Journal of Lie Theory Volume 4 (1994) 1–16 ◯^{C} 1994 Heldermann Verlag

Stabilizers of Lattices in Lie Groups

Richard D. Mosak and Martin Moskowitz

Communicated by K. H. Hofmann

Abstract. Let G be a connected Lie group with Lie algebra 𝔤, containing a lattice Γ. We shall write Aut(G) for the group of all smooth automorphisms of G. If A is a closed subgroup of Aut(G) we denote by Stab_{A}(Γ) the stabilizer of Γ in A; for example, if G is ℝ^{n}, Γ is ℤ^{n}, and A is SL(n,ℝ), then Stab_{A}(Γ)=SL(n,ℤ). The latter is, of course, a lattice in SL(n,ℝ); in this paper we shall investigate, more generally, when Stab_{A}(Γ) is a lattice (or a uniform lattice) in A.

Introduction

Let G be a connected Lie group with Lie algebra 𝔤, containing a lattice Γ; so Γ is a discrete subgroup and G/Γ has finite G-invariant measure. We shall write Aut(G) for the group of all smooth automorphisms of G, and M(G) = {α ∈ Aut(G): mod(α) = 1} for the group of measure-preserving automorphisms of G (here mod(α) is the common ratio measure(α(F))/measure(F), for any measurable set F ⊂ G of positive, finite measure). If A is a closed subgroup of Aut(G) we denote by Stab_{A}(Γ) the stabilizer of Γ in A, in other words {α ∈ A: α(Γ) = Γ}. The main question of this paper is:

When is Stab_{A}(Γ) a lattice (or a uniform lattice) in A? We point out that the thrust of this question is whether the stabilizer

Stab_{A}(Γ) is cocompact or of cofinite volume, since in any event, in all cases of interest, the stabilizer is discrete (see Prop. 1.1).

Our first result, for mixed groups, applies to all the classical groups. We prove (Cor. 1.9) that if the radical of G is a vector group V , and the Levi factor S has no compact part and acts irreducibly on the complexification V^{ℂ}, then Stab_{M(G)}(Γ) is a lattice in M(G) for any lattice Γ in G. The same is true if M(G) is replaced by I(G), since for this class of groups I(G) has finite index in M(G) (1.7). In the course of the proof we obtain a generalization (see Th. 1.6) of the fact that every derivation of a semisimple Lie algebra is inner.

In Section 3 we turn to nilpotent groups G. Here our main results concern the Iwasawa nilpotent parts of the simple, non-compact, real, rank-1 simple Lie groups: the real vector groups ℝ^{n}, their complex analogues, the Heisenberg groups N_{n}, their quaternionic analogues H_{n}, and the exceptional case n = 1 of the

ISSN 0940–2268 / $2.50 ◯^{C} Heldermann Verlag groups C_{n} built on the Cayley numbers. In all these cases, stabilizers of loglattices (lattices whose log is a lattice in the Lie algebra) in G are lattices in M(G). Nevertheless, there is an interesting dichotomy between ℝ^{n} and N_{n} on the one hand, and H_{n} and C_{1} on the other.^{1} In the case of real vector groups and Heisenberg groups, the stabilizer is a non-uniform lattice in M(G) (2.4). When G is H_{n} or C_{1}, on the other hand, stabilizers of log-lattices turn out actually to be uniform lattices in M(G) (see Prop. 2.5 for H_{n} with n ≤ 2 and for C_{1}, and the forthcoming thesis of Paolo Barbano for the general case).

We conclude with various classes of examples and counterexamples. First we consider a class of nilpotent groups G, and a connected subgroup A satisfying I(G) ⊂ A ⊂ M(G) (strict containment) in which Stab_{A}(Γ) is a uniform lattice in A whenever Γ is a compatible log-lattice (see Prop. 2.10). Then we give examples (2.12, 2.13) to show that our results on Heisenberg groups cannot be extended to arbitrary unitriangular groups or to arbitrary two-step nilpotent groups.

1. Lattices in Mixed Groups

In connection with the main question of the paper, stated in the Introduction, we shall actually only consider subgroups A of M(G), since M(G) is the largest subgroup of Aut(G) in which it makes sense to consider this question. More precisely, if A is a closed subgroup of Aut(G) and Stab_{A}(Γ) a lattice in A, then A is contained in M(G). Indeed, if α ∈ Aut(G) and α(Γ) = Γ, then α preserves the (finite) measure of a fundamental domain for Γ. Thus α is measure-preserving, so Stab_{Aut(G)}(Γ) ⊂ M(G). Therefore the lattice Stab_{A}(Γ) is contained in the closed normal subgroup A ∩ M(G) of A, so mod maps A/A ∩ M(G) onto a compact subgroup of ℝ_{+}^{×}. It follows that A ⊂ M(G).

We begin by fixing some notation. If G is a connected Lie group, then Aut(G) and M(G) are also Lie groups. If α^{∘} denotes the differential in Aut(𝔤) of α ∈ Aut(G), then the map α ↦ α^{∘} is an injection of Aut(G) to Aut(𝔤). We denote by M(𝔤) the group of Lebesgue-measure preserving automorphisms of Aut(𝔤): M(𝔤) = {β ∈ Aut(𝔤): | det(β)| = 1}. The differential map takes M(G) into M(𝔤). If G is simply connected, the differential map is surjective, and in particular, M(G)^{∘} = M(𝔤).

Relative to any basis of 𝔤, M(𝔤) is defined by the polynomials induced by the Lie algebra structure, together with the polynomial det^{2}, so M(𝔤) is a real linear algebraic group (the group of real points of a linear algebraic group defined over ℝ). Of course, if the Lie algebra 𝔤 has rational structure constants, then these polynomials have rational coefficients, so M(𝔤) is defined over ℚ. We denote by D(𝔤) the Lie algebra of derivations of 𝔤; D(𝔤) is the Lie algebra of Aut(𝔤). We further denote by D_{0}(𝔤) the subalgebra of derivations of trace 0; D_{0}(𝔤) is the Lie algebra of M(𝔤).

