Maximal sublattice index in Minkowski's Second Theorem

Question

Let $B$ be a (small) convex compact set in $\mathbb{R}^n$, symmetric around the origin. Let $\Gamma$ be a lattice in $\mathbb{R}^n$ of dimension $n$ (I'm almost sure we can just take $\mathbb{Z}^n$, but it may not always be the most natural example).

Now, we dilate $B$ by simply multiplying it by a growing $\lambda$. We define the successive minima $\lambda_1,...,\lambda_n$ as the smallest $\lambda$ such that $\lambda\cdot B$ contains $i$ linearly independent non-zero elements of $\Gamma$. Define by $v_i$ the corresponding elements of $\Gamma$ (if there are multiple, we can choose any).

$v_1,...,v_n$ generate a sublattice of $\Gamma$, but not necessarily the whole lattice. In fact, Minkowski's Second Theorem states that the maximal index of that sublattice is $n!$. It is easy to give an example with index $2^n$ - simply take a cube and $\Gamma=\mathbb{Z}^{n-1}$ and choose the vertices of the cube with side 2 as $v_i$s. Another natural example is the 5-dimensional sphere with $\Gamma = \mathbb{Z}^5\cup(\mathbb{Z}^5+(0.5,...,0.5))$ - the "half-points" are further than the standard $\mathbb{Z}^5$ points, giving index 2.

My question is: what is an example of set $B$ (and lattice $\Gamma$) that has the index equal to $n!$, even for a specific $n>2$? Of course it would be best if the example was easily generalized to any dimension. Or if there are no such examples, what is the largest possible index?

Answer 1

For the case of $B$ a Euclidean sphere, the paper On the Index System of Well-Rounded Lattices is probably state-of-the-art (the well-rounded assumption is known to be WLOG). It investigates bounding the exact index you are interested in (it actually describes the set of isomorphism classes of the quotients $\Lambda / \Lambda'$, so somewhat more), and:

1. Has exact bounds for dimensions $\leq 9$

2. Has exact characterizations for root lattices, of which the maximal index (parametrized as a function of $n$) is given by $\mathbb{D}_n$, of index $\leq 2^{\lfloor (n-1)/2\rfloor}$ (and it achieves this index).

One can generically bound the index by $\gamma_n^{n/2}$ for $\gamma_n$ the Hermite constant, which is known to satisfy the bound $\gamma_n < 1 + \frac{n}{4}$, so $\gamma_n^{n/2} = O(n^n) = O(n!)$, as you mention.

The appendix of On Classifying Minkowskian Sublattices additionally mentions that the Leech lattice has a quotient of the form you are interested in of index $2^{24}$. This is all to say that for $B$ a Euclidean sphere, there appears to be a gap.

A somewhat similar question is, for an $n$-dimensional lattice $L$, comparing the products:

$$\frac{\prod_i\lVert \vec e_i\rVert_2^2}{\det L}$$

over independent vectors $\vec e_i$ (denoted $M(L)$) and basis vectors $\vec e_i$ (denoted $H_b(L)$), where $\det L$ is the determinant of the Gram matrix of $L$. In Hermite vs. Minkowski, Martinet shows that:

$$Q_b(L) = \frac{H_b(L)}{M(L)} \leq (5/4)^{n-4}$$

for $n\geq 4$. I don't see how this might immediately imply an index bound, but it does show that on average one only has to inflate the size of each minimal independent vector by a constant factor to get a basis, which is perhaps interesting to you.