Espina de Pescado

simple algebraic groups such as G = SLd(R), d ≥ 3. In that respect, Bader–

Shalom’s NST 5.4and Margulis’s NST 5.6are complementary.

The strategy of the proof of Theorem 5.4follows Margulis’s remarkable idea. Let N C Γ be any nontrivial normal subgroup. In order to show that N has finite index in Γ, we will prove that the quotient group Γ/N has property (T) (see Theorem5.8) and is amenable (see Theorem5.15). Using Proposition 2.27, it will follow that Γ/N is finite. The rest of this chapter is devoted to proving the “property (T) half” and the “amenability half” of Theorem5.4.

Remark 5.7. Note that the assumption that (G1, G2) 6= (R, R) is nec-

essary. Indeed, regard Z2 < R2 as a uniform irreducible lattice via the irra- tional embedding ι : Z2 → R2_{: (m, n) 7→ (m−n}√_{2, m+n}√_{2). Then ZCZ}2_is

an infinite index nontrivial normal subgroup where Z ,→ Z2: m 7→ (m, m). 1. Property (T) half

The main result of this section provides a complete characterization of when factors of irreducible lattices in product groups have property (T). This is the “property (T) half” of Theorem5.4 which is due to Shalom.

Theorem 5.8 (Shalom [Sh99]). For every i ∈ {1, 2}, let Gi be any

compactly generated lcsc group. Set G = G1× G2 and denote by pi : G → Gi

the canonical factor map. Let Γ < G be any finitely generated L2-integrable weakly uniform irreducible lattice and N C Γ any normal subgroup.

Then Γ/N has property (T) if and only if the following two conditions are satisfied:

(i) For every i ∈ {1, 2}, Gi/pi(N ) has property (T).

(ii) Every continuous homomorphism ϕ : G → C that vanishes on N is identically zero.

It is easy to see that conditions (i) and (ii) are necessary. The heart of the proof will consist in showing that conditions (i) and (ii) are also sufficient. We will use results from Chapter 4.

Before proving Theorem 5.8, we need some preparation. For every i ∈ {1, 2}, let Gi be any lcsc group and set G = G1× G2. Let Γ < G be any

irreducible lattice. Let p1 : G → G1 be the canonical factor map.

Definition 5.9. Let π : Γ → U (Hπ) be any unitary representation with

Hπ separable. We say that ξ ∈ Hπ is G1-continuous if for every sequence

(γn)n in Γ such that p1(γn) → e in G1, we have limnkπ(γn)ξ − ξk = 0. We

denote by H1 ⊂ Hπ the subset of all G1-continuous elements.

The following result allows us to regard the subset H1 ⊂ Hπ of G1-

continuous vectors as the subspace (H

b

π)G2 ⊂ Hbπ of bπ(G2)-invariant vectors

in the induced G-space.

Proposition 5.10. Keep the same notation as above. The following assertions hold.

(i) The subset H1 ⊂ Hπ is a π(Γ)-invariant closed susbspace of Hπ.

(ii) The unitary representation π : Γ → U (H1) extends to a strongly

continuous unitary representation π : G → U (H1) for which G2

acts trivially.

(iii) Denote by π : G → U (H_b

b

π) the induced representation. Then the

linear map

W : H1→ (Hπb)

G2 _{: ξ 7→ (g 7→ π(g}−1_)ξ)

is a G-equivariant unitary operator in the sense that W (π(g)ξ) = b

π(g)W (ξ) for every g ∈ G and every ξ ∈ H1.

