Journal of Lie Theory Volume 12 (2002) 449–460 C ◯2002 Heldermann Verlag

Vanishing of the First Cohomologies for Lattices in Lie Groups

A. N. Starkov^{∗}

Communicated by E. B. Vinberg

Abstract. We prove the following “maximal” theorem on vanishing of the first cohomologies. Let G be a connected semisimple Lie group with a lattice Γ. Assume that there is no epimorphism ϕ:G→H onto a Lie group H locally isomorphic to SO(1,n) or SU(1,n) such that ϕ(Γ) is a lattice in H. Then H^{1}(Γ,ρ)=0 for any finite-dimensional representation ρ of Γ over ℝ. This generalizes Margulis’ Theorem on vanishing of the first cohomologies for lattices in higher rank semisimple Lie groups. Some applications for proving general results on the structure of lattices in arbitrary Lie groups, are given.

Introduction

In his in-depth study of lattices in semisimple Lie groups, Margulis proved the following theorem on vanishing of the first cohomologies:

Theorem 1. (See [4].) Let Γ be a lattice in a connected semisimple Lie group G. Assume that

a) G has no compact factors,

b) G has finite center,

c) Γ is irreducible in G,

d) rk_{ℝ} G ≥ 2.

Then H^{1}(Γ, ρ) = 0 for any finite-dimensional representation ρ of Γ over local field of characteristic 0.

We will avoid the restrictions a)–c) and relax d) thus proving a general result for representations over ℝ (and hence over ℂ). To formulate it, we give a definition. Let G be a connected Lie group with a lattice Γ ⊂ G. Let us say that the pair (G, Γ) satisfies condition (∗) if there is no epimorphism ϕ: G → H onto a Lie group H locally isomorphic to SO(1, n) or SU(1, n) such that ϕ(Γ) is a lattice in H.

ISSN 0949–5932 / $2.50 ^{_{C} ◯} Heldermann Verlag Theorem 2. Let G be a connected semisimple Lie group with a lattice Γ. Assume that the pair (G, Γ) satisfies condition (∗). Then H^{1}(Γ, ρ) = 0 for any finite-dimensional representation ρ of Γ over ℝ.

∗ The author was partially supported by the Russian Foundation for Basic Research, grant No 01-01-00067, and by the Leading Scientific School Grant, No 00-15-96107.

In a sense, this is the “maximal” theorem on vanishing of the first cohomologies for lattices (see §4 for the discussion).

Theorem 2, in particular, claims vanishing of the first cohomologies for lattices in rank one Lie groups Sp(1, n), n ≥ 2, and F_{4}^{−20} . This is a new result which we derive from old papers of Raghunathan [6,7] in combination with more recent Superrigidity and Arithmeticity Theorems obtained by Corlette [1] and Gromov–Schoen [3].

The paper consists of four sections. In §1 we give definitions and some immediate observations. Theorem 2 is reduced to the rank one case in §2. In §3 we study the rank one situation. In §4 we give some applications of the main result. These include: a general form of Superrigidity Theorem, a general form of the Levi–Mostow splitting of lattices in arbitrary Lie groups, and a criterion for vanishing of the first cohomologies (modulo a still open Millson problem).

Acknowledgements. The work was done in the course of the author’s visit to the University of Bielefeld in August 2000. It is a pleasure to thank the University for excellent conditions. The author is very grateful to E. Vinberg, H. Abels, G. Soifer and D. Witte for useful discussions. Special thanks are given to T. Venkataramana for the idea of combining Raghunathan’s paper [6] with Superrigidity Theorem of Corlette to handle uniform lattices in Sp(1, n) and F_{4}^{−20} .

§1. Definitions and easy observations

In what follows, we work with the real field ℝ. All linear representations considered in this paper are assumed to be continuous and finite-dimensional. Several standard results are included for the sake of completeness.

Let F be a topological group and ρ: F → GL(V) be a representation on a (real) vector space V . A continuous map α: F → V is said to be a cocycle over ρ if

α(gg′) = α(g) + ρ(g)(α(g′)), g, g′ ∈ F.

A cocycle α is said to be a coboundary (or α ∼ 0) if there exists x ∈ V such that α(g) = ρ(g)(x) − x. The additive group H^{1}(F, ρ) of all cocycles modulo coboundaries is called the first cohomology group of F over ρ.

Let τ_{1} stand for the trivial representation of F on ℝ. Then H^{1}(F, τ_{1}) is the additive group Hom(F, ℝ) of all homomorphisms F → ℝ.