Although we shall not use this fact below, we note in Proposition 1.1 below that in most cases of interest, the stabilizer of a lattice is always a discrete subgroup of Aut(G). Tits [18] introduced the notion of automorphism of bounded displacement: an automorphism α ∈ Aut(G) has bounded displacement if {α(g)g^{−1} : g ∈ G} has compact closure. Many groups have no non-trivial automorphisms of bounded displacement, for example, groups G whose radical is simply connected and of exponential type (so that the radical is a type E group), and whose Levi factor has no compact part [15, (1.1) and (1.3)]. For a simply connected nilpotent group, the proposition below is due to Malcev ([13, Th. 3]).

1This dichotomy is seen in other areas as well; for example, the corresponding simple groups satisfy Kazhdan’s property T iff the nilpotent parts are H_{n} and C_{1} (see, for example, the table on p. 47 of [2]).

Proposition 1.1. Let G be a connected Lie group with no non-trivial automorphisms of bounded displacement. If Γ is a lattice in G, then Stab_{Aut(G)}(Γ) is a discrete subgroup of Aut(G).

Proof. Let U be a neighborhood of 1 in G such that U ∩ Γ = (1), and let F = {γ_{1}, …, γ_{n}} be a finite^{2} generating set for Γ. Then W (F, U) = {α ∈ Aut(G): γ^{−1}αγ ∈ U for all γ ∈ F } is a neighborhood of 1 in Aut(G). If α ∈ W (F, U) ∩ Stab_{Aut(G)}(Γ), then γ_{i}^{−1}αγ_{i} ∈ U ∩ Γ for all i = 1, …, n, so α = 1 on Γ. Since Γ is a lattice it now follows from [8, Lemma 13] that α is of bounded displacement. Hence α = 1.

Our first result applies to the large class of groups G which satisfy the “density” condition B(G) = Z(G), where B(G) is the set of elements of G with bounded conjugacy classes; such groups were extensively studied in, for example, [8, 9, 15]. For these groups, we answer the main question of the paper for the group A = I(G) of inner automorphisms.

Theorem 1.2. Let G be a connected Lie group satisfying B(G) = Z(G). If Γ is a lattice in G, then Stab_{I(G)}(Γ) is a lattice in I(G).

Proof. Since I(G) is closed in Aut(G) [7, Th. 2], the differential α ↦ α^{∘}, which maps I(G) onto Ad(G), is an isomorphism, and the image of the stabilizer of Γ is clearly Ad(N_{G}(Γ)). Thus it suffices to prove that Ad(N_{G}(Γ)) is a lattice in Ad(G).

We first note that if G is a connected Lie group, then the condition B(G) = Z(G) implies the analogous condition for the simply connected covering group G̃: B(G̃) = Z(G̃) [9, p. 257]. Furthermore, by Corollary 5 of [9], Ad(Γ) is discrete in GL(𝔤), and in the closed subgroup Ad(G) [7, Th. 2]. Pushing the measure forward, we see that Ad(Γ) is a lattice in Ad(G). Let N denote the closed subgroup N_{G}(Γ). Since Ad(N) ⊃ Ad(Γ), to show that Ad(N) is a lattice it suffices to show that it is discrete.^{3}

2For the reader’s convenience we complete the sketch of the proof in [17, 6.18] that lattices in a connected Lie group are finitely generated. For a semisimple group, a sketch is given in [17, 13.21]. For an arbitrary Lie group G, let G = RCS where R = rad(G), C is the compact part of a Levi factor, and S is semisimple with no compact part. Then by an extension of a result of H.C. Wang and one of Mostow ([7, Prop. 8]), Γ∩RC is a uniform lattice in the normal subgroup RC. Hence by [5, chap. VII, §3, Lemme 3] Γ ∩ RC is finitely generated, while Γ mod RC is a lattice in a factor group of S, and therefore also finitely generated (as above). Thus Γ is itself finitely generated.

3If we assumed that G has no non-trivial automorphisms of bounded displacement, then we could already conclude from Proposition 1.1 that Ad(N) is discrete, since it is the stabilizer in Ad(G) of Ad(Γ).

Now since N normalizes Γ, Γ is a uniform lattice in N. Furthermore, the Euclidean component group N_{0} normalizes, hence centralizes, Γ. By [9, Th. 3], this implies that N_{0} ⊂ Z(G). Thus Z(G)Γ is an open subgroup of finite index in N. It now follows that Ad(Γ) has finite index in Ad(N), so in particular Ad(N) is discrete. Therefore Ad(N) is a lattice in Ad(G).

Corollary 1.3. Let G be a connected Lie group G whose radical is simply connected and of type E, and whose Levi factor has no compact part. If Γ is a lattice in G, then Stab_{I(G)}(Γ) is a lattice in I(G).

Proof. As noted above, G has no non-trivial automorphisms of bounded displacement, hence automatically satisfies B(G) = Z(G).

If G is itself semisimple, then I(G) has finite index in Aut(G), so we get:

Corollary 1.4. If G is a connected semisimple Lie group without compact part, then Stab_{Aut(G)}(Γ) is a lattice in Aut(G).

In the remainder of this section we shall consider a class of generalized motion groups. These groups will be constructed as semidirect products, as follows. Let 𝔥 be a Lie algebra, 𝔰 a subalgebra of D(𝔥) containing ad(𝔥), and 𝔤 the semidirect sum 𝔤 = 𝔥 ⊕ 𝔰, with bracketing given by

[(h, X), (k, Y)] = ([h, k] + Xk − Y h, [X, Y]). (1)

If H is a connected and simply connected Lie group with Lie algebra (isomorphic to) 𝔥, we can identify Aut(H) with Aut(𝔥) via the differential. If we denote by S the analytic subgroup of Aut(H) corresponding to 𝔰, then S ⊃ I(H), and 𝔤 is the Lie algebra of the connected Lie group G = H ×_{η}S (semidirect product). G is unimodular iff H and S are both unimodular, and 𝔰 ⊂ D_{0}(𝔥). In particular, G is unimodular if 𝔰 is semisimple.