Proof. (i) It is clear that H1 ⊂ Hπ is a subspace. Let ξ ∈ Hπ and

(ξk)k be any sequence in H1 such that limkkξ − ξkk = 0. Let (γn)n be any

sequence in Γ such that p1(γn) → e. Let ε > 0 and choose k ∈ N such that

kξ − ξkk ≤ ε. Since ξk∈ H1, we have limnkπ(γn)ξk− ξkk = 0. This implies

that lim sup_nkπ(γn)ξ − ξk ≤ 2ε. Since this is true for every ε > 0, it follows

that ξ ∈ H1 and hence H1 ⊂ Hπ is closed subspace. Let now ξ ∈ H1 and

γ ∈ Γ. Let (γn)n be any sequence in Γ such that p1(γn) → e. Then we have

p1(γγnγ−1) → e. Since ξ ∈ H1, we have

kπ(γn) π(γ)ξ − π(γ)ξk = kπ(γγnγ−1)ξ − ξk → 0.

This shows that H1⊂ Hπ is π(Γ)-invariant.

(ii) Set Λ =. (m, n) ∈ N2| m ≤ n . Let g ∈ G be any element and choose a sequence (γn)n in Γ so that p1(γn) → p1(g). Define the sequence

(γ(m, n))(m,n)∈Λ in Γ by γ(m, n) = γn−1γm for every (m, n) ∈ Λ. Then

p1(γ(m, n)) → e as (m, n) → (+∞, +∞). This implies that for every ξ ∈ H1,

we have

kπ(γn)ξ − π(γm)ξk = kξ − π(γ(m, n))ξk → 0 as (m, n) → (+∞, +∞)

and so the sequence (π(γn)ξ)n is Cauchy in H1. Since H1 ⊂ Hπ is a closed

subspace, there exists ξg ∈ H1 such that limnkξg − π(γn)ξk = 0. It is

immediate to check that ξg ∈ H1 does not depend on the choice of the

sequence (γn)n for which p1(γn) → p1(g). Define

ρ : G → U (H1) : g 7→ (ξ 7→ ξg) .

We show that ρ : G → U (H1) is a strongly continuous unitary representation

such that ρ|Γ= π and ρ|G2 = 1H1. Firstly, let g ∈ G and choose a sequence

(γn)n in Γ so that p1(γn) → p1(g). For all ξ, η ∈ H1, we have

hρ(g)ξ, ρ(g)ηi = hξg, ηgi = lim

n hπ(γn)ξ, ηgi = limn hξ, π(γn) ∗

ηgi = hξ, ηi.

Then ρ(g) ∈ U (H1). Next, let g1, g2 ∈ G and choose sequences (γ1,n)n

and (γ2,n)n in Γ so that p1(γ1,n) → p1(g1) and p1(γ2,n) → p1(g2). Then

p1(γ1,nγ2,n) → p1(g1g2). Then for all ξ ∈ H1, we have

kρ(g1g2)ξ − ρ(g1)ρ(g2)ξk = lim n kπ(γ1,nγ2,n)ξ − ρ(g1)ρ(g2)ξk = lim n kπ(γ2,n)ξ − π(γ1,n) ∗_ρ(g 1) ρ(g2)ξk = 0.

1. PROPERTY (T) HALF 103

Then ρ : G → U (H1) is a unitary representation. Let us prove that

ρ : G → U (H1) is continuous at e ∈ G. This will imply that ρ : G →

U (H1) is strongly continuous. Let ξ ∈ H1 be any element. Let (gk)k

be any sequence in G such that gk → e. By contradiction, assume that

limkkρ(gk)ξ − ξk 6= 0. Up to taking a subsequence, we may assume that

ε = inf {kρ(g. k)ξ − ξk | k ∈ N} > 0. For every k ∈ N, choose a sequence

(γk,n)n in Γ such that p1(γk,n) → p1(gk) as n → +∞. Choose a sequence

of open neighborhoods Om ⊂ G1 of e ∈ G1 such that T_m∈NOm = {e}.

Since p1(gk) → e as k → +∞, we can find increasing sequences (km)m and

(nm)m in N such that p1(gkm) ∈ Om, p1(γkm,nm) ∈ Om and kρ(gkm)ξ −

π(γkm,nm)ξk < ε/2 for every m ∈ N. Then we have p1(γkm,nm) → e as

m → +∞ and hence limmkξ − π(γkm,nm)ξk = 0. This further implies

that lim sup_mkρ(g_km)ξ − ξk ≤ ε/2, a contradiction. Thus, ρ : G → U (H1)

is a strongly continuous unitary representation. By construction, we have ρ(γ) = π(γ) for every γ ∈ Γ and ρ(g) = 1H1 for every g ∈ G2.