Lemma 1.1. 1) Let α be a cocycle over ρ and let ρ_{α}(g) = α(g)ρ(g). Then ρ_{α} is a homomorphism of F to the affine group Aff(V) = GL(V)·V , where the multiplication in Aff(V) is given by

gx = gxg^{−1}g = g(x)g, g ∈ GL(V), x ∈ V.

2) Conversely, let π: Aff(V) → GL(V) be the natural epimorphism and let ρ′: F → Aff(V) be a homomorphism covering ρ (i.e., ρ = π ∘ ρ′). Then the formula ρ′(g) = α(g)ρ(g) defines a cocycle over ρ.

3) Cocycle α is a coboundary iff ρ(F) and ρ_{α}(F) are conjugate by an element of V . Proof. 1) ρ_{a}(gg′) = α(gg′)ρ(gg′) = (α(g) + ρ(g)(α(g′)))ρ(gg′) = α(g)ρ(g)α(g′)ρ(g′) = ρ_{a}(g)ρ_{α}(g′).

2) α(gg′)ρ(gg′) = ρ′(gg′) = ρ′(g)ρ′(g′) = α(g)ρ(g)α(g′)ρ(g′) = (α(g) + ρ(g)(α(g′)))ρ(g)ρ(g′).

3) α ∼ 0 ⟺ there exists x ∈ V such that α(g) = ρ(g)(x) − x = x^{−1}ρ(g)xρ(g)^{−1} ⟺ ρ_{α}(g) = α(g)ρ(g) = x^{−1}ρ(g)x.

Let G ⊂ GL(n, ℝ) be a linear Lie group. We say that G is ℝ-algebraic group if there exists an algebraic group 𝔾 ⊂ GL(n, ℂ) defined over ℝ with G = 𝔾_{ℝ}. Clearly, given an n-dimensional vector space V , the group GL(V) can be identified with GL(n, ℝ) (and hence becomes ℝ-algebraic) after choosing a basis in V . Also, Aff(V) viewed as a subgroup of GL(n + 1, ℝ) is ℝ-algebraic.

Given a subgroup G ⊂ GL(n, ℝ) we will let A(G) ⊂ GL(n, ℝ) denote the algebraic hull of G, i.e. the smallest ℝ-algebraic subgroup of GL(n, ℝ) that contains G.

Note that any ℝ-algebraic group has finite number of connected components. Intersection of two ℝ-algebraic subgroups is ℝ-algebraic. Factorgroup 𝔾_{ℝ}/ℍ_{ℝ} of an ℝ-algebraic group 𝔾_{ℝ} over its normal ℝ-algebraic subgroup ℍ_{ℝ} is a finite index subgroup of the ℝ-algebraic group (𝔾/ℍ)_{ℝ} (Example: SL(2, ℝ)/ℤ_{2} ≃ SO(1, 2)_{0} ⊂ SL(3, ℝ)).

Lemma 1.2. Let F be a topological group. Then all the first cohomology groups of F are trivial iff given any homomorphism ϕ from F to an ℝ-algebraic group H, the group A(ϕ(F)) is reductive. Proof. ⇒. Let A(ϕ(F)) = M·U ⊂ H be a Levi decomposition, where M is reductive and U is the unipotent radical. Let V = U/[U, U], and let q: M → GL(V) be the representation given by the action of M on U/[U, U] by conjugations. Take the composite homomorphism ρ: F → M·U → q(M)·V . Let π: q(M)·V → q(M) be the projection. Since H^{1}(F, π ∘ ρ) = 0, it follows that ρ(F) is conjugate to π(ρ(F)) ⊂ q(M). But ρ(F) is Zariski dense in q(M)·V . We derive that V is trivial and hence so is U.

⇐. Take ρ: F → GL(V) and let ρ′: F → Aff(V) be a covering homomorphism, i.e., ρ = π ∘ ρ′. The groups A(ρ′(F)) and A(ρ(F)) are reductive and V is the unipotent radical of A(ρ′(F))·V = A(ρ(F))·V . Hence A(ρ′(F)) and A(ρ(F)) are conjugate by an element of V .

Remark. In the statement of the Lemma one can say more about M = A(ϕ(F)). It is either a finite group or a semisimple group (with finite number of connected components). In fact, M is an almost direct product of semisimple group L and abelian group A, i.e. L and A commute, M = LA, and the intersection L ∩ A is discrete (hence finite). Note that the image of ϕ(Γ) in M/L = A/A∩L is Zariski dense. On the other hand, it should be finite (otherwise H^{1}(Γ, τ_{1}) ≠ 0). Hence A is finite.