For later use we need to determine the derivations of 𝔤 which leave 𝔥 invariant. Consider a linear map D ∈ 𝔤𝔩(𝔤) which leaves 𝔥 stable. For such a map we can write

D = (\begin{matrix}D_{𝔥} & φ \\ 0 & D_{𝔰} \end{matrix})

where D_{𝔥} = D|_{𝔥} and D_{𝔰} = D|𝔥 are endomorphisms of 𝔥 and 𝔰, respectively, and φ is a linear map of 𝔰 to 𝔥.

The following proposition can be proven by means of formal calculations.

Proposition 1.5. A linear map D which leaves 𝔥 invariant is a derivation of 𝔤 if and only if both of the following conditions hold:

1. D_{𝔥} and D_{𝔰} are derivations, and φ is a 1-cocycle;

2. [D_{𝔥}, X] = D_{𝔰}(X) + ad_{𝔥}(φ(X)) for all X ∈ 𝔰.

The next result can be viewed as a generalization of the classical fact that every derivation of a semisimple Lie algebra is inner.

Theorem 1.6. Suppose 𝔰 is a semisimple algebra which acts irreducibly on 𝔥^{ℂ} (in the case of real scalars) or on 𝔥 (in the case of complex scalars). Then every derivation D of trace 0 of 𝔤 which leaves 𝔥 invariant is inner.

Proof. By semisimplicity of 𝔰, D_{𝔰} = ad_{𝔰}(X_{0}) for some X_{0} ∈ 𝔰, while φ(X) = −X(h_{0}) for some h_{0} ∈ 𝔥 (Whitehead’s lemma). It follows that

\begin{aligned}[D_{𝔥}, X] & = D_{𝔰}(X) + ad_{𝔥}(φ(X)) \\ & = [X_{0}, X] + [ad(h_{0}), X] \end{aligned} (2)

for all X ∈ 𝔰. Thus L = D_{𝔥} − X_{0} − ad(h_{0}) centralizes 𝔰, so by Schur’s lemma is a multiple of the identity. On the other hand, since tr(D_{𝔥}) = tr(D) − tr(D_{𝔰}) and 𝔰 is semisimple and contains ad(𝔥), the trace of each term in the sum L is 0. Thus D_{𝔥} = X_{0} + ad(h_{0}), from which one sees easily that D = ad(h_{0}, X_{0}).

Corollary 1.7. Under the same hypotheses, if 𝔥 is characteristic in 𝔤, then I(G) has finite index in M(G) (so M(G)_{0} = I(G)).

Proof. The algebra D_{0}(𝔤) is the Lie algebra of M(𝔤), so Theorem 1.6 implies that M(𝔤)_{0} = Ad(G). Since M(𝔤) is a real algebraic group, it has only finitely many components, so Ad(G) has finite index in M(𝔤). Now the differential map α ↦ α^{∘} is injective, so I(G) has finite index in M(G).

Remark 1.8. These results hold, for example, whenever 𝔥 = V is a vector space, since then 𝔥 is characteristic in 𝔤. It may be noted, however, that if the scalars are real, and 𝔰 merely acts irreducibly on V , then the conclusions need not be true. For example, let V be the real vector space ℍ = ℝ^{4}, regarded as the algebra of quaternions, and let S be the semisimple group SU(2), identified with the unit sphere in ℍ. Since S acts transitively on itself by left multiplication, both it and its Lie algebra 𝔰 act irreducibly on ℍ. But any right multiplication by an element of S commutes with 𝔰, so we can find nontrivial linear maps ρ with trace 0 which commute with 𝔰. Then ( \begin{matrix}ρ & 0 \\ 0 & 0 \end{matrix} ) is an outer derivation of trace 0 on 𝔤.

Corollary 1.9. Let G = V ×_{η}S, where V is a real vector space, and S is a semisimple subgroup of GL(V) without compact factors, which acts irreducibly on V^{ℂ}. Then Stab_{M(G)}(Γ) is a lattice in M(G) for all lattices Γ in G.

Proof. ¿From Corollary 1.3 we know that Stab_{I(G)}(Γ) is a lattice in I(G). But by Corollary 1.7, I(G) has finite index in M(G).

2. Lattices in Nilpotent Groups

In this section we consider stabilizers of log-lattices in nilpotent groups. If G is a linear algebraic group defined over the field k, we denote by X_{k}(G) the group of k-rational characters of G, by G_{u} the unipotent radical of G, and by G^{0} the Zariski connected component. If 𝔤 is a linear Lie algebra ⊂ 𝔤𝔩(V), we denote by 𝔤_{n} its nilradical, that is, the operators in rad(𝔤) which are nilpotent on V . Similarly, if G is a linear Lie group, we denote by G_{u} its unipotent radical, in other words, the normal analytic subgroup whose Lie algebra is 𝔤_{n}, and by G_{0} its Euclidean connected component. If G is defined over ℝ, and G = G_{ℝ}, then the component group G/G_{0} is finite ([17, p. 10]).

Our first result in this section is a consequence of the celebrated theorems of Borel–Harish-Chandra [4] and Mostow–Tamagawa[16].

Theorem 2.1. Let G be a linear algebraic group, defined over ℚ, and let 𝔤 be the Lie algebra of the group of real points G_{ℝ}. If rad(𝔤) = 𝔤_{n}, then G_{ℝ}/G_{ℤ} has finite invariant measure. If in addition 𝔤/𝔤_{n} is of compact type, then G_{ℝ}/G_{ℤ} is compact.

Proof. For the proof, we first need some observations about Lie groups and real algebraic groups. Let G be a linear Lie group, and 𝔤 its Lie algebra. The hypothesis that rad(𝔤) = 𝔤_{n} (or equivalently that 𝔤/𝔤_{n} is semisimple) is equivalent to the assertion that G_{0}/G_{u} is semisimple. Furthermore, the additional hypothesis that 𝔤/𝔤_{n} be of compact type is then equivalent, by Weyl’s theorem, to the assertion that G_{0}/G_{u} is compact.