(iii) We adopt the second viewpoint on induction. Fix a Haar measure mG on G and a Borel fundamental domain F ⊂ G so that G = F · Γ

and 0 < mG(F ) < +∞. Recall that we view Hbπ as the Hilbert space of

mG-equivalence classes of all measurable functions η : G → Hπ that satisfy

• For m_G-almost every g ∈ G and every γ ∈ Γ, η(gγ−1) = π(γ)η(g). • R

Fkη(g)k2dmG(g) < +∞.

The map W : H1 → (Hbπ)

G2 _{clearly preserves inner products and is G-}

equivariant in the sense that W (π(g)ξ) = bπ(g)W (ξ) for every g ∈ G and every ξ ∈ H1. It remains to prove that W is surjective. Let η ∈ (Hbπ)

G2 _be

any element.

Claim 5.11. Every essential value of η is an element of H1.

Indeed, let ξ ∈ Hπ be any essential value of η and choose a sequence

(γn)n∈N in Γ such that p1(γn) → e in G1. We show that limnkπ(γn)ξ −

ξk = 0. We may find elements hn ∈ G2 such that γnhn → e in G. Let

ε > 0. Choose a Borel probability measure µ ∈ Prob(G) such that µ ∼ mG and hence the right translation action G y (G, µ) is nonsingular. By

assumption, the measurable subset B = {g ∈ G | kη(g) − ξk < ε} satisfies. µ(B) > 0. Since γnhn → e in G, for any n ∈ N sufficiently large, we have

µ(B ∩ B · (γnhn)−1) > 0. As an element of (Hbπ)

G2_{, the function η : G → H}

π

is left G2-invariant (so right G2-invariant as well since G2C G is a normal

subgroup) and right Γ-equivariant. Thus for every n ∈ N and almost every g ∈ G, we have η(gγnhn) = η(gγn) = π(γn−1)η(g). So for any n ∈ N

sufficiently large, choosing g ∈ B ∩ B · (γnhn)−1, we obtain

kξ−π(γn)ξk ≤ kξ−η(g)k+kη(g)−π(γn)ξk ≤ kξ−η(g)k+kη(gγnhn)−ξk ≤ 2ε.

As ε > 0 can be arbitrarily small, this finishes the proof of Claim 5.11. Using Claim5.11, we may modify η ∈ (H_bπ)G2 on a mG-null set if neces-

sary and regard η : G → H1. Then the measurable function G → H1 : g 7→

Since the subset G2· Γ is dense in G, this implies that the above measurable

function is right G-invariant and hence mG-almost everywhere constant. If

we denote by ξ ∈ H1 its unique essential value, we find that ξ ∈ H1 and

η = W (ξ). 

The next dichotomy result will be one the key ingredients in the proof of Theorem 5.8.

Proposition 5.12 (Shalom [Sh99]). For every i ∈ {1, 2}, let Gi be any

compactly generated lcsc group and set G = G1× G2. Let π : G → U (Hπ) be

any strongly continuous unitary representation. At least one of the following assertions holds:

(i) H1(G, π) = 0.

(ii) There exists i ∈ {1, 2} such that π|Gi is not ergodic.

Proof. For every i ∈ {1, 2}, choose a Borel probability measure µi ∈

Prob(Gi) as in Terminology 4.6. Set µ = µ1⊗ µ2 ∈ Prob(G). Assume that

π|Gi is ergodic for every i ∈ {1, 2}. In order to show that H

1

(G, π) = 0, using Theorem 4.7, it suffices to show that any µ-harmonic 1-cocycle b ∈ Harµ(G, π) is identically zero.

