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ONLY COUNTABLY MANY SIMPLY-CONNECTED LIE GROUPS ADMIT LATTICES

JÖRG WINKELMANN

A lattice in a locally compact topological group G is a discrete subgroup Γ ⊂ G such that the quotient space G/Γ admits a G-invariant probability measure. This includes all discrete subgroups Γ ⊂ G with compact quotient G/Γ. An important class of examples of lattices is given by arithmetic subgroups in linear-algebraic groups.

Lattices tend to strongly reflect the properties of the ambient group. For instance, if Γ_{i} (i = 1, 2) are lattices in simply-connected semisimple or nilpotent real Lie groups G_{i} and Γ_{1} ≃ Γ_{2}, then G_{1} ≃ G_{2}. Moreover, any lattice Γ in a complex linear-algebraic group G is dense in the Zariski-topology.

In this spirit it is natural to assume that are only countably many non-isomorphic Lie groups admitting lattices, because lattices in Lie groups are finitely presentable and evidently there are only countably many non-isomorphic finitely presentable groups. We prove that indeed there are only countably many simply-connected real or complex Lie groups containing lattices. Since there do exist continuous families of non-isomorphic simply-connected Lie groups, this underlines that the existence of a lattice is a rather special property.

Theorem. Let k = ℝ or k = ℂ. Then up to isomorphism as k-Lie groups there exist only countably many simply-connected k-Lie groups admitting lattices (i.e. discrete subgroups such that the quotient admits an invariant probability measure).

Remarks. – The assumption of the Lie group being simply-connected is necessary, since every compact complex torus is an example of a complex Lie group containing the trivial subgroup {e} as lattice.

– There do exist isomorphic lattices in non-isomorphic Lie groups. For instance, Γ = (ℤ^{3}, +) may be embedded as a lattice in both G_{1} = (ℝ^{3}, +) and

G_{2} = \{( \begin{matrix}1 & t \\ & 1 \\ & & cos t & sin t & x \\ & & - sin t & cos t & y \\ & & & & 1 \end{matrix} ) : t, x, y ∈ ℝ\}

.

1. Reduction to the real case

For a complex Lie group the existence of a lattice has nothing to do with the complex structure on G. If a complex Lie group G contains a lattice, so does every complex Lie group which is isomorphic to G as a topological group. Hence it is natural to start our considerations by considering the following question: Given

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a complex Lie group G, how many non-isomorphic complex Lie groups can there exist which are isomorphic to G as topological groups ?

Proposition 1. Let G be a simply-connected real Lie group. Then there exist only finitely many non-isomorphic complex Lie groups (H_{i})_{i∈I} which are isomorphic to G as topological groups (resp. real Lie groups).

Proof. We begin the proof by the following auxiliary result on direct summands of Lie algebras. (A Lie subalgebra h of a Lie algebra g is called a direct summand if there exists another Lie subalgebra i such that g = h ⊕ i.)

Lemma 1. Let k be a field and g a finite-dimensional k-Lie algebra.

Then there exist only finitely many direct summands up to isomorphism as Lie algebra.

Proof. Let d ∈ ℕ. The set of all d-dimensional Lie subalgebras of g carries the structure of a closed subvariety V of a Grassmann variety in a natural way. For every direct summand h there is a projection π : g → h with a kernel ker π = c. If i is a Lie subalgebra of g with dim h = dim i and i ∩ c = {0}, then π induces a Lie algebra isomorphism between h and i. Thus for every direct summand h there is an open neighbourhood U of [h] in V such that i ≃ h for every [i] ∈ U. Now the lemma follows from quasi-compactness of Zariski topology.

We now show that this lemma implies the proposition. By the assumption of simply-connectedness it suffices to prove the corresponding statement for Lie algebras. Let g_{i} be a family of complex Lie algebras such that the underlying real Lie algebras are all isomorphic to a given real Lie algebra h. In this case h⊗_{ℝ}ℂ ≃ g_{i}⊕\overline{g_{i}} for every i. In particular g_{i} is a direct summand of h ⊗_{ℝ} ℂ. Hence we may invoke the lemma in order to conclude the proof of the proposition.

2. Reduction to the solvable case

The reduction of the general case to the solvable case consists of two steps. First we show that if a Lie group admits a lattice, so does its radical. Second we show that there are only countably many non-isomorphic Lie groups with prescribed radical.

Proposition 2. Let G be a connected real Lie group with radical R. Assume that G contains a lattice.

Then R also contains a lattice.

