IV. ON THE STRUCTURE OF II 1 FACTORS OF NEGATIVELY CURVED GROUPS
IV.6 A proof of Proposition IV.2.2
The aim of this section is to establish that Γ =Z2oSL(2,Z) belongs to the classQHreg. Appealing to Theorem A then furnishes an alternate proof of the solidity ofLΓ, the main result of [57]. As in [57], our proof will make use of the amenability of the natural action of SL(2,Z) on SL(2,R)/T ∼= RP1, whereT is the group of upper-triangular 2×2 real matrices.
To begin, note that Γ0 = SL(2,Z) admits a proper cocycle b: Γ0 →`2(Γ0) with respect to the left regular representation. Let π be the representation of Γ on `2(Γ0) obtained by pulling the left regular representation of Γ0 back along the quotient ΓΓ/Z2 ∼= Γ0, so thatπ is weakly contained in the left regular representation of Γ. Let p : Z2 \ {(0,0)} →RP1 be the projection defined by p((x, y)) = x/y, and note that p is equivariant with respect to the natural actions of SL(2,Z) on Z2 and RP1.
Given a sequence of continuous mapsξn:RP1 →`2(Γ0) satisfying Definition IV.5.1, define the maps ξn0 :Z2 →`2(Γ0) by
ξn0(z) =σ(z)ξn(p(z)),
forz = (z1, z2)∈Z2\ {(0,0)}, and ξ0n(z) = 0, otherwise. Here σ(z) = 1, if−π/2<arg(z)≤π/2, and σ(z) =−1, otherwise. Note that for any a∈Z2 we have
lim sup
z→∞
kξn0(z)−ξn0(z+a)k2 = 0, (IV.6.1)
for all n∈N.
Now, consider finite, symmetric generating subsets S0 ⊂Γ0 and S00⊂Z2. DefineS1 =S0∪S00 and Sk+1 =Sk∪(S1)k+1 for allk∈N. By equations IV.5.1 and IV.6.1, there exists an increasing sequence of finite, symmetric subsetsF1 ⊂F2 ⊂ · · · ⊂Fk⊂ · · · ⊂Z2 such that S∞
k=1Fk=Z2 and a subsequence (nk) such that
sup
s∈Sk
sup
g∈Z2\Fk
kπs(ξn0k(g))−ξn0k(s·g)k2 ≤ 1
2k, (IV.6.2)
where s·g is the natural Γ-action onZ2. Define a map ∂ :Z2 → `2(N;`2(Γ0)) =H by ∂(z)(k) = ξn0k(z), if z 6∈ Fk, and 0, otherwise. It is then straightforward to check that ∂ is proper, anti-
symmetric, and boundedly Γ-equivariant. For (z, γ)∈Z2oSL(2,Z) we define the mapq((z, γ)) = b(γ)⊕∂(z) ∈ `2(Γ0)⊕ H. It is easy to see that q is an array into the weakly `2 representation π⊕π⊕∞. Thus, Z2oSL(2,Z)∈ QHreg and we are done.
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