$ (may depend on the lattices, but not on the scale parameters $N_1, N_2$) such that $$ D(N_1,L_{2,\min})\le \frac{K}{N_1}, $$ where $L_{2,\min}$ is the length of the shortest vector of $\Lambda_2$. The proof is easy. Fix a short non-zero vector $\vec{v}_2$ from $\Lambda_2$. The orthogonal projections of the vectors from $\Lambda_1\cap B(N_1/2)$ on to the line $\perp\vec{v}_2$ have lengths $\le N_1/2$. As there are $O(N_1^2)$ points in $\Lambda_1\cap B(N_1/2)$, some of those projections are within $O(1/N_1)$ of each other. Let $\vec{v}_1$ be the difference vector. As it is the difference between two vectors of length $\le N_1/2$, we have $\vec{v}_1\le N_1$.

selecting a pair of lattice points with nearby projections

In the above image the short arrow is the vector $\vec{v}_2$. The black line is the subspace generated by it, and the grey line is the orthogonal complement. The dots mark the points of the lattice $\Lambda_1$. Two points with nearby projections have been singled out by using larger dots. As is their difference vector $\vec{v}_1$, whose projection is then predictably very short. I also drew another copy (= the other arrow) of $\vec{v}_2$ to begin at the endpoint of $\vec{v}_1$. The two copies of $\vec{v}_2$ are a pair of parallel edges of the parallelogram. The small parallelogram has base $=|\vec{v}_2|$ and height = the length of the projection of $\vec{v}_1$ on the grey line. When drawing this image, I was selecting the lattice points from a square shaped region as opposed to a disk. The effects of that change can be absorbed into the constant $K$, and can IMHO be ignored.

Using real quadratic number fields and Liouville's theorem it is easy to construct lattices $\Lambda_1$ and $\Lambda_2$ in such a way that we always have $$ D(N_1,N_2)\ge \frac{K'}{N_1N_2}. $$ In this sense the pigeon hole bound from the previous paragraph has the correct order, when $N_2=L_{2,\min}$. In light of this I think a natural main question would be

Is their a known upper bound $D(N_1,N_2)\le f(N_1,N_2)$ where both $N_1$ and $N_2$ appear in an essential role?

With the word essential I mean that a bound should decrease, when we increase either $N_1$ or $N_2$. For example, in the above application of the pigeon hole principle we can trivially switch their roles to get a symmetric function $D(N_1,N_2)\le K\max\{N_1,N_2\}^{-1}$, but this is not useful :-)

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.