Let b ∈ Harµ(G, π) be any µ-harmonic 1-cocycle. Recall that we have

b(µ)=. R_Gb(g) dµ(g) = 0. For every i ∈ {1, 2}, set b(µi)=.

R Gib(gi) dµi(gi) ∈ Hπ and π(µi) =. R_G iπ(gi) dµi(gi) ∈ B(Hπ). Since G = G1 × G2 and µ = µ1⊗ µ2, we have π(µ)=. Z G π(g) dµ(g) = Z G1×G2 π(g1) π(g2) d(µ1⊗ µ2)(g1, g2) = π(µ1) π(µ2).

Using the 1-cocycle relation, for every (g1, g2) ∈ G1× G2, we have

(5.1) b(g1) + π(g1) b(g2) = b(g1g2) = b(g2g1) = b(g2) + π(g2) b(g1).

By integrating (5.1) against µ = µ1⊗ µ2 ∈ Prob(G) and since b(µ) = 0, we

obtain

b(µ1) + π(µ1) b(µ2) = 0 = b(µ2) + π(µ2) b(µ1).

This implies that

b(µ1) = −π(µ1) b(µ2) = −π(µ1) (−π(µ2) b(µ1)) = π(µ) b(µ1).

Since π is ergodic, we have ker(1 − π(µ)) = {0} and so b(µ1) = 0. Likewise,

we have b(µ2) = 0.

Next, for every (g1, g2) ∈ G1× G2, rewriting (5.1) as

(5.2) (1 − π(g1)) b(g2) = (1 − π(g2)) b(g1).

By integrating (5.2) against µ1∈ Prob(G1), for every g2 ∈ G2, we obtain

(1 − π(µ1)) b(g2) = (1 − π(g2)) b(µ1) = 0.

Since π|G1 is ergodic, we have ker(1 − π(µ1)) = {0} and so b(g2) = 0 for

every g2 ∈ G2. Likewise, we have b(g1) = 0 for every g1 ∈ G1. The 1-cocycle

1. PROPERTY (T) HALF 105

The next result allows to extend homomorphisms ϕ : Γ → C to contin- uous homomorphisms ϕ : G → C.

Proposition 5.13 (Shalom [Sh99]). For every i ∈ {1, 2}, let Gi be any

compactly generated lcsc group and set G = G1× G2. Let Γ < G be any

finitely generated L2-integrable weakly uniform irreducible lattice. Then any homomorphism ϕ : Γ → C extends continuously to ϕ : G → C.

Proof. We can identify the vector space Hom(G, C) of all continuous homomorpshisms G → C with H1(G, 1G) and the vector space Hom(Γ, C)

with H1(Γ, 1Γ). We claim that the restriction map

Hom(G, C) → Hom(Γ, C) : ϕ 7→ ϕ|Γ

is injective. Indeed, let ϕ ∈ Hom(G, C) such that ϕ|Γ= 0. Set H = ker(ϕ)

and observe that H C G is a closed normal subgroup and Γ < H. The factor map G/Γ → G/H : gΓ 7→ gH is well-defined and G-equivariant. Since Γ < G is a lattice, the group G/H carries a G-invariant Borel prob- ability measure and so G/H is a compact group. Considering the well- defined continuous group homomorphism G/H → C : gH 7→ ϕ(g), we have ϕ(G) ∼= G/H and so ϕ(G) is a compact subgroup of C. Since the only compact subgroup of C is {0}, we necessarily have ϕ(G) = 0 and so ϕ ≡ 0. Next, the space Hom(Γ, C) is finite dimensional since Γ is finitely gener- ated. Since Hom(G, C) → Hom(Γ, C) : ϕ 7→ ϕ|Γ is injective, Hom(G, C)

is also finite dimensional. Thus, in order to show that the restriction map Hom(G, C) → Hom(Γ, C) is an isomorphism, it suffices to show that there exists an injection H1(Γ, 1Γ) ,→ H1(G, 1G).