Specializing now to the case G = G_{ℝ}, we note that in this case G_{u} = (G_{ℝ})_{u} actually equals (G_{u})_{ℝ}. In fact, (G_{u})_{ℝ} is connected and unipotent [17, p. 10], hence contained in the radical of G_{ℝ}, and therefore in (G_{ℝ})_{u}. Conversely, rad(G_{ℝ}) ⊂ rad(G), so (G_{ℝ})_{u} = rad(G_{ℝ})_{u} ⊂ rad(G)_{u} = G_{u}; thus (G_{ℝ})_{u} ⊂ (G_{u})_{ℝ}. Now by [17, p. 11], if G = G_{u}H (semidirect), with H reductive, then (G_{ℝ})_{0} is the semidirect product (G_{u})_{ℝ}(H_{ℝ})_{0} = (G_{ℝ})_{u}(H_{ℝ})_{0}. Thus the hypothesis rad(𝔤) = 𝔤_{n} is in this case equivalent to semisimplicity of (H_{ℝ})_{0} (or H_{ℝ}), and the additional hypothesis that 𝔤/𝔤_{n} be of compact type is equivalent to compactness of (H_{ℝ})_{0} (or H_{ℝ}).

One further general observation about G is that the Euclidean connected components of (G^{0})_{ℝ} and of G_{ℝ} are the same: ((G^{0})_{ℝ})_{0} = (G_{ℝ})_{0} (so that G_{ℝ} and (G^{0})_{ℝ} have the same Lie algebra). In fact, the group (G_{ℝ})_{0} is Zariski connected, hence contained in G^{0}; since it consists of real points, it must be contained in (G^{0})_{ℝ}; but since it is Euclidean connected, it must therefore be contained in ((G^{0})_{ℝ})_{0}. The reverse inclusion is obvious.

Now suppose rad(𝔤) = 𝔤_{n}. To prove that G_{ℝ}/G_{ℤ} has finite invariant measure, we need to show that G^{0} has no non-trivial ℚ-rational characters [4, Th. 9.4]. If χ ∈ X_{ℂ}(G^{0}), then χ must be trivial on (G_{ℝ})_{0} = (G_{ℝ})_{u}(H_{ℝ})_{0}, since a rational character kills unipotent groups and connected semisimple groups. Since [(G^{0})_{ℝ}: ((G^{0})_{ℝ})_{0}] is finite, χ sends (G^{0})_{ℝ} to a finite subgroup of ℂ^{×}. Now by a result of Rosenlicht, (G^{0})_{ℝ} is Zariski dense in G^{0} ([17, p. 10]), and finite subgroups are Zariski closed, so χ maps the connected group G^{0} into a finite subgroup. Therefore χ is trivial, proving the first assertion of the Proposition.

Next suppose that 𝔤/𝔤_{n} is semisimple and of compact type. To prove that G_{ℝ}/G_{ℤ} is compact, we must show that all unipotent elements in G_{ℤ} are actually in G_{u} ([4, Th. 11.8], [16, Th., p. 453]). From the previous remarks, we know that (G_{ℝ})_{0}/(G_{ℝ})_{u} = (H_{ℝ})_{0} is compact, so also G_{ℝ}/(G_{ℝ})_{u}. Therefore if u is any unipotent in G_{ℤ}, or even in G_{ℝ}, the image of u by the quotient morphism is a unipotent element of a compact group. Hence the image of u is both unipotent and semisimple, and therefore u ∈ (G_{ℝ})_{u}.

The converses in 2.1 are not true. For a counterexample to the first assertion, we may take the ℚ-group G = SO(2, ℂ); then G_{ℝ} = SO(2, ℝ) is a compact abelian group, so G_{ℤ} is a (uniform) lattice, but 𝔤_{n} = (0). For a counterexample to the second assertion we may take a semisimple algebraic ℚ-group G, such that G_{ℝ} is not compact (so 𝔤 is not of compact type), but G_{ℝ}/G_{ℤ} is compact [17, Th. 14.2].

We shall now consider the case of a simply connected nilpotent Lie group G that contains a lattice Γ. If 𝔤_{ℚ} denotes the ℚ-span of log(Γ), then by Malcev’s theorem ([17, p. 34]), 𝔤_{ℚ} generates 𝔤 over ℝ, and the structure constants of 𝔤, relative to any ℚ-basis of 𝔤_{ℚ}, are rational. In particular, 𝔤_{ℚ} is a ℚ-structure for 𝔤, that is, a ℚ-Lie subalgebra which generates 𝔤 over ℝ. Thus every lattice in G generates a ℚ-structure of 𝔤. Our results in this section are valid for special lattices called log-lattices, that is, lattices Γ for which log(Γ) is a lattice in (𝔤, +). ¿From [14, Th. 2] it follows if G contains a lattice, then it must actually contain a log-lattice.

As we have observed in the beginning of the paper, if we choose any basis in 𝔤_{ℚ}, then the group of differentials M(G)^{∘} of the measure-preserving automorphisms of G is the real linear algebraic group M(𝔤) (defined over ℚ, in our case). M(𝔤) is the set of real points of a linear algebraic group M = M(𝔤). The ℚ-algebraic group structure on M is independent of the choice of basis in 𝔤_{ℚ}, since any two such bases are conjugate by an element of GL(n, ℚ). Furthermore, as we noted earlier, the Lie algebra of M(𝔤) is the algebra D_{0}(𝔤) (or simply D_{0}) of derivations of trace 0. More generally, if A is a closed subgroup of M(G), and A = {α^{∘}: α ∈ A} denotes the set of differentials, then the Lie algebra of A is an algebra of derivations 𝔞 ⊂ D_{0}.

