Lie groups as permutation groups: Ulam’s problem in the nilpotent case

A group is countably representable if it admits a faithful action on some countable set.

Proposition 1 (Schreier–Ulam)

The group 𝐑 is countably representable.

Problem (Ulam)

Is every connected Lie group countably representable?

If the answer to Ulam’s problem is affirmative, then every locally compact second countable group is countably representable.

Proposition 3 (de Bruijn [6])

Every abelian group is countably representable as long as its cardinality does not exceed 𝔠.

Theorem 4 (Thomas [29])

Every linear Lie group is countably representable.

Every nilpotent connected Lie group is countably representable.

Any homomorphism from McKenzie’s or Churkin’s nilpotent group to any separable Hausdorff topological group is trivial on the centre.

The same statement holds for homomorphisms to Lindelöf Hausdorff topological groups.

Let G = SL n ⁢ ( R ) , and let G ~ be its universal cover. Is G ~ countably representable?

Ulam did not explicitly assume connectedness, but this is only a matter of terminology. Under the convention that Lie groups are second countable, they have at most countably many connected components, and hence we are immediately reduced to the connected case (Lemma 11 below).

If on the other hand no such restriction is made, then we can trivially obtain disconnected Lie groups that are not countably representable by considering a suitably large group as zero-dimensional Lie group.

The class of countably representable groups is closed under finite and countable products.∎

The class of countably representable groups is closed under taking inverse limits of countable inverse systems.∎

Proof of Proposition 1

By Lemma 8, the countable power Q N is countably representable. Since this group is uniquely divisible, it is a 𝐐-vector space. Its dimension is 𝔠, and therefore it is isomorphic to 𝐑. ∎

The circle group S 1 is countably representable.

Proof

Viewing 𝐑 again as a 𝐐-vector space, we split off 𝐐 and obtain an isomorphism R / Z ≅ Q / Z ⊕ R ; the result follows from Proposition 1 (and Lemma 8). ∎

If 𝐺 admits a countably representable cocountable subgroup, then 𝐺 is itself countably representable.

Proof

Suppose that the group 𝐺 admits a cocountable subgroup G 0 < G which acts faithfully on a countable set 𝑋. Define Y = ( G × X ) / G 0 , where G 0 acts diagonally on G × X , from the left on 𝐺. Then 𝑌 is countable since it admits a bijection with ( G 0 \ G ) × X . We endow 𝑌 with a 𝐺-action by letting 𝐺 act by right multiplication on 𝐺 and trivially on 𝑌, noting that this action on G × X descends indeed to 𝑌. The resulting action is faithful. ∎

The congruence subgroups SL m ⁢ ( V C ; L n ) are cocountable in SL m ⁢ ( C ) .

Proof

As recalled above, SL m ⁢ ( V C ) is cocountable in SL m ⁢ ( C ) by [29, Theorem 2.5]. Therefore, it suffices to show that SL m ⁢ ( V C ; L n ) is in turn cocountable in SL m ⁢ ( V C ) , i.e. that the ring V C / L n is countable. This is the case because any given element of this ring is represented by a finite sum

for some choice of denominator q ∈ N > 0 and elements λ j ∈ Q ¯ indexed by j ∈ N ; here 𝑡 is the formal variable of Puiseux series. ∎

Let C < C be a finitely generated subgroup, and let z ∈ C be an element not in 𝐶. Then there is some 𝑛 such that L n does not meet the coset z + C .

Proof

Viewing everything in 𝐾 rather than in 𝐂, the claim is that 𝑣 is uniformly bounded over z + C ; note that this coset does not contain 0. Letting 𝐷 be the group generated by 𝐶 and 𝑧, it suffices to prove the following claim: for every finitely generated subgroup D < K , the valuation 𝑣 is uniformly bounded over the non-zero elements of 𝐷.

The group 𝐷 is free abelian of finite rank r ≥ 1 . We shall prove by induction on 𝑟 that 𝑣 takes at most r + 1 distinct values on 𝐷. The base case r = 1 holds because v ⁢ ( p ⁢ x ) = v ⁢ ( x ) for all x ∈ K and all non-zero p ∈ Z ; this gives the two values v ⁢ ( x ) and v ⁢ ( 0 ) = ∞ on 𝐷 in rank one.