Denote by ν ∈ Prob(G/Γ) the unique G-invariant Borel probability measure and by κG/Γ : G → U (L2(G/Γ, ν)0) the restriction of the quasi-

regular representation λG/Γ to the orthogonal complement L2(G/Γ, ν)0 =

L2(G/Γ, ν) C1G/Γ. Since Γ < G is a weakly uniform lattice, κG/Γ does

not have almost invariant vectors. Then Proposition4.4implies that H1(G, κG/Γ) = H

1

(G, κG/Γ).

Next, we claim that H1(G, κG/Γ) = 0. Indeed, otherwise, applying Propo-

sition 5.12 to π = κ_G/Γ and up to permuting the indices, there exists a unit vector ξ ∈ L2(G/Γ, ν)0 such that κ_G/Γ(g)ξ = ξ for all g ∈ G1. Since

Γ < G1× G2 is irreducible, the pmp action G1y G/Γ is ergodic and so the G1-representation κG/Γ|G1 is ergodic. This is a contradiction.

Therefore, we have H1(G, κG/Γ) = H 1

(G, κG/Γ) = 0. Writing λG/Γ =

κG/Γ⊕ 1G with respect to the orthogonal sum L2(G/Γ, ν) = L2(G/Γ, ν)0⊕

C1G/Γ, it follows that H1(G, λG/Γ) = H1(G, 1G). Next, applying Theo-

rem 4.22 to the trivial Γ-representation π = 1Γ, we have bπ = λG/Γ and H1(Γ, 1Γ) ,→ H1(G, λG/Γ) is injective, which in turn implies that the map

We are ready to prove Theorem 5.8.

Proof of Theorem 5.8. Assume that Γ/N has property (T). Let i ∈ {1, 2}. Since N C Γ is normal and since pi(Γ) < Gi is dense, pi(N ) C Gi is a

closed normal subgroup. Moreover, the well-defined group homomorphism Γ/N → Gi/pi(N ) has dense range. Therefore, Gi/pi(N ) has property (T)

by Proposition2.26. Let now ϕ : G → C be any continuous homomorphism such that ϕ|N = 0. Since Γ/N has property (T) and since C is abelian,

the closure of ϕ(Γ/N ) is compact. Since the only compact subgroup of C is {0}, it follows that ϕ(Γ/N ) = 0. The well-defined group homomorphism Γ/N → C : γN 7→ ϕ(γ) must be identically zero. Thus, ϕ|Γ = 0. As

explained in the proof of Proposition5.13, this further implies that ϕ ≡ 0. Conversely, assume that assumptions (i) and (ii) are satisfied. By contra- diction, assume that Γ/N does not have property (T). Choose a symmetric finitely supported probability measure µ ∈ Prob(Γ) whose support supp(µ) generates Γ. Denote by µ ∈ Prob(Γ/N ) the pushforward measure under the factor map Γ → Γ/N . Then µ ∈ Prob(Γ/N ) satisfies the assumption of Terminology 4.6 for the factor group Γ/N . By Theorems 4.7 and 4.10, there exist a unitary representation π : Γ/N → U (Hπ) and a nonzero µ-

harmonic cocycle b ∈ Harµ(Γ/N, π). Replacing Hπ with the closure of the

linear span of b(Γ/N ), we may assume that Hπ is separable. We regard π

as a Γ-unitary representation such that π|N = 1Hπ and b ∈ Harµ(Γ, π) as a

µ-harmonic cocycle for π : Γ → U (Hπ) such that b|N = 0.

Step 1: One may assume that π is ergodic.