Theorem 2.2. Suppose G is a simply connected nilpotent group whose Lie algebra 𝔤 has a ℚ-structure 𝔤_{ℚ}, and Γ is a log-lattice in G such that log(Γ) ⊂ 𝔤_{ℚ}. Let A be a closed subgroup of M(G), and assume that A^{∘} is the set of real points of a linear algebraic group A ⊂ M defined over ℚ (for example, A = M(G) satisfies this hypothesis). Then

1. If rad(𝔞) = 𝔞_{n}, then Stab_{A}(Γ) is a lattice in A.

2. If in addition 𝔞/𝔞_{n} is of compact type, then Stab_{A}(Γ) is uniform in A.

Proof. We shall show first that the map α ↦ α^{∘} induces an equivariant diffeomorphism of A/ Stab_{A}(Γ) → A_{ℝ}/A_{ℤ}. Indeed, if S denotes the stabilizer of log(Γ) in A, then the isomorphism α ↦ α^{∘} of M(G) onto M(𝔤) takes A onto A and Stab_{A}(Γ) to S, because of the relation α ∘ exp = exp ∘ α^{∘}. Now if β ∈ 𝔤𝔩(𝔤), then β stabilizes log(Γ) iff it maps the basis elements of log(Γ) to another basis of log(Γ). If β ∈ A, then det β is a unit, so β stabilizes log(Γ) iff it has integer matrix entries (with respect to a fixed basis of log(Γ)). Thus S = A_{ℤ} = A_{ℤ}, and by hypothesis, A = A_{ℝ}. Now the result follows from 2.1.

Corollary 2.3. If D(𝔤) ⊂ [𝔤, 𝔤] for every D ∈ rad(𝔞), then Stab_{A}(Γ) is a lattice in A. If the inclusion actually holds for every D ∈ 𝔞, then Stab_{A}(Γ) is a uniform lattice.

Proof. If D is a derivation such that D(𝔤) ⊂ [𝔤, 𝔤], then D^{i}(𝔤) ⊂ 𝔤^{i} for each term 𝔤^{i} in the descending central series. Since 𝔤 is a nilpotent Lie algebra, it follows that D is a nilpotent transformation. Therefore the first hypothesis implies that rad(𝔞) = 𝔞_{n}. The second hypothesis implies that every derivation in 𝔞 is nilpotent, so 𝔞 is a nilpotent Lie algebra. The corresponding analytic group A_{0} is therefore nilpotent, so all its lattices are uniform [13]. Since A_{0} has finite index in A, all lattices in A and therefore A are also uniform.

Our next results concern the Iwasawa nilpotent parts of the simple, noncompact, real, rank-1 simple Lie groups. These nilpotent parts have been classified in [6, Th. 1.1 and Th. 4.2], and consist of the real vector groups ℝ^{n}, the Heisenberg groups N_{n}, their quaternionic analogues H_{n}, and the exceptional case n = 1 of the groups C_{n} built on the Cayley numbers. In the class of real vector groups, G = ℝ^{n}, stabilizers in M(G) of lattices in the group are of course non-uniform lattices in M(G). In the Heisenberg groups, we shall see below (2.4) that stabilizers of loglattices are also non-uniform lattices in M(G) . Turning to the last two classes of groups, we prove in Proposition 2.5 that for G = H_{n} with n = 1 or n = 2, and for the exceptional case C_{1}, stabilizers of log-lattices are actually uniform lattices in M(G).

Proposition 2.4. Let G be a Heisenberg group. If Γ is any log-lattice in G, then the stabilizer S = Stab_{M(G)}(Γ) is a non-uniform lattice in M(G).

Proof. Assuming G is 2n + 1-dimensional, we shall write elements of G as matrices

(\begin{matrix}1 & x_{1} & … & x_{n} & z \\ & 1 & & 0 & y_{1} \\ & & ... & & ⋮ \\ & 0 & & 1 & y_{n} \\ & & & & 1 \end{matrix})

with x_{i}, y_{i}, z ∈ ℝ. Then 𝔤 has a basis consisting of {X_{1}, …, X_{n}, Y_{1}, …, Y_{n}, Z} where X_{i} is defined by the condition x_{i} = 1 and all other off-diagonal matrix entries are 0 (similarly for Y_{i} and Z). The only non-trivial bracketing relations are given by [X_{i}, Y_{i}] = Z. A direct calculation shows that relative to this basis, the general derivation D of trace 0 has matrix given by

D = (\begin{matrix}A & B & 0 \\ C & −A^{t} & 0 \\ v & w & 0 \end{matrix})

where C and B are symmetric, and v and w are vectors in ℝ^{n}. Thus the algebra D_{0} is isomorphic to the semidirect sum ℝ^{2n} ⊕ 𝔰𝔭(n, ℝ). 𝔰𝔭(n, ℝ) is semisimple of non-compact type, and the radical of D_{0} consists of the nilpotent operators

(\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ v & w & 0 \end{matrix}) .

By Corollary 2.2 the stabilizer of Γ in M(G) is a lattice.

To see that this stabilizer is non-uniform, we consider first the lattice Γ_{k} defined by the conditions x_{i}, y_{i} ∈ ℤ, and z ∈ \frac{1}{k}ℤ, with k even. Then log Γ_{k} is a lattice, generated over ℤ by the basis B_{k} = \{X_{1}, …, X_{n}, Y_{1}, …, Y_{n}, \frac{1}{k}Z\}. From the description of D_{0} given above, it is clear that the identity component of M(𝔤) is the set of matrices of the form

β = ( \begin{matrix}σ & 0 \\ v w & 1 \end{matrix} )

with σ ∈ SP(n, ℝ) and v, w ∈ ℝ^{n}. Furthermore, as in the proof of Corollary 2.2, β is in the stabilizer of log Γ_{k} iff it is in (M(𝔤)_{0})_{ℤ}. Hence the orbit of log Γ_{k} under M(𝔤)_{0} is diffeomorphic to (SP(n, ℝ)/ SP(n, ℤ))×T^{2n}. This orbit is of course noncompact, since SP(n, ℤ) has many non-trivial unipotents. Since M(𝔤) has only finitely many components, it follows that M(𝔤)/ Stab_{M(𝔤)}(log Γ_{k}) is not compact either.

