# Do lattices of small co-volume always exist in rational, connected, simply connected, nilpotent Lie groups?

Given a connected, simply connected, rational, nilpotent Lie group $$G$$, is there a lattice of arbitrarily small co-volume in $$G$$? If $$G$$ is Carnot, the answer is "yes" by applying a dilatation to the "integer" lattice. If $$G$$ is not Carnot, I seem to be encountering two conflicting sources, one of which references the other as its justification.

At the start of section 2.3 of this paper, the claim is made that "a nilpotent Lie group need not have a lattice, and may only have lattices of large co-volume". This paper makes the running assumption that all Lie groups are connected, and simply connected. I am fine with the fact that a lattice may not exist.

For the claim that (if lattices do exist) such groups "may only have lattices of large co-volume" the reference may be found here.

However, in reading this referenced paper, I cannot seem to find the claim mentioned. In fact, it appears to me that the opposite is proved: In Theorem 2, it is claimed that if $$D \subseteq G$$ is a rational, discrete subgroup of a finite dimensional, nilpotent, simply connected, nilpotent Lie group $$G$$, then $$G$$ can be "approximated" by discrete subgroups $$D_m$$ containing $$D$$. Moreover, these discrete subgroups are constructed in such a way that their co-volume seems it should be quite small. The discrete group $$D_m$$ is generated by a set of elements $$\{d_1,\dots,d_n\}$$, and in the inductive step the newly constructed generator $$d_n$$ is of the form $$d_n = (\frac{1}{M},0,0,\dots,0)$$ where $$M \in \mathbb{Z}$$ can be made arbitrarily large. Accordingly, if this subgroup is a lattice, the co-volume should be able to be made arbitrarily small.

Under the assumption that we start with a nilpotent, connected, simply connected Lie group $$G$$ containing a lattice $$D$$, we know that $$G$$ must be rational by Malcev's correspondence. Therefore, by Theorem 2 of this second paper, any of the "approximating discrete subgroups" (which I believe may have small co-volume) ought also to be a lattice. After all, if $$D_m$$ were not a lattice, as a discrete subgroup of a nilpotent Lie group, $$D_m$$ would have to be contained in a proper connected subgroup of $$G$$. But then so would be $$D$$, since $$D \subseteq D_m$$, and this contradicts the fact that $$D$$ is a lattice.

It seems to me that I have misapplied or otherwise misunderstood the claim from the first paper (or perhaps I've just made some other illogical leaps along the way about lattices in nilpotent Lie groups). Hence the question: do lattices of small co-volume always exist in rational, connected, simply connected, nilpotent Lie groups?

References

“Separated Nets in Nilpotent Groups.” Indiana University Mathematics Journal, vol. 67, no. 3, 2018, pp. 1143–83. JSTOR, http://www.jstor.org/stable/45010323. Accessed 3 Nov. 2023.

On Everywhere Dense Imbedding of Free Groups in Lie Groups. Nagoya Mathematical Journal 2 (1951): 63-71.

• The paper you're quoting (arxiv.org/abs/1608.08572 for people who don't have billions to pay JSTOR) is mistaken. I wrote an answer with the simple proof.
– YCor
Commented Nov 3, 2023 at 7:29

Yes, in a simply connected nilpotent Lie group with a $$\mathbf{Q}$$-form (i.e. with a lattice), there exist lattices approximating the whole group. In particular, if the group has positive dimension, there are lattices of arbitrary small covolume.

To prove it, recall that in a Lie algebra, a strong Lie subring (see Definition A.2 in this paper) is a Lie subring that is stable, for each $$i$$, under $$(x_1,\dots,x_i)\mapsto \frac1{m_i}[x_1,\dots,x_i]$$, where $$m_i$$ is the least common denominator of the terms of degree $$i$$ in the Baker-Campbell-Hausdorff formula.

If $$G$$ is simply connected nilpotent Lie group and $$\mathfrak{g}$$ its Lie algebra, then for every strong Lie subring $$\mathfrak{h}\subset\mathfrak{g}$$, its exponential $$\exp(\mathfrak{h})$$ is a subgroup. It is a lattice if and $$\mathfrak{h}$$ is an (additive) lattice in $$\mathfrak{g}$$.

Now fix a rational structure on $$\mathfrak{g}$$. Then it is easy to check that for every lattice $$\Lambda$$ of $$\mathfrak{g}$$ contained in $$\mathfrak{g}_\mathbf{Q}$$, the strong Lie subring it generates is also a lattice. In particular, $$\mathfrak{g}$$ has strong subrings approximating $$\mathfrak{g}$$. Passing to the exponential, we obtain lattices in $$G$$ approximating $$G$$.

• Thank you so much. This is essentially what I had hoped/expected to be true, but the original paper made me second guess myself.
– Kyle
Commented Nov 3, 2023 at 13:00
• By the way, whenever there is a derivation of nonzero trace, one can obtain small covolume lattices from a single one by iterating an automorphism of determinant $<1$. This applies when there is a contracting automorphism (which is more common than being Carnot), and in other cases as well. But not always (cf. characteristically nilpotent Lie algebras).
– YCor
Commented Nov 3, 2023 at 13:21