Lemma 1.3. Let F be a topological group, F′ ⊂ F be a subgroup of finite index in F, and ρ: F → GL(V) be a representation. Assume that H^{1}(F′, ρ) = 0. Then H^{1}(F, ρ) = 0. Proof. Let ϕ: F → H be a homomorphism to an ℝ-algebraic group. Note that A(ϕ(F′)) is of finite index in A(ϕ(F)). Now the claim follows from Lemma 1.2.

Remark . The converse statement is wrong. Take, for instance F = PSL(2, ℤ). It is well known that F is the free product of finite groups ℤ_{2} and ℤ_{3} . Hence H^{1}(F, τ_{1}) = 0. On the other hand, F contains a subgroup F′ of index 6 which is a free group with two generators. Hence H^{1}(F′, τ_{1}) ≃ ℝ^{2} .

Lemma 1.4. Let C be a finite normal subgroup in a topological group F, and let F′ = F/C. Then all the first cohomology groups of F′ are trivial iff all the first cohomology groups of F are trivial. Proof. ⇒. Let ϕ: F → H be a homomorphism to an ℝ-algebraic group. The group A(ϕ(F))/ϕ(C) is a finite index subgroup of some ℝ-algebraic group H′. Since the image of F′ in H′ is Zariski dense, it follows that H′ is reductive. Since ϕ(C) is finite, it follows that A(ϕ(F)) is also reductive.

⇐. Obvious.

Lemma 1.5. Let F be a product of two commuting subgroups F_{1} and F_{2} . Assume that all the first cohomology groups of F_{1} and F_{2} are trivial. Then all the first cohomology groups of F are trivial. Proof. Let ϕ: F → H be a homomorphism to an ℝ-algebraic group. Then A(ϕ(F) = A(ϕ(F_{1}))A(ϕ(F_{2})). Since the groups A(ϕ(F_{1})) and A(ϕ(F_{2})) are reductive and commute, it follows that A(ϕ(F)) is also reductive.

Lemma 1.6. Let Z be a central subgroup of F and let F′ = F/Z. Assume that all the first cohomology groups of F′ are trivial and that H^{1}(F, τ_{1}) = 0. Then all the first cohomology groups of F are trivial. Proof. Let ϕ: F → H be a homomorphism to an ℝ-algebraic group. We know that the group A(ϕ(F))/A(ϕ(Z)) is a finite index subgroup of a reductive ℝ-algebraic group and hence is reductive itself. It follows that the unipotent radical U of A(ϕ(F)) lies in A(ϕ(Z)). Hence A(ϕ(F)) is the direct product of U with a reductive subgroup. Assume that U is nontrivial. The projection of ϕ(F) into U is Zariski dense therein, and hence is an infinite finitely generated nilpotent group. This contradicts the equality H^{1}(F, τ_{1}) = 0.

§2. Reduction to the rank one case

Now we are ready to apply our observations to the study of the first cohomology group for lattices in semisimple Lie groups. From now on, L is a connected semisimple Lie group and L = SK is the almost direct decomposition into noncompact and compact parts (i.e., K is compact and all simple components of S are noncompact).

Proposition 2.1. Let Γ be a lattice in a connected semisimple Lie group L = SK, p: L → S/S ∩ K be the projection, and S/S ∩ K = ∏_{1}^{n} S_{i} be the (almost direct) decomposition into irreducible components with respect to p(Γ). Assume that all the first cohomology groups of p(Γ) ∩ S_{i} are trivial for all i. Then all the first cohomology groups of Γ are trivial.

Proof. By Lemma 1.5, all the first cohomology groups of ∏_{1}^{n}(p(Γ) ∩ S_{i}) are trivial. But ∏_{1}^{n}(p(Γ) ∩ S_{i}) is a normal subgroup of finite index in p(Γ). Hence by Lemma 1.3, all the first cohomology groups of p(Γ) are trivial. On the other hand, p(Γ) = Γ/Γ ∩ K, where Γ ∩ K is a finite normal subgroup of Γ. Now by Lemma 1.4, all the first cohomology groups of Γ are trivial.

Now we need the following nontrivial result which can be derived from papers of Deligne [2] and Raghunathan [8] and Margulis Arithmeticity Theorem.

Theorem 2.2. (See [4, Assertion IX.6.18 (B)]). Let S be a connected semisimple Lie group without compact factors, and let Γ be an irreducible lattice in S. Assume that rk_{ℝ} S ≥ 2. Then the group Γ/[Γ, Γ] is finite and hence H^{1}(Γ, τ_{1}) = 0.

Corollary 2.3. Let S be a connected semisimple Lie group without compact factors and let Γ be an irreducible lattice in S. Assume that rk_{ℝ} S ≥ 2. Then all the first cohomology groups of Γ are trivial.