Now in general, by [1], any lattice Γ in G is isomorphic under an automorphism α of G to a lattice of the form Γ_{k,r⃗} for some k ∈ ℤ^{+} and some r⃗ = (r_{1}, …, r_{n}) ∈ (ℤ^{+})^{n} satisfying r_{1} = 1, r_{i}|r_{i+1} for all i: Γ_{k,r⃗} is defined by the conditions x_{i} ∈ r_{i}ℤ, y_{i} ∈ ℤ, and z ∈ \frac{1}{k}ℤ. Since M(G) is normal and conjugation by α carries Stab_{M(G)}(Γ) to Stab_{M(G)}(Γ_{k,r⃗}), the quotient spaces are diffeomorphic. Therefore it is enough to prove that the stabilizer of Γ_{k,r⃗} is not uniform when Γ_{k,r⃗} is a log-lattice, that is, when k is even. In this case log Γ_{k,r⃗} is the lattice generated over ℤ by the basis B_{k,r⃗} = \{r_{1}X_{1}, …, r_{n}X_{n}, Y_{1}, …, Y_{n}, \frac{1}{k}Z\}. If ϕ is the isomorphism of 𝔤 taking r_{i}X_{i} to X_{i} and leaving all other elements of B_{k,r⃗} fixed (so ϕ takes B_{k,r⃗} to B_{k}), then conjugation by ϕ is a ℚ-rational isomorphism Φ of the ℚ-group M(𝔤) relative to the basis B_{k}, to the ℚ-group M(𝔤) relative to the basis B_{k,r⃗} . Furthermore, since by [17, Cor. 10.14.ii] Φ takes arithmetic subgroups to arithmetic subgroups, Φ(Stab_{M(𝔤)}(log Γ_{k})) is commensurable with Stab_{M(𝔤)}(log Γ_{k,r⃗}), since the stabilizers are the sets of ℤ-points of M(𝔤) relative to the B_{k} and B_{k,r⃗}, respectively. Since we have already shown that Stab(log Γ_{k}) is a non-uniform lattice, the same is true of Stab(log Γ_{k,r⃗}). This completes the proof.

Next we shall examine the groups G = H_{n}, which are the Iwasawa nilpotent parts of the symplectic rank-1 groups, and G = C_{1}, the Iwasawa nilpotent part of the exceptional simple rank-1 group. We begin with H_{n} and its Lie algebra 𝔥_{n}.

For the construction, we let ℍ denote the algebra of quaternions over ℝ, and P denote the subspace of pure quaternions, generated by i, j, and k. Then P = Im(ℍ), with Im(q) = (q − q̄)/2. For any n ≥ 1, we define

𝔤 = 𝔥_{n} = P ⊕ ℍ^{n},

a vector space of dimension 4n + 3, with operations given as follows:

[(p, v), (q, w)] = (Im⟨v, w⟩, 0),

where

⟨v, w⟩ = v̄_{1}w_{1} + … v̄_{n}w_{n}

for v = (v_{1}, …, v_{n}) and w = (w_{1}, …, w_{n}) in ℍ^{n}. We note that if ℍ is replaced by ℂ, then the analogous construction gives the Heisenberg algebra of dimension 2n+1 (if ℍ is replaced by ℝ, then we get ℝ^{n+1}!). It is easy to see that P = [𝔤, 𝔤], and is the center of 𝔤, so that 𝔤 is two-step nilpotent.

Similarly, we can define two-step nilpotent groups C_{n} by analogy with the groups H_{n}, by replacing the associative algebra ℍ by the non-associative algebra 𝕂 of Cayley numbers. C_{1} is the Iwasawa nilpotent part of the exceptional rank-1 simple group.

Proposition 2.5. Let G = H_{n}, with n = 1 or n = 2, or G = C_{1}, and let Γ be a log-lattice in G. Then the stabilizer Stab_{M(G)}(Γ) is a uniform lattice in M(G).^{4}

Proof. For each of the three groups mentioned, one can exhibit an explicit decomposition D_{0}(𝔤) = (D_{0})_{n} ⊕ 𝔰 (semidirect sum), where (D_{0})_{n} is the nilradical, and 𝔰 is a compact semisimple algebra^{5}. The result then follows from Corollary 2.2.

We turn now to another class of semidirect products, in which the group of automorphisms A is nilpotent, but in contrast with the examples discussed earlier, here A turns out to be essentially different from both I(G) and M(G). We shall prove that for these groups, the A-orbit spaces of log-lattices are compact (see Proposition 2.10).

The groups G under consideration are of the form G = V ×_{η}S, where V is an n-dimensional vector space with n > 1, and S is a 1-parameter group in SL(V), say S = (e^{tC})_{t∈ℝ} with tr C = 0. G is unimodular since tr C = 0. As usual, we write 𝔰 (= ℝC) for the Lie algebra of S, and we identify V with its Lie algebra. Thus 𝔤 = V ⊕ 𝔰 (semidirect sum). We define an algebra 𝔞 of derivations of trace 0 on 𝔤, as follows:

𝔞 = {D ∈ D_{0}(𝔤): D(V) ⊂ V, tr_{V} (D_{V}) = 0}. (3)

Thus 𝔞 consists of the derivations of trace 0 which preserve V , and have trace 0 on V .^{6}

(\begin{matrix}A & v \\ w^{t} & δ \end{matrix}) ,

4This result has been proved in general by P. Barbano, and will appear in his forthcoming thesis.

5In fact, one can show that 𝔥_{1}/(𝔥_{1})_{n} = 𝔰𝔭(1)⊕𝔰𝔭(1), 𝔥_{2}/(𝔥_{2})_{n} = 𝔰𝔭(2)⊕𝔰𝔭(1) and 𝔠_{1}/(𝔠_{1})_{n} = 𝔰𝔬(7).

6If the degree of nilpotence of C is greater than 2, then the condition D(V) ⊂ V in the definition of 𝔞 is redundant, since V is actually a characteristic ideal of 𝔤. In fact, one can show that in the present situation, if we describe a linear map D ∈ End(𝔤) as a block matrix

Lemma 2.6. If D ∈ 𝔤𝔩(𝔤), then D ∈ 𝔞 iff D equals

D(L, φ) = (\begin{matrix}L & φ \\ 0 & 0 \end{matrix}) (4)

where L ∈ 𝔰𝔩(V), and [L, C] = 0.