Consider the orthogonal decomposition Hπ = (Hπ (Hπ)Γ)⊕(Hπ)Γinto

π(Γ)-invariant subspaces. We claim that p_(Hπ)Γ◦ b = 0. Indeed, otherwise

there exists a nonzero vector ξ ∈ (Hπ)Γ, for which the homomorphism ϕ :

Γ → C : γ 7→ hp(Hπ)Γ(b(γ)), ξi is nontrivial and satisfies ϕ|N = 0. Using

Proposition 5.13, ϕ : Γ → C extends to a nontrivial homomorphism ϕ : G → C. This contradicts assumption (ii). Thus, up to replacing (π, Hπ)

with (π, Hπ (Hπ)Γ) and b with pHπ (Hπ)Γ◦ b, we may assume that π is

ergodic and b 6= 0. Then the induced representation _bπ : G → U (H

b π) is

ergodic by Proposition 2.13.

Step 2: One may assume that π extends to a strongly contin- uous G-unitary representation.

Using Proposition 5.10(i), for every j ∈ {1, 2}, denote by Hj ⊂ Hπ the

π(Γ)-invariant closed subspace of Gj-continuous elements. Using Proposi-

tion 5.10(ii), the unitary representation π : Γ → U (H1) (resp. π : Γ →

U (H₂)) extends to a strongly continuous unitary representation π : G → U (H1) (resp. π : G → U (H2)) for which G2 (resp. G1) acts trivially. More-

over, Proposition 5.10(iii) shows that W1 : H1 → (Hbπ)

G2 _{: ξ 7→ (g 7→}

π(g−1)ξ) (resp. W2 : H2→ (Hπb)

G1 _{: ξ 7→ (g 7→ π(g}−1_{)ξ)) is a G-equivariant}

1. PROPERTY (T) HALF 107

We claim that H1 and H2 are orthogonal in Hπ. Indeed, for every

ξ1 ∈ H1, every ξ2∈ H2 and every g ∈ G, we have

hξ₁, ξ2i = hπ(g−1)ξ1, π(g−1)ξ2i.

This implies that hW1(ξ1), W2(ξ2)iH b π = Z F hπ(g−1)ξ1, π(g−1)ξ2i dmG(g) = mG(F ) · hξ1, ξ2i.

Sinceπ is ergodic and G = Gb 1×G2, the closed subspaces (Hbπ)

G1 _{and (H} b π)G2 are othogonal in H b π. Thus, hξ1, ξ2i = _mG1_{(F )}hW1(ξ1), W2(ξ2)iH b π = 0 and so

H1 and H2 are orthogonal in Hπ.

Denote by K ⊂ Hπ the orthogonal complement of H1⊕ H2 in Hπ and

observe that K is π(Γ)-invariant. We have the following orthogonal de- composition Hπ = H1⊕ H2 ⊕ K into π(Γ)-invariant subspaces. This fur-

ther yields the following orthogonal decomposition H

b

π = bH1 ⊕ bH2 ⊕ bK

into _bπ(G)-invariant subspaces. Note that Proposition 5.10(iii) implies that (H

b

π)G2 = ( bH1)G2 ⊂ bH1 and (H_bπ)G1 = ( bH2)G1 ⊂ bH2. Then Proposition 5.12 applied to (bπ, bK) implies that H

1

(bπ, bK) = 0. Then Theorem 4.22 im- plies that pK◦ b = 0. Therefore, up to replacing (π, Hπ) with (π, H1⊕ H2)

and b with pH1⊕H2◦ b, we may assume that Hπ = H1⊕ H2 and b 6= 0. Up to

permuting the indices and replacing (π, Hπ) with (π, H1) and b with pH1◦ b,

we may further assume that Hπ = H1 and b 6= 0.

Step 3 : Extending Γ-cocycles to G-cocycles for π.

We may now regard the ergodic unitary representation π : Γ → U (Hπ)

as an ergodic strongly continuous unitary representation π : G → U (Hπ)

such that π|G2 = 1Hπ. By Example 2.12(iii), bπ is unitarily equivalent to π ⊗ λG/Γ.

Observe that (H

b

π)G1 ⊂ bH2= 0 hence bπ is G1-ergodic. Set c = p_(H b π)G2◦ bb ∈ Z 1_(G, b π, (Hπ_b)G2).

