Does every connected Lie group act faithfully on a countable set? In other words: is every Lie group a subgroup of \(\mathrm{Sym}(\mathbb{N})\)? This question is sometimes called *Ulam’s problem* and there is recent progress in a paper of Nicolas Monod. Monod proves that every nilpotent connected Lie group acts faithfully on a countable set. It is a really nice paper, which also explains some background and sketches previously known results.

We will say that a group is *countably representable*, if it acts faithfully on a countable set. A group is countably representable if and only if it admits a decreasing sequence of subgroups of countable index which intersect trivially. Schreier-Ulam proved that \(\mathbb{R}\) is countably representable and Thomas proved that every *linear* Lie group is countably representable.

I think that these results are quite stunning! How can a connected Lie group act on a countable set? A central observation is that the field of complex numbers \(\mathbb{C}\) is isomorphic to a field of Puiseux series and thus admits a “nice” valuation \(\nu\colon \mathbb{C} \to \mathbb{Q}\) such that the valuation ring \(O = \{ z \in \mathbb{C} \mid \nu(z) \geq 0\}\) has the property that the ideals \(I_n = \{z \in \mathbb{C} \mid \nu(z) > n\}\) are of countable index in \(O\). This already suffices to see that \(\mathbb{C}\) is countably representable since the subgroups \(O \supseteq I_1 \supseteq I_2 \supseteq I_3 \dots\) are of countable index in \(\mathbb{C}\) and intersect trivially. To obtain the result for linear Lie groups it is sufficient to consider the groups \(\mathrm{SL}_n(\mathbb{C})\) — indeed, every subgroup of a countably representable group is countably representable. Here a similar argument works using the congruence subgroups in \(\mathrm{SL}_n(O)\), i.e. the kernels of the reduction maps \(\mathrm{SL}_n(O) \to \mathrm{SL}_n(O/I_n)\).

In general Ulam’s problem is still open. For example, it is unknown if the universal covering of \(\mathrm{SL}_n(\mathbb{R})\) is countably representable.