Proof. If D is a linear operator on 𝔤 leaving V invariant, and we decompose D as in Proposition 1.5 (replacing 𝔥 by V ), then D_{𝔰} is given by D_{𝔰}(X) = λX where λ is a constant (since 𝔰 is 1-dimensional). If in addition tr D = 0, then λ = − tr_{V} (D_{V}). Now it is easy to see from Proposition 1.5 that if D leaves V invariant and has trace 0, then D is a derivation iff

[D_{V} , C] = − tr(D_{V})C.

Therefore such a D is in 𝔞 iff [D_{V} , C] = 0.

Henceforth we shall assume that C is a nilpotent operator. In this case, the algebra 𝔞 is essentially different from both ad(𝔤) and D_{0}(𝔤).

Proposition 2.7. If C is a non-zero nilpotent operator, then both of the inclusions ad(𝔤) ⊂ 𝔞 ⊂ D_{0}(𝔤) are strict.

Proof. It is easy to see that ad(v, X) = D(X, v̂), where v̂ is the map Y ↦ −Y (v) on 𝔰. Since X is a multiple of C, X has trace 0 and commutes with C. Therefore ad(𝔤) ⊂ 𝔞. Now dim ad(𝔤) = dim 𝔤 − dim 𝔷(𝔤), and an easy calculation shows that 𝔷(𝔤) = ker C. Since C is nilpotent, we have

dim ad(𝔤) = n + 1 − dim ker C ≤ n.

On the other hand,

dim(𝔞) = dim{D(L, φ): L ∈ 𝔰𝔩(V), [L, C] = 0, φ ∈ Hom_{ℝ}(𝔰, V)}.

Since D(L, φ) = 0 iff L = φ = 0, and 𝔰 is 1-dimensional, dim 𝔞 = dim Z_{𝔰𝔩(V)}(C)+n (Z denotes the centralizer). Since C centralizes itself, dim 𝔞 ≥ n + 1. Therefore ad(𝔤) is a proper subset of 𝔞.

For the other inclusion, we observe that by the Jacobson–Morozov theorem [11, p. 100], there is an H ∈ 𝔤𝔩(V) with [H, C] = 2C. Set H_{1} = H + λI with λ chosen so that tr(H_{1}) = −2. Then [H_{1}, C] = − tr(H_{1})C, so by Lemma 2.6 and its proof, the map

D = (\begin{matrix}H_{1} & 0 \\ 0 & − tr(H_{1}) \end{matrix})

is a derivation of 𝔤 with trace 0 (leaving V invariant, in fact), but is not in 𝔞.

with A ∈ End(V), v, w ∈ V , and δ ∈ ℝ, then D is a derivation iff w ⊥ im C, (w · y)Cx = (w · x)Cy for all x, y ∈ V , and [A, C] = δC. If C has degree of nilpotence > 2, then we can find y ∈ im C but not in ker C. The second condition then shows that if D is a derivation, then w = 0, so D(V) ⊂ V .

¿From now on we assume that C is of the form

C = (\begin{matrix}0 & c_{12} & … & ∗ \\ 0 & 0 & ... & ∗ \\ ⋮ & ⋮ & & c_{n−1,n} \\ 0 & 0 & … & 0 \end{matrix}) , c_{k−1,k} ≠ 0 for k = 1, …, n (5)

with respect to a fixed basis {e_{1}, …, e_{n}} of V . Since C is a nilpotent operator, the group G = V ×_{η}(e^{tC}) is nilpotent. We denote by 𝔫_{n} ⊂ 𝔰𝔩(V) the Lie algebra of upper niltriangular operators on V with respect to this basis.

Lemma 2.8. Let C be of the form (5). If X ∈ 𝔰𝔩(V) and [X, C] = 0 then X ∈ 𝔫_{n}.

Proof. Because [X, C] = 0 it follows that [X, C^{k}] = 0 for all positive k and hence X leaves each ker C^{k} invariant. But {ker C^{k}: k = 1, …, n} form a flag. This means that X = (x_{ij}) is upper triangular, so for each k,

Xe_{k} = x_{kk}e_{k} mod < e_{1}, …, e_{k−1} > .

Thus from the hypothesis on C we get

CXe_{k} = c_{k−1,k}x_{kk}e_{k−1} mod < e_{1}, ..., e_{k−2} >,

while

XCe_{k} = Xc_{k−1,k}e_{k−1} = c_{k−1,k}x_{k−1,k−1}e_{k−1} mod < e_{1}, …, e_{k−2} >.

Because [X, C] = 0 and c_{k−1,k} ≠ 0, we see that x_{kk} = x_{k−1,k−1} for all k. Since tr X = 0, each x_{kk} = 0.

Lemma 2.9. Let C be of the form (5). If the entries c_{ij} are rational for all i, j, then G has a log-lattice which intersects V in a lattice of V .

Proof. The map

(v, e^{tC}) ↦ (\begin{matrix}e^{tC} & v \\ 0 & 1 \end{matrix})

is an isomorphism of G onto a subgroup H of the unitriangular group in GL(n + 1, ℝ). H is a real algebraic group defined over ℚ, because C is niltriangular with rational entries. Furthermore, the Zariski closure H of H is unipotent, so H_{ℝ}/H_{ℤ} is compact. Now H = H_{ℝ} ([17, p. 9]). Furthermore, H_{ℤ} = H_{ℤ} intersected with the image of V is the integer points of this image, and is therefore a lattice. Pulling back H_{ℤ} to G gives a lattice satisfying the condition.

We are now ready for the main result for this class of nilpotent semidirect product groups.

Proposition 2.10. Let V be a real n-dimensional vector space (n > 1), S = (e^{tC}) with C of the form (5) with c_{ij} rational for all i, j. Let G = V ×_{η}S, and A be the subgroup of Aut(G) defined by

A = {α ∈ M(G): α(V) ⊂ V, α|_{V} ∈ M(V)}.