Proof. Let Z(S) be the center of S. Then Z = Γ ∩ Z(S) is of finite index in Z(S). Hence Γ′ = Γ/Z is an irreducible lattice in the group S′ = S/Z with finite center. Now the claim follows from Theorems 1 and 2.2 and Lemma 1.6.

Corollary 2.4. Let L be a connected semisimple Lie group with a lattice Γ. Assume that there is no epimorphism ϕ: L → H onto a semisimple Lie group H such that rk_{ℝ} H = 1 and ϕ(Γ) is a lattice in H. Then all the first cohomology groups of Γ are trivial.

Proof. Let p: L = SK → S/S ∩ K be the projection along the compact part, and let S/S ∩ K = ∏_{1}^{n} S_{i} be the (almost direct) decomposition into irreducible components with respect to p(Γ). By assumption, rk_{ℝ} S_{i} ≥ 2 for all i. Now the claim follows from Proposition 2.1 and Corollary 2.3.

§3. Rank one case

Here we prove the following result.

Theorem 3.1. Let G be either Sp(1, n), n ≥ 2, or F_{4}^{−20} , and let Γ be a lattice in G. Then H^{1}(Γ, ρ) = 0 for any representation ρ of Γ.

It is well known that G as above is simply connected, ℝ-algebraic (in the natural linear presentation) and of ℝ-rank one. The center of Sp(1, n) is isomorphic to ℤ_{2} , and that of F_{4}^{−20} is trivial. Besides, both G and Γ are Kazhdan groups. In particular, H^{1}(Γ, τ_{1}) = 0 (see [10] or [4]).

To prove Theorem 3.1, we use Superrigidity and Arithmeticity Theorems of Corlette [1] and Gromov and Schoen [3] and combine them with results of Raghunathan [6, 7].

Theorem 3.2. (Superrigidity), [1] Let G and Γ be as in Theorem 3.1. Assume that

1) H is a connected semisimple Lie group,

2) H is center-free,

3) H has no connected compact normal subgroups,

and let ρ: Γ → H be a homomorphism such that ρ(Γ) is Zariski dense in H. Then there exists an epimorphism R: G → H that extends ρ.

Theorem 3.3. (Arithmeticity), [3] Let G and Γ be as in Theorem 3.1. Then there exists a ℚ-defined semisimple subgroup ℍ ⊂ SL(k, ℂ) and an epimorphism ϕ: ℍ_{ℝ} → G with compact kernel such that ϕ(ℍ_{ℤ}) and Γ are commensurable.

Now we formulate two results of Raghunathan which (modulo the Arithmeticity Theorem) essentially say that H^{1}(Γ, ρ) = 0 whenever ρ can be extended to a representation of G.

Theorem 3.4. [6] Let G be a connected Lie group which has no components locally isomorphic to SO(1, n) and SU(1, n), and let R: G → GL(V) be a nontrivial irreducible representation. Let Γ be a uniform lattice in G. Then H^{1}(Γ, R) = 0.

Corollary 3.5. Let G = S × K, where S is either Sp(1, n), n ≥ 2, or F_{4}^{−20} , and K is a compact semisimple Lie group. Then given a uniform lattice Γ ⊂ G and a representation R: G → GL(V) we have H^{1}(Γ, R) = 0. Proof. Since Γ is a Kazhdan group, it follows that H^{1}(Γ, τ_{1}) = 0. Now the claim follows by decomposition of R into irreducible representations.

Theorem 3.6. [7] Let 𝔾 ⊂ GL(n, ℂ) be a ℚ-defined ℚ-simple algebraic Lie group with rk_{ℚ} 𝔾 = 1. Assume that 𝔾 has no components locally isomorphic to SL(n, ℂ) and 𝔾_{ℝ} has no components locally isomorphic to SO(1, n). Assume that Γ ⊂ 𝔾_{ℝ} is an arithmetic group (i.e., Γ is commensurable with 𝔾_{ℤ}). Let R: 𝔾 → GL(N, ℂ) be a nontrivial irreducible rational representation. Then H^{1}(Γ, R) = 0.

