On successive minima and basis of a lattice

Question

Let $\Lambda$ be a lattice in $\mathbb{R}^n$. For $\bar{x} \in \mathbb{R}^n$, let $\| \bar{x} \| = max_{1 \leq i \leq n} \{ |x_i| \}$, i.e. the sup norm. Let $\lambda_1, ..., \lambda_n$ be a successive minima of $\Lambda$ with respect to the sup norm. Let $\lambda_{j} \leq U < \lambda_{j+1}$.

If we consider the set $S = \{ \bar{x} \in \Lambda : \| \bar{x} \| < U \}$, then by the definition of successive minima it follows that every point in $S$ must be a linear combination of $v_1, ..., v_j$. Since points achieving successive minima aren't necessarily a basis for the lattice, there can be points in $S$ where the coefficients in front of $v_i$'s are not integers. I am sure there is an example of a lattice $\Gamma$, where this happens, but I could not come up with one and I am interested in seeing an actual example. I would appreciate any assistance/comments! Thank you!

PS This question is related to my previous question, which was resolved Counting number of points in a lattice with bounded sup norm

Answer 1

You can read about these things in Siegel-Chandrasekharan: Lectures on the geometry of numbers (Springer, 1988). In particular, Section X.5 contains an example for $n=3$ where the minimizing vectors do not form a basis of the lattice. The same section and the next one also explain that for practical purposes the successive minima can be replaced by slightly larger quantities such that the corresponding vectors do form a lattice basis.

Answer 2

In dimension 3 or 4, each lattice must have a Minkowski reduced basis. Thus, you won't find any example in these dimensions. But for dimension 5 or higher, there are lattices which don't have any Minkowski bases. One such example is the lattice $C_n$, which is generated by $e_1, \ldots, e_n$ and $(e_1 + \cdots e_n)/2$, where $e_1, \ldots, e_n$ are the standard basis vectors for $\mathbb R^n$.

You may read more in Martinet's book "Perfect lattices in Euclidean spaces".

Answer 3

I was thinking about the same question and just ran into this thread. As I was unable to find any source for this answer, I think it might benefit others to provide a construction here.

Consider the 3-dimensional lattice in $\mathbb{R}^4$ generated by $(1,1,1,1),(1,-1,1,-1), \frac{1}{2}(3, 1, 1, -1)$. Then the vectors achieving the successive minima w.r.t. the sup norm are $v_1=(1,1,1,1), v_2=(1,-1,1,-1)$ and $v_3=(1,1,-1,-1)$.

Now for every $(a,b,c)\in \langle(1,0,0),(0,1,0),(0,0,1),\frac{1}{2}(1,1,1)\rangle$, if $a,b,c>0$ then we see that the first coordinate of $av_1+bv_2+cv_3$ is $a+b+c\geq 3\cdot \frac{1}{2}$. By a similar argument for all other sign patterns, we can show that $||av_1+bv_2+cv_3||_{\infty}\geq \frac{3}{2}$ when $a,b,c$ are all nonzero. Then it pretty much follows that the three vectors are indeed the minimzers.