Then Stab_{A}(Γ) is a uniform lattice in A, for every log-lattice Γ in G which intersects V in a lattice of V .

Proof. We observe first by (3) that the Lie algebra of A = A^{∘} is 𝔞, since as one sees easily, A is defined by the conditions

β ∈ M(𝔤), β(V) ⊂ V, det_{V}^{2} β = 1.

Now by Lemma 2.8, Z_{𝔰𝔩(V)}(C) ⊂ 𝔫_{n}. It follows from Lemma 2.6 that 𝔞 ⊂ 𝔫_{n+1}, so that A is a subgroup of the unipotent group of order n + 1. The conditions defining A show that it is a real linear algebraic ℚ-group relative to the ℚstructure defined by any log-lattice which intersects V in a lattice. The result now follows from Corollary 2.2, since 𝔞 is nilpotent.

Our final results furnish examples of nilpotent groups where the stabilizer Stab_{M(G)}(Γ) is not a lattice at all. These examples show that there are significant limitations on the possibility of extending Corollary 2.2. The first example (2.12) is the unitriangular group (n ≥ 4), and the second (2.13) is even two-step nilpotent. We begin with a lemma.

Lemma 2.11. Let G ⊂ GL(V) be a linear algebraic group, defined over ℚ. If 𝔤 is the Lie algebra of G_{ℝ}, and 𝔤 has a ℚ-rational weight when acting on V_{ℝ}, then G^{0} has a non-trivial ℚ-rational character.

Proof. We assume given a ℚ-form V_{ℚ} ⊂ V_{ℝ} of V , and (by hypothesis) also a non-zero semi-invariant vector v ∈ V_{ℚ} satisfying Xv = λ(X)v for all X ∈ 𝔤, with λ non-trivial. If v = v_{1} is extended to a basis of V_{ℚ}, then the matrices (X_{ij}) of elements of 𝔤 satisfy X_{i1} = 0 for i > 1. By the proof of Theorem 2.1, 𝔤 is the Lie algebra of (G^{0})_{ℝ}. Exponentiation therefore shows that the matrices (g_{ij}) of elements of (G^{0})_{ℝ} also satisfy g_{i1} = 0 for i > 1. But (G^{0})_{ℝ} is Zariski dense in G^{0}, so the same relation holds for the latter group. Thus the map g ↦ g_{11} is a non-trivial ℚ-rational character of G^{0}.

Example 2.12. Let G be the group of n × n real unitriangular matrices, for n ≥ 4. We shall show that there are log-lattices in G whose stabilizer in M(G) is not a lattice, i.e. is a discrete subgroup, but has infinite covolume (of course, such lattices cannot exist when n = 3). We believe that the underlying reason that these stabilizers are not lattices is that M(G) is not unimodular, but we have not proven that. Instead, we give a direct proof.

The Lie algebra 𝔤 of G consists of all n × n upper triangular matrices with 0’s on the diagonal, and so has a basis

B = {E_{12}, E_{23}, …, E_{n−1,n}, E_{13}, …, E_{n−2,n}, …, E_{1n}}.

Here E_{ij} is the matrix with a single 1 in the i, j position, and 0’s elsewhere. The Lie brackets in 𝔤 are as follows:

[E_{ij}, E_{kl}] = δ_{jk}E_{il} − δ_{il}E_{kj}

for i < j and k < l. In particular, the basis B generates a ℚ-structure for 𝔤. Now the center of 𝔤 is 1-dimensional, generated by E_{1n}, so for any D ∈ D, and Z ∈ 𝔷(𝔤), D(Z) = λ(D)Z, with λ(D) ∈ ℝ. In view of Lemma 2.11 applied to the Lie algebra D_{0}, and the Borel–Harish-Chandra theorem [4], to prove that not all stabilizers of log-lattices are themselves lattices, it is sufficient to show that for n ≥ 4, λ(D) ≠ 0 for some D ∈ D_{0}.

To see this, choose n − 1 arbitrary real numbers d_{1}, …, d_{n−1}, and define a derivation D as follows:

D(E_{ij}) = (∑_{p=i}^{j−1} d_{p}) E_{ij}

for 1 ≤ i < j ≤ n. It is easy to see that D is a derivation, for any choice of the d_{i}’s. Furthermore,

λ(D) = ∑_{p=1}^{n−1} d_{p},

while

tr(D) = (n − 1)d_{1} + 2(n − 2) ∑_{p=2}^{n−2} d_{p} + (n − 1)d_{n−1}.

For n ≥ 4 these two forms are clearly linearly independent, so we can choose the d_{i}’s so that tr(D) = 0 and λ(D) ≠ 0.

The Iwasawa nilpotent parts of rank-1 simple groups are two-step nilpotent (or abelian). In our last example, we show that it is not simply the low degree of nilpotence that causes stabilizers of log-lattices to be lattices themselves.

Example 2.13. Let G be the simply connected two-step nilpotent group whose Lie algebra 𝔤 is the direct sum of the Heisenberg Lie algebra and ℝ. Let Z span [𝔤, 𝔤], {Z, W } span Z(𝔤), and let {Z, W, X, Y } be a basis for 𝔤 with [X, Y] = Z. It is easy to see that with respect to this basis any D ∈ D_{0} is of the form

D = (\begin{matrix}\begin{matrix}α & β \\ 0 & −2α \end{matrix} & ∗ \\ 0 & D̃ \end{matrix})

where α = tr D̃ and D̃ is an arbitrary element of 𝔤𝔩(2, ℝ). The same argument as in Example 2.12 shows that G contains a log-lattice whose stabilizer in M(G) is not a lattice.

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16 Mosak and Moskowitz

Department of Mathematics and Computer Science Lehman College, CUNY Bedford Park Boulevard West Bronx, NY 10468 rmosak@lcvax.lehman.cuny.edu Department of Mathematics Graduate Center, CUNY 33 West 42 Street New York, NY 10036 mamgc@cunyvm.cuny.edu

Received March 23, 1994