We turn to the induction step for r ≥ 2 . Let x 1 , … , x r be a basis of 𝐷. We order the basis so that, for some 1 ≤ s ≤ r , we have

Given any element x = p 1 ⁢ x 1 + ⋯ + p r ⁢ x r in 𝐷 (with p i ∈ Z ), write

and consider its residue ω ∈ Q ¯ at v ⁢ ( x 1 ) , i.e. the coefficient of t v ⁢ ( x 1 ) in x ′ . We have v ⁢ ( x ) = v ⁢ ( x ′ ) = v ⁢ ( x 1 ) unless 𝜔 vanishes. However, ω = 0 defines a subgroup D 0 of 𝐷; this subgroup is still free abelian but of smaller rank since x ↦ ω defines a non-trivial torsion-free quotient of 𝐷. Therefore, the induction hypothesis limits at 𝑟 the number of distinct values on D 0 , which completes the induction step once we add the value v ⁢ ( x 1 ) . ∎

End of proof of Theorem 5

Let 𝐺 be a nilpotent connected Lie group. We refer to [15, § 11.2] (especially [15, Theorem 11.2.10]) for background on the structure of 𝐺. We recall in particular the following. The fundamental group of 𝐺 is a free abelian group of finite rank d ≥ 0 . Moreover, denoting by G ~ the universal cover of 𝐺, there is a discrete central subgroup Γ < G ~ isomorphic to π 1 ⁢ ( G ) with G ~ → G ~ / Γ ≅ G implementing the covering morphism. In addition, Γ is a lattice in a central subgroup Z ≅ R d of G ~ so that T = Z / Γ is a central 𝑑-dimensional torus in 𝐺.

Our proof is by induction on 𝑑. The base case d = 0 holds by Thomas’s result since 𝐺 is then simply connected and hence linear [15, § 16.2.7].

The case d = 1 is the main case because the general case is easily reduced to it; hence we treat it separately. Since 𝑍 is connected, the quotient H = G / T ≅ G ~ / Z is simply connected. Therefore, by the case d = 0 , there is a countable family of cocountable subgroups H n < H such that every non-trivial element of 𝐻 is outside some H n . Taking pre-images gives cocountable subgroups G n < G such that every element of 𝐺not in T lies outside G n for some 𝑛. It therefore suffices to produce another sequence of cocountable subgroups K n < G such that every non-trivial element of 𝑇 lies outside some K n .

Since G ~ is simply connected, it is linear; more precisely, by the Lie–Kolchin theorem, we can identify it with a Lie subgroup of the upper unitriangular subgroup U m of SL m ⁢ ( R ) for some 𝑚. We define K ^ n < G ~ by taking the intersection of G ~ with the congruence subgroup SL m ⁢ ( V C ; L n ) under such an embedding; thus K ^ n is cocountable in G ~ by Lemma 12. Let K n < G be the image of K ^ n in 𝐺, which is cocountable.

Let us reformulate for K ^ n < G ~ the desired property of K n : we need to show that every element ζ ∈ Z < G ~ which is not in Γ lies outside the group Γ ⁢ K ^ n for some 𝑛. Indeed, Γ ⁢ K ^ n is the pre-image of K n in G ~ . Equivalently, we need to prove that ζ ⁢ Γ ∩ K ^ n = ∅ .

The lower central series U m i of U m is defined by U m 0 = U m , U m i + 1 = [ U m i , U m ] . Concretely, the ( p , q ) -entry of a matrix in U m i vanishes when p < q ≤ p + i . Consider the largest 𝑖 such that ζ ∈ U m i ; since 𝜁 is non-trivial, we have i ≤ m - 2 . Moreover, the evaluation map ϵ : U m → R at a ( p , p + i + 1 ) -entry is a group homomorphism on U m i , namely one of the canonical coordinates of

By the maximality of 𝑖, we can choose 𝑝 such that ϵ ⁢ ( ζ ) is non-zero. Recalling that 𝑍 has dimension d = 1 , it follows that 𝜖 restricts to an isomorphism ϵ | Z : Z → R . In conclusion, z = ϵ ⁢ ( ζ ) lies outside C = ϵ ⁢ ( Γ ) . By Lemma 13, there is some 𝑛 such that z + C does not meet L n . This implies that ζ ⁢ Γ does not meet K ^ n , as was to be shown.

Finally, we perform the induction step for d ≥ 2 . We choose two distinct elements γ 1 , γ 2 of some basis of Γ, write Γ i = ⟨ γ i ⟩ for the group generated by γ i and Z i < Z for the one-parameter subgroup containing γ i . The quotient T i = Z i / Γ i is a central circle group in 𝐺, and thus we can form H i = G / T i . Then the fundamental group of H i is isomorphic to Γ / Γ i , e.g. because G ~ / Z i is a universal cover of H i since Z i is connected.