Then Proposition 5.12applied to (_bπ, H

b π (Hbπ) G2_{) implies that} d= b. b − c ∈ B1(G,π, H_b b π (Hbπ) G2_).

Recall that the linear mapping W : Hπ → (Hbπ)

G2 _{: ξ 7→ (h 7→ π(h}−1_{)ξ) is a}

unitary operator such that W π(g)W∗=π(g)|b (H

b

π)G2 for every g ∈ G. Define

the mapping bc : G → Hπ : g 7→ W∗c(g) and observe that bc ∈ Z1(G, π).

Consider the transfer operator T : Z1(G,π) → Z_b 1(Γ, π) associated with a relatively compact subset C ⊂ F as in Claim 4.24.

Claim 5.14. Regarding bc∈ Z1(Γ, π), we have b − bc∈ B1(Γ, π).

Indeed, by Claim4.24, we have T (bb) = b and T (d) ∈ B1(Γ, π). Next, a simple calculation shows that for every γ ∈ Γ, we have

T (c)(γ) = 1 mG(C)2

Z

C2

= 1 mG(C)2 Z C2 π(g−1) bc(gγh−1) dm⊗2G (g, h) = 1 mG(C)2 Z C2 (−bc(g−1) + bc(γh−1)) dm⊗2_G (g, h) = 1 mG(C)2 Z C2 (−bc(g−1) + bc(γ) + π(γ)bc(h−1)) dm⊗2_G (g, h) = bc(γ) + π(γ)ξ − ξ where ξ = _m 1 G(C) R Cbc(g −1_{) dm}

G(g) ∈ Hπ. This implies that T (c) = bc+

∂πξ. Therefore, we obtain

b − bc= T (bb − c) + ∂πξ = T (d) + ∂πξ ∈ B1(Γ, π).

This finishes the proof of Claim 5.14.

Since b|N = 0 and π|N = 1Hπ, Claim 5.14implies that bc|N = 0. More-

over, for every g = (g1, g2) ∈ G1× G2= G, we have

bc(g1) + π(g1)bc(g2) = bc(g1g2)

= bc(g2g1) = bc(g2) + π(g2)bc(g1)

= bc(g2) + bc(g1).

This implies that bc(g2) ∈ Hπ is π(G1)-invariant and so bc(g2) = 0 for every

g2∈ G2. This further implies that bc|G2 = 0 and hence bc|_{p1(N )} = 0.

We may now regard π : G1/p1(N ) → U (Hπ) as a strongly continuous

unitary representation of the factor group G1/p1(N ). Moreover, we may

regard bc ∈ Z1(G1/p1(N ), π) as a continuous cocycle for the strongly con-

tinuous unitary representation π : G1/p1(N ) → U (Hπ). Since G1/p1(N )

has property (T), we have bc ∈ B1(G1/p1(N ), π) and hence bc : G → Hπ

is bounded. Finally, Lemma 4.3 and Claim 5.14 imply that b ∈ B1(Γ, π). Since by assumption b ∈ Harµ(Γ, π), it follows that b ≡ 0 by Theorem 4.7.

This is a contradiction. Therefore, Γ/N has property (T). This finishes the

proof of Theorem 5.8. 

2. Amenability half

The main result of this section provides a complete characterization of when factors of irreducible lattices in product groups are amenable. This is the “amenability half” of Theorem 5.4which is due to Bader–Shalom.

Theorem 5.15 (Bader–Shalom [BS04]). For every i ∈ {1, 2}, let Gi be

any lcsc group. Set G = G1× G2 and denote by pi : G → Gi the canonical

factor map. Let Γ < G be any irreducible lattice and N C Γ any normal subgroup.

Then Γ/N is amenable if and only if for every i ∈ {1, 2}, Gi/pi(N ) is

In document EL CRM COMO HERRAMIENTA ALTERNATIVA DEL PLAN DE MEJORAMIENTO FRENTE AL ENVIO DE LAS COMUNICACIONES EN LA EPS-S SALUD TOTAL (página 32-66)