# Size of reduced bases of orthogonal lattice

I consider the following setting. Let $$L$$ be a lattice of rank $$d$$ in $$\mathbb{Z}^m$$ ($$d\leq m$$). The orthogonal lattice of $$L$$, denoted by $$L^{\perp}$$, is defined as the intersection of the orthogonal to the $$\mathbb{Q}$$-vector space generated by $$L$$ with $$\mathbb{Z}^m$$.

My question is in the following context: if I know that the determinant of $$L$$ is bounded from above, say $$\det(L) \leq B$$ for some $$B$$, what can I then say about the successive minima of the lattice $$L^{\perp}$$?

Some papers I came across suggest that $$L^{\perp}$$ has a reduced basis consisting of very short vectors compared to a reduced basis of $$L$$, and thus one expects that $$L^{\perp}$$ has a reduced basis consisting of vectors with norm around $$\sqrt{m-d} \det(L^{\perp})^{1/(m-d)}$$. (see for instance page 36 between Theorem 1 and 2 in the paper http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.90.7518&rep=rep1&type=pdf).

Concerning this heuristic, I would like to know how strong thedge the Gaussian heuristic applies to the length of the shortest non-zero vector and not to an entire basis of reduced vectors. Is there a generalised heuristic to explain this?

• The PDF you linked has only 16 pages but you referenced page 36. – AleksanderRas Dec 6 '18 at 10:19
• I am referring to the page numbers in the headings of the paper. The heuristic I mention is described between Theorem 1 and 2. – Luca Notarnicola Dec 6 '18 at 10:32
• Let $u = (1, 0, k)$ and $v = (0, 1, 0)$, where $k \in \mathbb{Z}$ is a variable. Then let $L = \{x_1 u + x_2 v : x_1, x_2 \in \mathbb{Z} \}$. We can see that the shortest vector of $L$ is $v$, however, the one of $L^\bot$ has length at least $k$, which we can make as big as we want. Thus, we can not relate the successive minima of $L$ and $L^\bot$ in general. So, the answer to your first question would probably be "nothing" (?). Don't you have a distribution or some bound to the entries of the vectors defining $L$? – Hilder Vítor Lima Pereira Dec 8 '18 at 18:49
• And I didn't understand your second question :s – Hilder Vítor Lima Pereira Dec 8 '18 at 18:52