Corollary 3.7. Let G be either Sp(1, n), n ≥ 2, or F_{4}^{−20} , let Γ ⊂ G be a non-uniform lattice, and let R: G → GL(V) be a representation. Then H^{1}(Γ, R) = 0. Proof. Since Γ is non-uniform and G is simply connected, by Arithmeticity Theorem 3.3 there exists a ℚ-defined group ℍ ⊂ GL(k, ℂ) such that ℍ_{ℝ} ≃ G and Γ is commensurable with ℍ_{ℤ} (compact factors of ℍ_{ℝ} can be avoided as in the higher rank case, cf. [12, Corollary 6.1.10]). Clearly, R can be extended from ℍ_{ℝ} to a rational representation R: ℍ → GL(V_{ℂ}). Since Γ is a Kazhdan group, it follows that H^{1}(Γ, τ_{1}) = 0. Now decomposing R into irreducible representations we derive from Theorem 3.6 that H^{1}(Γ, R) = 0. Note that R|_{Γ} is isomporphic to the direct sum of two copies of R|_{Γ} . Hence H^{1}(Γ, R) = 0.

Now we proceed to the proof of Theorem 3.1. It suffices to prove that the Zariski closure F = A(ϕ(Γ)) ⊂ H is reductive for any homomorphism ϕ: Γ → H to an ℝ-algebraic group. Let F = (LA)·U be a Levi decomposition, where U is the unipotent radical, L is a semisimple subgroup (with finite number of connected components), and A is a finite abelian subgroup (see Remark after Lemma 1.2). Let L_{0} stand for the identity component of L. Replacing Γ with its finite index subgroup (see Lemma 1.3), we can assume that A is trivial and ϕ(Γ) ⊂ L_{0} .

If L_{0} is trivial then ϕ(Γ) ⊂ U. But Γ is a Kazhdan group and hence ϕ(Γ) is trivial and there is nothing to prove. So we may assume that L_{0} is nontrivial. We have to prove that U is trivial.

Let V = U/[U, U], and let q: L → GL(V) be the natural representation. By α: F = L·U → q(L)·V we denote the composite epimorphism and by π: q(L)·V → q(L) the projection. Let ρ′ = α ∘ ϕ and ρ = π ∘ ρ′. Then ρ′ is a covering homomorpshism for the representation ρ of Γ on the vector space V .

We know that ρ′(Γ) is Zariski dense in q(L)·V , and ρ(Γ) is Zariski dense in q(L). Assume that H^{1}(Γ, ρ) = 0. Then ρ(Γ) is conjugate to ρ′(Γ) and hence V is trivial, so U is also trivial. Now it suffices to prove that H^{1}(Γ, ρ) = 0.

Let q(L) = SK be the almost direct decomposition of q(L) into noncompact and compact parts.

Assume first that S is trivial and hence q(L) = K. The group Γ̂ = {(γ, ρ(γ)), γ ∈ Γ} is a lattice in G × K that projects densely into K. Hence Γ̂ is a uniform lattice in G × K. Let ρ̂: Γ̂ → GL(V) be given by ρ̂(γ, ρ(γ)) = ρ(γ), γ ∈ Γ. Note that ρ̂ extends to a representation G × K → GL(V) given by the projection onto the second factor. By Corollary 3.5 it follows that H^{1}(Γ, ρ) = H^{1}(Γ̂, ρ̂) = 0, and we are done.

Now we assume that S is nontrivial. Let Z(S) be the center of S, σ: q(L) = SK → Ad S × K be the natural epimorphism, and p_{S}: Ad S × K → Ad S be the projection. Then p_{S} ∘ σ ∘ ρ(Γ) is Zariski dense in Ad S and by Superrigidity Theorem 3.2, it follows that p_{S} ∘ σ ∘ ρ extends to an epimorphism R: G → Ad S. It follows that R(Γ) = p_{S} ∘ σ ∘ ρ(Γ) is a lattice in Ad S. Since G has finite center, the kernel of R is finite and hence the intersection K ∩ σ ∘ ρ(Γ) is finite. It follows that σ ∘ ρ(Γ) is a lattice in Ad S × K and hence ρ(Γ) is a Zariski dense lattice in q(L).

Let i: q(L) → GL(V) be the inclusion. We know that ρ has finite kernel. Now it suffices to prove that H^{1}(ρ(Γ), i) = 0.

Assume that ρ(Γ) is a uniform lattice in q(L). Since the natural representation of ρ(Γ) on V extends to the natural representation of q(L) on V , we can apply Corollary 3.5 and conclude that H^{1}(ρ(Γ), i) = 0 (to apply the Corollary directly it suffices to lift the representation of q(L) = SK to that of G × K).

Assume now that ρ(Γ) is a nonuniform lattice in q(L). Then q(L) = S and by Corollary 3.7, H^{1}(ρ(Γ), i) = 0 as well.

Proof of Theorem 2. Any rank one simple Lie group is locally isomorphic to one of the following: Sp(1, n), F_{4}^{−20} , SO(1, n), SU(1, n). Now Theorem 2 can be derived from Theorems 1 and 3.1 in the same manner as Corollary 2.4.