Thus the fundamental group of H i has rank d - 1 . By the induction hypothesis, there are two countable families H i , n of cocountable subgroups of H 1 respectively H 2 such that any non-trivial element of H i lies outside H i , n for some 𝑛. Taking pre-images, we obtain cocountable subgroups G i , n < G such that every g ∉ T i lies outside some G i , n . Since T 1 ∩ T 2 is trivial, this completes the induction step. ∎

Consider the question of whether the double cover G ~ of G = SL 3 ⁢ ( R ) is countably representable. Since 𝐺 is countably representable, it all amounts to finding a cocountable subgroup of G ~ which meets the kernel of the covering map trivially. Equivalently, the question is: Does 𝐺 itself contain a cocountable subgroup that can be lifted to G ~ ?

This cannot be proved by using, as above, a cocountable congruence subgroup G n = G ∩ SL 3 ⁢ ( V C ; L n ) . Indeed, we claim that G n does not lift to G ~ .

The reason is as follows. The set 1 + L n meets 𝐑 densely because L n ∩ R is a cocountable subgroup of 𝐑. Therefore, 1 + L n contains a negative real number 𝑥. It follows that G n contains the diagonal matrices with entries ( x , x - 1 , 1 ) and ( x , 1 , x - 1 ) (noting that 1 + L n is a multiplicative subgroup of V C × ). The classical relation between Steinberg symbols and central extensions of 𝐺, as explained e.g. in Milnor’s book [23, p. 95], shows that these two diagonal matrices do not admit commuting lifts in G ~ . This is because the Steinberg symbol for G ~ is the Hilbert quadratic residue symbol (see [23, p. 104]), but x < 0 means that the Hilbert symbol is non-trivial. This excludes a minori any lift of the group G n .

Proof of Proposition 2

Let 𝐺 be a locally compact second countable group. Denoting the neutral component of 𝐺 by G 0 , the quotient G / G 0 is locally compact and totally disconnected; therefore, van Dantzig’s theorem [34, p. 18] implies that G / G 0 contains a compact-open subgroup 𝐾. We denote by G 1 < G the pre-image of 𝐾 in 𝐺; thus G 1 / G 0 is compact.

Since 𝐺 is second countable, we can choose a countable base 𝒰 of neighbourhoods U ⊆ G 1 of the identity in G 1 . By the solution of Hilbert’s fifth problem (see e.g. [24, § 4.6]), there is a compact normal subgroup N U ⊲ G 1 contained in 𝑈 such that G 1 / N U is a Lie group. Note that G 1 / N U has finitely many connected components; therefore, since we assumed that the answer to Ulam’s problem is affirmative, G 1 / N U is countably representable.

On the other hand, G 1 is the inverse limit of the system ( G 1 / N U ) U ∈ U . Since 𝒰 is countable, Lemma 9 implies that G 1 is countably representable. Finally, we use again the fact that 𝐺 is second countable to deduce that G / G 1 is countable since G 1 is open in 𝐺. Applying Lemma 11, we conclude that 𝐺 itself is countably representable. ∎

Analytic proof of Proposition 3

Let 𝐺 be an abelian group with | G | ≤ c . We consider 𝐺 as a discrete group so that its Pontryagin dual G ^ = Hom ⁡ ( G , S 1 ) is a compact group. Then the topological weight of G ^ , namely the smallest cardinality of a base for its topology, is also ≤ c ; see [16, § 7.76]. It now follows from Engelking’s theorem [7, Theorem 10] that G ^ is separable; the fact that Engelking’s theorem applies to compact groups is due to the fact that they are dyadic spaces; see e.g. [16, § 10.40]. We can therefore choose a sequence of characters f n : G → S 1 that is dense in G ^ .

Since S 1 is countably representable by Corollary 10, there is a sequence of morphisms h m from S 1 to countable quotients h m ⁢ ( S 1 ) such that the intersection of the kernels of all h m is trivial. It now suffices to verify that, for every non-trivial g ∈ G , there is a pair ( n , m ) such that h m ⁢ ( f n ⁢ ( g ) ) is non-trivial.

To this end, Pontryagin duality ensures that there is a character f ∈ G ^ such that f ⁢ ( g ) is non-trivial. By density of ( f n ) , we can fix 𝑛 such that f n ⁢ ( g ) is non-trivial in S 1 , and it remains only to choose h m in terms of f n ⁢ ( g ) . ∎

Algebraic proof of Proposition 3

Let 𝐺 be an abelian group with | G | ≤ c . The injective hull 𝐷 of 𝐺 contains 𝐺, and its construction shows that 𝐷 still has size at most 𝔠 (see e.g. the proof of [10, III, Theorem 17]). It therefore suffices to consider divisible abelian groups such as 𝐷. By the structure theory of divisible abelian groups (see e.g. [9, Theorem 23.1]), 𝐷 is a sum of a 𝐐-vector space and of 𝑝-torsion parts, where 𝑝 ranges over all primes. It suffices to consider each piece individually because of Lemma 8. The 𝐐-vector space is handled by the argument of Proposition 1 since it has dimension at most 𝔠.

