I was wondering how random unit lattices in number fields are. To make this more precise:

If $K$ is a number field with embeddings $\sigma_1, \dots, \sigma_n, \overline{\sigma_{r+1}}, \dots, \overline{\sigma_n} \to \mathbb{C}$ (so we have $r$ real embeddings and $2 (n - r)$ complex embeddings), let $\mathcal{O}_K$ be the ring of integers and $\Lambda_K := \{ (\log |\sigma_1(\varepsilon)|^{d_1}, \dots, \log |\sigma_n(\varepsilon)|^{d_n}) \mid \varepsilon \in \mathcal{O}_K^\ast \}$ be the unit lattice, where $d_i = 1$ if $\sigma_i(K) \subseteq \mathbb{R}$ and $d_i = 2$ otherwise.

Then $\Lambda_K$ is always contained in $H := \{ (x_1, \dots, x_n) \in \mathbb{R}^n \mid \sum_{i=1}^n x_i = 0 \}$, and $\det \Lambda_K$ is the regulator $R_K$ of $K$. Let us normalize $\Lambda_K$ by $\hat{\Lambda}_K := \frac{1}{\sqrt[n]{R_K}} \Lambda_K$; then $\det \hat{\Lambda}_K = 1$.

Now my question is: can we say something on how random the lattices $\hat{\Lambda}_K$ are among all lattices in $H$ of determinant 1? (For example, for fixed signature $(r, n-r)$ of $K$.)

Since these lattices are not completely random (they consist of vectors of logarithms of algebraic numbers), it is maybe better to ask something like this:

  • Given $\varepsilon > 0$ and a lattice $\Lambda \subseteq H$ with determinant 1, does there exists a number field $K$ of signature $(r, n - r)$ such that there is a basis $(v_i)_i$ of $\hat{\Lambda}_K$ and a basis $(w_i)_i$ of $\Lambda$ such that $\|v_i - w_i\| < \varepsilon$ for all $i$?

  • And if this exists, can one bound the discriminant of $K$ (or any other invariant of $K$) in terms of $\varepsilon$?

(Of course, this question is only interesting when $n > 2$.)

I assume that this is a very hard problem, so I'd be happy about any hint on whether something about this is known, whether someone is working on this, how one could proof such things, etc.

  • $\begingroup$ What's non-interesting about $n=2$? I was going to comment that much was known, at least in terms of continued fraction expansions, about fundamental units of real quadratic fields, enough that one could likely probabilistically differentiate between such a lattice and a random one. $\endgroup$ Oct 10, 2010 at 22:58
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    $\begingroup$ In case $n = 2$, $\dim H = 1$, and $\hat{\Lambda}_K = \mathbb{Z}$ due to the normalization. So the question will only be interesting for $n = 2$ if we consider $\Lambda_K$ itself :) $\endgroup$
    – felix
    Oct 10, 2010 at 23:09
  • $\begingroup$ The same question occurred to me a while back. I couldn't find anything in the literature but I didn't do an exhaustive search. Even the totally real cubic case seems very interesting. $\endgroup$ Oct 10, 2010 at 23:20
  • $\begingroup$ Aha, I had forgotten about your normalization. Fair enough. $\endgroup$ Oct 11, 2010 at 0:26
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    $\begingroup$ Nice question! $ $ $\endgroup$
    – JBorger
    Oct 11, 2010 at 6:21

1 Answer 1


This question was certainly discussed over past years, with no proven results though (as far as I am aware). I learned it from M. Gromov about 15 years ago (probably after he discussed it with G. Margulis). Here how I would formulate it:

Let us fix the signature $(r, n-r)$ (for example, $(3,0)$ for totaly real cubic fields -- the simplest non-trivial example which I researched numerically to some extend). Let $F_{(r,n-r)}$ be the set of all such number fields. For a field $K\in F_{(r,n-r)}$, consider the unit lattice $\mathcal{O}_K^*$, and its normalized (by the volume) logarithmic embedding $\hat{\Lambda}_K\subset \mathbb{R}^n$ as above.

Hence we obtain for each field of fixed signature, a unimodular lattice in $ \mathbb{R}^{n-1}$. We are interested in the type of such a lattice. It is reasonable to consider lattices up to isometries of the ambient $\mathbb{R}^{n-1}$.

Hence we obtain for each field $K\in F_{(r,n-r)}$, a point $x_K$ in the the moduli space of unimodilar lattices in $ \mathbb{R}^{n-1}$ up to an isometry. I call this the conformal type of the unit lattice of the field.

This moduli space is the familiar space $X_{n-1}=SL(n-1,\mathbb{Z})\setminus SL(n-1,\mathbb{R})/SO(n-1)$. For $n=3$ this is the modular curve. The space $X_{n-1}$ has a distinctive probability measure $\mu$ (coming from the right invariant measure on $SL(n-1,\mathbb{Z})\setminus SL(n-1,\mathbb{R})$; for $n=3$ this is the (volume one) hyperbolic measure on the modular curve).

The natural question would be what is the behavior of the set of points $x_K\in X_{n-1}$, $K\in F_{(r,n-r)}$ with respect to the measure $\mu$, or the geometry of $X_{n-1}$ (recall that the space $X_{n-1}$ is not compact, and there are cusps).

To formulate conjectures/questions, one need to introduce an order on the set of fields of fixed signature. I am aware of $3$ (probably inequivalent) natural orderings:

Arithmetic order: order the the set $F_{(r,n-r)}$ by the discriminant $d_K$ of the field.

Geometric order: order the the set $F_{(r,n-r)}$ by the regulator $R_K$ of the field.

Dynamical order: order the the set $F_{(r,n-r)}$ by the shortest unit $\epsilon_K$ of the field.

Numerical experiments (with tables provided by PARI) suggest the following conjecture:

Conjecture: The set of points $x_K$, $K\in F_{(r,n-r)}$ becomes equidistributed in $X_{n-1}$ with respect to $\mu$ when $F_{(r,n-r)}$ is ordered arithmetically.

As far as I understand, Margulis expects that when ordered dynamically, points in $F_{(r,n-r)}$ escape to infinity (i.e., to the cusp) with the probability $1$ (certainly some points may stay low, e.g., coming from Galois fields). Probably one should expect the same with respect to the geometric ordering (i.e., by the regulator). It is very difficult to have numerical data for these orderings.

The question about density of points $x_K$ in $X_{n-1}$ would follow from equidistribution, but of course is somewhat separate. In particular even if Margulis is right, this would not mean that $x_K$'s are not dense (its quite possible that this is still true). I do not know about (effective) approximation of a lattice by the unit lattice.

I also would like to mention that the additive analog of this question for totally real cubic fields (ordered by the discriminant) was solved by D. Terr (PhD, Berkeley, 1997, unpublished).

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    $\begingroup$ Terr's result is for general cubic fields. The rings of integers of totally real cubic fields (i.e. cyclic cubic fields) all have the same hexagonal "shape". It is the S3-cubic fields whose rings of integers have equidistributed lattices. One can still say that the "shapes" of the rings of integers in a general cubic field are equidistributed since luckily the C3-cubic fields contribute an insignificant amount to the total count of cubic fields. $\endgroup$
    – Rob Harron
    Feb 21, 2011 at 16:38

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