Recall that the $i$-th successive minimum of $L\in \mathcal L$ (space of full rank lattices in $\mathbb R^d$), denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.

Let $\Lambda$ be a unimodular lattice in $\mathbb R^d$. For $d\le 4$, it is possible to construct a basis $\{v_1,\dots, v_d\}$ of $\Lambda$ such that $\lambda_i(\Lambda)=\|v_i\|$ for any $i$.

If $d>4$, then the example $\frac{1}{2^{1/d}}L_0$, where $L_0:=Span_{\mathbb Z} \{e_1,\dots, e_{d-1},\frac{1}{2}(e_1+\cdots+e_d) \}$ shows it may be impossible to find such a basis. The coefficients $\frac{1}{2^{1/d}}$ is used to scale $L_0$ to a unimodular lattice.

It seems to me that such examples are really rare, but how rare?

Consider the set $S$ of unimodular lattices whose sucesssive minima cannot be attained by a basis of lattice. Recall the space of unimodular lattices are be identified with the homogeneous space $SL(d,\mathbb R)/SL(d,\mathbb Z)$. Can we see that $S$ is contained in (countable union) of submanifold/subvariety of dimension strictly lower than the full dimension? Or show it has Haar measure zero?

If it does not have measure zero, can we give an estimate of its measure?

(1) To begin with can anyone construct any examples different from this $\frac{1}{2^{1/d}}L_0$? (As Alison commented below small perturbation of this $\frac{1}{2^{1/d}}L_0$ example may work)

(2) Can any one construct any examples significantly different from this $\frac{1}{2^{1/d}}L_0$. For example, with $\lambda_1(\Lambda)$ close to zero? Such lattices are often called "unbounded".

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    $\begingroup$ I'd expect that small perturbations of your example of $2^{-1/d}L_0$ also fail to be spanned by their shortest vectors. $\endgroup$ Nov 8, 2022 at 0:08
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    $\begingroup$ @AlisonMiller Right, that is also what I am thinking but it is hard to see the set of those perturbations could be of "full dimension" or "positive measure". $\endgroup$
    – taylor
    Nov 8, 2022 at 0:19
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    $\begingroup$ It's an open neighborhood of a point in the homogeneous space, so I'd expect it to have positive measure! $\endgroup$ Nov 8, 2022 at 0:41
  • $\begingroup$ @AlisonMiller , obviously it cannot hold that both of us are right, as you will have two open sets. My idea is simply - given a short basis which spaces the lattice, `small’’ unipotent elements (smaller than the minima, ratios, etc), should still gives a short basis. And enough short unipotents form an open set. For the ``bad’’ lattice, how do these perturbation idea work for a small Cartan element? It will be a statement in diophantine approximation… $\endgroup$
    – Asaf
    Nov 11, 2022 at 16:09
  • $\begingroup$ @Asaf I agree they can't both be open -- but I'm claiming a weaker statement: that if $e'_1, \dotsc, e'_d$ is a small perturbation of the standard basis $e_1, \dotsc, e_d$, then the lattice ${\mathrm Span}_{\mathbb Z} \{e_1,\dots, e_{d-1},\frac{1}{2}(e_1+\cdots+e_d)$.will still have successive minima given by $\|e_1\|' \dotsc, \|e_d\|'$ in some order. I haven't thought too hard about the boundary of this set but I expect it to occur where there are multiple choices of sets of vectors representing the successive minima, where one choice gives a basis but the other doesn't. $\endgroup$ Nov 11, 2022 at 19:05

2 Answers 2


Edit 2 - Definitely the set of lattices spanned by their shortest vectors is of positive measure. For each such lattice $\Lambda \in SL_{n}(\mathbb{R})/SL_{n}(\mathbb{Z})$, there exists $\epsilon=\epsilon(\Lambda)>0$ such that $G_{\epsilon}.\Lambda$ consists of lattices with such a short basis (where $G_{\epsilon}$ is appropriate small neighborhood of unity). Hence it is of positive measure. This also shows that the technique described below is specialized to well-rounded lattices and cannot be improved.