The 𝑝-torsion part is of the form ⨁ κ Z ⁢ ( p ∞ ) for some cardinal 𝜅, where Z ⁢ ( p ∞ ) is the Prüfer 𝑝-group (see again [9, Theorem 23.1]); upon enlarging it, we can assume κ = c = 2 ℵ 0 . It is therefore contained in Z ⁢ ( p ∞ ) ℵ 0 ; see [9, § 23, Example 3]. We conclude again by Lemma 8 since Z ⁢ ( p ∞ ) is countable. ∎

Theorem 15 (McKenzie, Churkin)

Every homomorphism from 𝐺 or from 𝑀 to Sym ⁢ ( ℵ 0 ) factors through 𝐴, respectively 𝐵.

Churkin’s proof of Theorem 15

Let 𝑋 be a countable set with a 𝐺-action. We need to show that 𝑐 fixes any given x ∈ X . Consider the map I → X , i ↦ a i ⁢ x . Since 𝐼 is uncountable, it contains an uncountable subset 𝐽 such that a i ⁢ x is constant over i ∈ J . For the same reason, there is an uncountable subset K ⊆ J such that b i ⁢ x is constant over i ∈ K . Choose three distinct indices i , j , k all in 𝐾. Since both a j - 1 ⁢ a i and b k - 1 ⁢ b i fix 𝑥, so does [ a j - 1 ⁢ a i , b k - 1 ⁢ b i ] . On the other hand, this commutator is 𝑐 because a j commutes with a i , b k , b i and b k commutes with a j , a i , b i . The same proof works for 𝑀. ∎

There exists a nilpotent group 𝐻 given by a central extension

which is not countably representable.∎

Proof of Theorem 6

We keep the notation introduced above for Churkin’s group 𝐺 and for its explicit construction as central extension 𝐸 (the proof for 𝑀 is the same).

We claim that if 𝐺 is covered by countably many left translates of some subset W ⊆ G , then the central generator 𝑐 is a commutator of two elements of the product set W - 1 ⁢ W .

Indeed, there is an uncountable subset J ⊆ I such that all a j with j ∈ J belong to a fixed translate of 𝑊, say r ⁢ W for some r ∈ G . Likewise, there is some infinite K ⊆ J and s ∈ G such that b k ∈ s ⁢ W for all k ∈ K . As in Theorem 15, we observe that c = [ a j - 1 ⁢ a i , b k - 1 ⁢ b i ] whenever i , j , k are distinct. Taking all three indices in 𝐾, both a j - 1 ⁢ a i and b k - 1 ⁢ b i belong to W - 1 ⁢ W , confirming the claim.

Consider now a group homomorphism π : G → H to some separable or Lindelöf topological group 𝐻. It suffices to show that π ⁢ ( c ) belongs to every neighbourhood U ⊆ H of the identity in 𝐻.

To that end, choose a symmetric neighbourhood 𝑉 of the identity in 𝐻 such that V 16 ⊆ U . There exists a sequence { h n } in 𝐻 such that { h n ⁢ V } covers 𝐻. Indeed, in the Lindelöf case, this holds by definition of that property. In the separable case, we can take any sequence { h n } that is dense in 𝐻 since then, for any h ∈ H , the sequence h n - 1 ⁢ h must meet 𝑉.

In particular, π ⁢ ( G ) is covered by { h n ⁢ V } , and upon extracting, we assume, moreover, that h n ⁢ V meets π ⁢ ( G ) for every 𝑛. We then choose g n ∈ G with π ⁢ ( g n ) ∈ h n ⁢ V . It now follows that

Thus π ⁢ ( G ) is covered by the sequence π ⁢ ( g n ) ⁢ V 2 , and we conclude that 𝐺 is covered by countably many left translates of π - 1 ⁢ ( V 2 ) .

We now apply the initial claim to W = π - 1 ⁢ ( V 2 ) and deduce that 𝑐 is a commutator of two elements of W - 1 ⁢ W = π - 1 ⁢ ( V 4 ) . Therefore, π ⁢ ( c ) is the product of four elements of V 4 and thus belongs indeed to V 16 ⊆ U , completing the proof. ∎

Our proof makes almost no distinction between the separable and the Lindelöf cases, but they are formally independent. On the one hand, there obviously are non-separable Lindelöf groups since there is no limit to the cardinality of a compact group (while separable regular spaces have cardinality at most 2 c ; see [12, a-3]). The reverse question is more delicate and was notably asked by Wilansky in 1968 for normal groups [28]. An example conditional on the continuum hypothesis was provided in [11], and then an unconditional example in [13].