§4. Applications of the main result

Superrigidity. As before, given a semisimple Lie group L, we denote by L = SK an almost direct decomposition of L into noncompact and compact parts. Clearly, a lattice Γ ⊂ G is Zariski dense in G (in the Ad-representation) iff G = \overline{SΓ} (here and in what follows the horizontal bar stands for the closure of a subset in a topological space).

We recall Margulis Superrigidity Theorem for irreducible lattices in higher rank semisimple Lie groups.

Theorem 4.1. [4] Let L be a connected semisimple Lie group with rk_{ℝ} L ≥ 2. Let Γ ⊂ L be an irreducible Zariski dense lattice, and let ϕ: Γ → H be a homomorphism such that

1) H is a connected semisimple Lie group,

2) H is center-free,

3) H has no connected compact normal subgroups.

Then ϕ uniquely extends to a continuous homomorphism ϕ̃: L → H whenever ϕ(Γ) is Zariski dense in H.

Remark . Assume that H is an ℝ-algebraic group and ϕ(Γ) is Zariski dense in H. Then H is either finite or semisimple because all the first cohomologies of Γ vanish. So condition 1) can be replaced by saying that H is an ℝ-algebraic group and ϕ(Γ) ⊂ H_{0} . Conditions 2) and 3) cannot be omitted.

To generalize this Theorem, we use a terminology from [11]. We will say that a connected Lie group G is algebraically simply connected if G admits a faithful finite-dimensional representation, and no proper covering group of G admits such a representation. Note that SL(2, ℝ) is algebraically simply connected but not (topologically) simply connected. Clearly, a semisimple Lie group L is algebraically simply connected iff it decomposes into direct product of simple algebraically simply connected normal subgroups.

We say that a homomorphism ϕ: Γ → H virtually extends to a continuous homomorphism ϕ̃: G → H if there exists a finite index subgroup Γ′ ⊂ Γ such that ϕ̃|_{Γ′} ≡ ϕ|_{Γ′}.

Finally, we say that a lattice Γ in a connected Lie group G is superrigid if every homomorphism ϕ: Γ → H, where H is an ℝ-algebraic group containing no connected simple compact normal subgroups, virtually extends to a continuous homomorphism ϕ̃: G → H whenever ϕ(Γ) is Zariski dense in H.

In these terms we can formulate the following Superrigidity Theorem.

Theorem 4.2. Let L be an algebraically simply connected semisimple Lie group, and Γ be a lattice in L such that (L, Γ) satisfies (∗). Then Γ is superrigid in L.

Remark. If Γ is a Zariski dense irreducible lattice in L and rk_{ℝ} L ≥ 2, the result is contained in [4, Theorem IX.5.12].

Proof. Since L is algebraically simply connected, it follows that L = S × K, where S is the noncompact part, and K is the compact part. Let p_{S}: L → S be the projection along K. Then p_{S}(Γ) is a lattice in S, and Γ ∩ K is finite normal subgroup in Γ.

Let ϕ: Γ → H be a homomorphism to an ℝ-algebraic group as in the definition of superrigidity. Then it follows from Theorem 2 that H_{0} is either trivial or semisimple group with finite center Z(H). Replacing Γ with its finite index subgroup (if necessary), we can assume that ϕ(Γ) ⊂ H_{0} . Then ϕ(Γ∩K) ⊂ Z(H). Let Ad_{H}: H → Ad H = H/Z(H) be the adjoint representation and define ϕ′ = Ad_{H} ∘ϕ. Since p_{S}(Γ) ≃ Γ/Γ ∩ K and Γ ∩ K ⊂ Ker(ϕ′), there exists a homomorphism ϕ′′: p_{S}(Γ) → Ad H such that ϕ′ = ϕ′′ ∘ p_{S}.

Let S = ∏_{i} S_{i} be the decomposition of S into irreducible components with respect to p_{S}(Γ), and Γ_{i} = p_{S}(Γ)∩S_{i}. Again, replacing Γ by its finite index subgroup, we can assume that ∏_{i} Γ_{i} = p_{S}(Γ). Notice that Ad H is a center-free semisimple Lie group with no connected compact normal subgroups. Applying Theorem 4.1 to the homomorphism ϕ′′: Γ_{i} → Ad H if rk_{ℝ} S_{i} ≥ 2 or Theorem 3.2 if rk_{ℝ} S_{i} = 1, we obtain a continuous homomorphism ϕ̃′′: S → Ad H that extends ϕ′′.