Proof. Let Γ be a lattice in G. Let C denote the maximal normal connected compact semisimple Lie subgroup of G. The projection map π : G → G/C is proper, hence π(Γ) is a lattice in G/C. The intersection R∩C is finite, hence π restricts to a finite map from R onto π(R). By construction G/C fulfills the condition of Cor. 8.28 of [3] ^{1}. Hence π(Γ) ∩ π(R) is a lattice in π(R). It follows that R ∩ π^{−1}(π(Γ)) is a lattice in R.

Proposition 3. Let R be a solvable real Lie group. Then there exist only countably many non-isomorphic simply-connected real Lie groups with R as radical.

1The statement of Cor. 8.28 in [3] is correct, although the proof is not. See [4] for details.

Proof. If G is a simply-connected Lie group with radical isomorphic to R, then G is isomorphic to a semidirect product of a semisimple real Lie group S and the radical R which is given by a Lie group homomorphism ρ : S → AutR (Levi-Malcev-decomposition). Let U be a maximal connected semisimple Lie subgroup of AutR. We may assume ρ(S) ⊂ U, since the maximal connected semisimple Lie subgroups of a given Lie group are conjugate to one another. Now the proposition follows from the fact that there are only countably many non-isomorphic semisimple Lie groups and that there are only countably many non-conjugate Lie group homomorphisms between two given semisimple real Lie groups.

3. The real solvable case

For a simply-connected solvable real Lie group our theorem follows from the proposition below which in turn is based on a arithmeticity result going back to Mostow.

Proposition 4. Let G be a simply-connected solvable real Lie group and Γ a lattice in G.

Then there exists a natural number n and finitely generated subgroups Λ_{1}, Λ_{2} ⊂ GL(n, ℤ) with the following properties:

1. Λ_{1} is a normal subgroup of Λ_{2},

2. Let U denote the group of ℝ-rational points of the Zariski-closure of Λ_{1} and

H the connected component of the normalizer of U · Λ_{2} in GL(n, ℝ).

Then U · Λ_{2} is closed and the quotient Lie group H/(U · Λ_{2}) contains a compact Lie subgroup K such that G ≃ τ^{−1}(K) where τ : H → H/(U · Λ_{2}) is the natural projection.

Corollary. Up to Lie group isomorphism there exist only countably many simplyconnected solvable real Lie groups admitting a lattice.

Proof. Evidently there are only countably many choices for n, Λ_{1} and Λ_{2}. Thus it suffices to show that there are only countably many choices for K.

Claim. Let G be a real Lie group. Up to conjugacy there are only countably many compact connected solvable Lie subgroups.

This holds, because every compact connected solvable Lie group is a torus, all maximal tori are conjugate and a torus contains only countably many subtori.

Proof. We start with the following result of Mostow-Raghunathan.

Claim. Let G be a simply-connected solvable real Lie group, N its nilradical and Γ a lattice in G.

Then there exists an injective Lie group homomorphism i : G → GL(n, ℝ) (for some n) such that i(Γ) ⊂ GL(n, ℤ). Moreover every element in i(N) is unipotent.

This is Th.4.34 on p.71 of [3] except for the claim on the unipotency of i(N). But this claim follows easily from the details of the proof as given in pp.72–76 in [3].

We define Λ_{2} = i(Γ) and Λ_{1} = i(Γ ∩ N). Now Γ ∩ N is a lattice in N. Hence both Λ_{i} are finitely generated. Since i(N) is unipotent, it furthermore follows that U = i(N) is algebraic and equals the Zariski-closure of i(N ∩ Γ) = Λ_{1}. The group U · Λ_{2} is closed, because the orbits of the natural left N-action on G/Γ are compact and therefore are in particular closed. Thus the normalizer H_{0} of U ·Λ_{2} in GL(n, ℝ) is a closed real Lie subgroup. Evidently U = i(N) contains the commutator group of i(G). Hence i(G) ⊂ H_{0}. Moreover i(G)/U · Λ_{2} is compact, because G/Γ ≃ i(G)/Λ_{2} is compact. Hence the proposition.

References

[1] Margulis, G.A.: Discrete Subgroups of Semisimple Lie Groups. Springer Berlin Heidelberg New York 1989

[2] Mostow, G.D.: Arithmetic subgroups of groups with radical. Ann.- Math. 93, 409–438 (1971)

[3] Raghunathan, M.S.: Discrete subgroups of Lie groups. Erg. Math. Grenzgeb. 68 , Springer (1972)

[4] Wu, T.S.: A Note on a Theorem on lattices in Lie Groups. Canad. Math. Bull. 31(2), 1988