Edit - the answer below is about well-rounded lattices, which are particular kind of lattices spanned by their shortest vectors...

It is well know by McMullen's work that that any compact $A$ orbit (where $A$ is the split Cartan of $SL_{n}(\mathbb{R})$) contains a well-rounded lattice. This has been generalized by Weiss and Shapira for every closed $A$ orbit.

The following holds for generic well-rounded lattices -

For the set of generic well-rounded lattices, one can show that for each such lattice, there exists an open neighborhood such that the intersection of the open neighborhood with the set of well-rounded lattices is a submanifold (actually subvarity) of positive co-dimension. Picking $\epsilon>0$, throwing away the elements high up in the cusp by throwing out a whole cuspidal region (say of measure $<\epsilon/2$), you end up with a precompact subset of the generic well-rounded lattices. Using compactness and the theorem I just mentioned, this set is covered by a finite amount of submanifolds (actually, subvarities) of proper codimension. Hence of zero Haar measure. This shows that the set of generic well-rounded lattices is of zero measure as $\epsilon$ was arbitrary.

One can probably push this strategy further (by taking a sequence of shrinking cuspidal regions) to show that one can cover the generic well-rounded lattices by a union of submanifolds of proper co-dimension, hence also the Hausdorff dimension of them is not full.

I need to think about the answer regarding non-generic well rounded lattices, but they must be rather special...

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    $\begingroup$ It looks like you're answering a slightly different question from the one that was asked here: as I understand it, "well-rounded" lattices have a basis of vectors which all have length equal to the minimal length of the lattice, while the question is allowing all successive minima of the lattice (which need not be equal) as lengths. $\endgroup$ Nov 7, 2022 at 18:12
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    $\begingroup$ @AlisonMiller , you are right, thank you, I clarified that in an edit... $\endgroup$
    – Asaf
    Nov 7, 2022 at 18:50

To begin with can anyone construct any examples different from this $(1/\sqrt[d]{2})L_0$?

My (somewhat imperfect) undertsanding of this is that the answer to this question is no. Questions related to this are studied in various works by Martinet (and others), see for example this and this, though this second paper is more relevant.

The setup is as follows. One starts with a lattice $\Lambda$. Then one considers its Minkowskian Sublattice $\Lambda'$, which has as its basis some set of linearly independent representatives of length $\lambda_i(\Lambda)$. One can then consider the quotient $\Lambda / \Lambda'$, and try to classify possible pairs $(\Lambda, \Lambda')$. This is done by letting $d\mathbb{Z}$ be the annihilator of $\Lambda/\Lambda'$, and then writing

$$\Lambda = \mathsf{span}_{\mathbb{Z}}\langle \Lambda, f_1,\dots, f_k\rangle$$

for $f_i = \frac{\sum_i \alpha_i^{(j)} e_i}{d}$, and $\alpha^{(j)} = (\alpha_1^{(j)},\dots, \alpha_n^{(j)})\in(\mathbb{Z}/d\mathbb{Z})^n$ certain words of a $\mathbb{Z}/d\mathbb{Z}$-linear code. In this way, the example of $L_0$ corresponds to applying construction A to the (binary) repetition code $[1,1,\dots,1]$. But one can obtain other pairs $(\Lambda, \Lambda')$ (starting in dimension 9) by instead lifting other codes. The second link classifies the codes that can appear (in small dimensions). Each code leads to a (somewhat) different analogue of $L_0$, though the linked work does not contain unbounded lattices. Instead it contains the comment

It should be noted that the two mentioned applications make use only of results for well rounded lattices, that is, for lattices with minimal vectors spanning $E$. In other words, these lattices have equal successive minima $\lambda_1(L),\dots, \lambda_n(L)$. Indeed, a deformation argument (see [Mar01, Theorem 1.5]) shows that all codes can be realized using well rounded lattices.

As their primary goal is to classify the codes which can arise from the above construction, they reduce their analysis to the case of well-rounded lattices (which are not unbounded in your sense).

None of this is useful for your overall of quantifying how "large" this set of lattices is, but may be useful for additional explicit examples of analogues of $L_0$.


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