# Lie groups acting on countable sets

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.