Since S is algebraically simply connected, it follows that there exists a continuous homomorphism ϕ̃: S → H such that Ad_{H} ∘ϕ̃ = ϕ̃′′. Now we extend ϕ̃ onto L = S ×K defining it trivially on K. Then Ad_{H} ∘ϕ̃|_{Γ} ≡ Ad_{H} ∘ϕ. Define a map σ: Γ → Z(H) by σ(g) = ϕ̃^{−1}(γ)ϕ(γ), γ ∈ Γ. Then σ is a homomorphism and hence ϕ̃ agrees with ϕ on finite index subgroup Ker σ ⊂ Γ.

Now we proceed to general Lie groups. Given a connected Lie group G, we denote by R its radical and by L its Levi subgroup. Then G = LR is a Levi decomposition of G. In general, L and R may intersect in a discrete central subgroup of G.

Let K be the compact part of Levi subgroup L ⊂ G. Then for any lattice Γ ⊂ G the product KRΓ is closed (see, e.g., [10]) and hence Γ ∩ KR is a uniform lattice in KR. Moreover, the product RΓ is closed whenever G = \overline{SRΓ}. The last condition is obviously verified whenever Ad(Γ) is Zariski dense in Ad G.

Remark. Let α: G → G/KR be the factor map. Then α(Γ) is a lattice in semisimple Lie group G/KR. Let ϕ: G → H be an epimorphism, where H is locally isomorphic to SO(1, n) or SU(1, n), such that ϕ(Γ) is a lattice. Factorizing H over its center if necessary, we can assume that H is center-free. Clearly, ϕ(Γ ∩ KR) is a normal amenable subgroup of the lattice ϕ(Γ). Since H is center-free, it follows that ϕ(Γ∩KR) is trivial. We derive that (G, Γ) satisfies (∗) iff (G/KR, α(Γ)) does so.

One knows that G is algebraically simply connected iff the radical R of G is (topologically) simply connected and G = L·R, where L is an algebraically simply connected Levi subgroup. Recall that Witte [11] proved superrigidity of Zariski dense lattices in solvable Lie groups. Now we can derive the following general result.

Theorem 4.3. Let Γ be a lattice in an algebraically simply connected Lie group G. Assume that 1) Ad(Γ) is Zariski dense in Ad G,

2) (G, Γ) satisfies (∗). Then Γ is superrigid in G.

Proof. From 1) it follows that Γ∩R is a lattice in R. ¿From 2) and Theorem 4.2 it follows that the image of Γ in G/R is superrigid. This together with condition 1) enables one to conclude that Γ is superrigid in G (see [11, Corollary 9.10]).

Levi–Mostow Splitting. Now we apply Theorem 2 to obtain a general result on Levi–Mostow splitting of lattices. Recall a result of Mostow [5] presented as in [11].

Theorem 4.4. Cf [11]. Let G be a connected Lie group with a lattice Γ. Assume that

1) G is algebraically simply connected Lie group,

2) G has no connected compact semisimple normal subgroups,

3) Ad(Γ) is Zariski dense in Ad G,

4) the image of Γ in G/R is a superrigid lattice. Then there exists a Levi subgroup L ⊂ G such that (Γ ∩ L)(Γ ∩ R) has finite index in Γ.

We will prove more general result.

Theorem 4.5. Let G be a connected Lie group with a lattice Γ. Assume that (G, Γ) satisfies (∗). Then there exists a Levi subgroup L = SK ⊂ G such that (Γ∩L)(Γ∩KR) has finite index in Γ. Moreover, if G = \overline{SRΓ} then (Γ∩L)(Γ∩R) has finite index in Γ.

Proof. It suffices to find a Levi subgroup L that intersects Γ in a lattice. Define G′ = (\overline{SRΓ})_{0} . Then G′ is a connected Lie group such that G′ = L′R′, where L′ = SK′ is a Levi subgroup, R′ = T R is the radical, K′ ⊂ K is a compact semisimple subgroup, and T ⊂ K is a torus. Clearly, Γ′ = Γ ∩ G′ is a lattice in G′ such that (G′, Γ′) satisfies (∗) and G′ = \overline{SR′Γ′}.

Therefore, we can assume that G = \overline{SRΓ}. Moreover, passing to the universal cover of G if necessary, we can assume that G is simply connected. Then G = L·R and L = S ×K. Let C ⊂ K be the maximal connected compact semisimple normal subgroup of G. Then G = G′ × C, where G′ = L′·R and L′ ⊂ L. Let p: G → G′ be the projection along C, and Γ′ = p(Γ). Then (G′, Γ′) satisfies (∗) and hence with no loss of generality we can assume that G contains no connected compact semisimple normal subgroups.

From the condition G = \overline{SRΓ} it follows that Γ ∩ R is a lattice in R. It is fairly well known (see, e.g., [11, Proposition 8.2]) that one can “straighten” the radical R to make Ad(Γ ∩ R) Zariski dense in Ad(R). More precisely, there exists a finite index subgroup Γ′ ⊂ Γ and a simply connected Lie group G′ = L·R′ such that Γ′ is a lattice in G′ and Ad(Γ′ ∩ R′) is Zariski dense in Ad(R′) (moreover, the nilradical of R lies in the nilradical of R′).

Hence with no loss of generality we can assume that Ad(Γ∩R) is Zariski dense in Ad(R). Since G = \overline{SRΓ}, it follows that Ad(Γ) is Zariski dense in Ad G.

Hence the product of Γ with the center Z(G) is closed. Let L′ be the linearizer of L, i.e. the maximal algebraically simply connected factor of L. Then L′ = L/Z, where Z ⊂ L ∩ Z(G).

In general, the product ZΓ need not be closed. Clearly, M = (\overline{ZΓ})_{0} is a connected central subgroup of G and hence M ⊂ Z(G) ∩ R. Let G′ = G/ZM, and α: G → G′ = L′·R′ be the factor map, where R′ = R/M. Then Γ′ = α(Γ) is a Zariski dense lattice in G′ such that (G′, Γ′) satisfies (∗). Moreover, G′ is algebraically simply connected, and it follows from Theorem 4.2 that G′ and Γ′ satisfy all the conditions 1)–4) of Theorem 4.4. Hence there exists a choice of L′ such that (Γ′ ∩ L′)(Γ′ ∩ R′) has finite index in Γ′.

It follows that Γ ∩ LM is a lattice in LM. Let p: L × M → L be the projection. Then p(Γ ∩ LM) is a superrigid lattice in L. Hence [Γ ∩ LM, Γ ∩ LM] = [p(Γ ∩ LM), p(Γ ∩ LM)] has finite index in p(Γ ∩ LM). It follows that Γ ∩ L is a lattice in L and we are done.

Remark. One can bypass few steps in the proof (in particular, “straightening” of R) using a result saying that for any lattice Γ in a connected Lie group G, the intersection Γ ∩ LN is a lattice in LN, N ⊂ R being the nilradical (cf. [9]). So one can concentrate on the Lie group LN.

Necessity of condition (∗) for vanishing of the first cohomologies. It is well known that for any n ≥ 2 there exists a lattice Γ in SO(1, n) with its factorgroup Γ/[Γ, Γ] infinite. Hence the group H^{1}(Γ, τ_{1}) = Hom(Γ, ℝ) is nontrivial. The same is valid for SU(1, n).

Moreover, there is a conjecture (called the Millson problem) saying that given any lattice Γ in SO(1, n) or SU(1, n), there exists a finite index normal subgroup Γ′ ⊂ Γ such that the group Γ′/[Γ′, Γ′] is infinite (see [10] for more details). It implies that the factor group F = Γ/[Γ′, Γ′] is an infinite (finitely generated) virtually abelian group, i.e. F contains a normal subgroup A ≃ ℤ^{k} for some k ≥ 1, of finite index. It is easy to construct an embedding to an ℝalgebraic group ϕ: F → H = C·U, where U ≃ ℝ^{k} is the unipotent radical and C ≃ F/A, such that ϕ(F) is Zariski dense in H. According to Lemma 1.2, this means that one can construct a representation ρ of Γ with nontrivial H^{1}(Γ, ρ).

We conclude the following.

Proposition 4.6. Let Γ be a lattice in a connected semisimple Lie group L. If the Millson problem has positive solution, then the following conditions are equivalent:

1) H^{1}(Γ, ρ) = 0 for any representation ρ of Γ,

2) Γ admits no epimorphism onto an infinite virtually abelian group,

3) (L, Γ) satisfies (∗).

Using Theorem 4.5 one can easily derive the following.

Proposition 4.7. Let Γ be a lattice in a connected Lie group G. If the Millson problem has positive solution, then the following conditions are equivalent:

1) H^{1}(Γ, ρ) = 0 for any representation ρ of Γ,

2) Γ admits no epimorphism onto an infinite virtually abelian group, 3) (G, Γ) satisfies (∗), Levi subgroup L ⊂ G is normal, and the factor-

group G/L is compact.

References

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A. N. Starkov All-Russian Institute of Electrotechnics 143500, Istra, Moscow Region Russia RDIEalex@istra.ru and Dept. of Mechanics and Mathematics Moscow State University, 117234 Moscow Russia.

Received May 10, 2001 and in final form November 10, 2001