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Publicly Available Published by De Gruyter May 7, 2019

On the Haagerup and Kazhdan properties of R. Thompson’s groups

Arnaud Brothier EMAIL logo and Vaughan F. R. Jones
From the journal Journal of Group Theory

Abstract

A machinery developed by the second author produces a rich family of unitary representations of the Thompson groups F, T and V. We use it to give direct proofs of two previously known results. First, we exhibit a unitary representation of V that has an almost invariant vector but no nonzero [F,F]-invariant vectors reproving and extending Reznikoff’s result that any intermediate subgroup between the commutator subgroup of F and V does not have Kazhdan’s property (T) (though Reznikoff proved it for subgroups of T). Second, we construct a one parameter family interpolating between the trivial and the left regular representations of V. We exhibit a net of coefficients for those representations which vanish at infinity on T and converge to 1 thus reproving that T has the Haagerup property after Farley who further proved that V has this property.

Introduction

Let FTV be the usual Thompson groups acting on the unit interval [2]. In [7], it was shown that certain categories 𝒞 with a privileged object 1𝒞 give rise to a group of fractions G𝒞 and that a functor Φ:𝒞𝒟 provides an action of the group G𝒞. Similar ideas in the context of semigroups were developed by Ore; see for instance [9]. Thompson’s groups F,T,V can be constructed in this way using various categories of forests written ,𝒜,𝒮. If 𝒟=Hilb is the category of Hilbert spaces with isometries for morphisms, then the functor Φ gives us a unitary representation of G𝒞. In this article, we consider functors Φ:Hilb such that Φ(n)=n and Φ(fi,n)=idi-1Ridn-i, where R: is a fixed isometry and fi,n is the forest with n trees all of which are trivial except the ith one that has two leaves. Since any tree is a composition of fi,n, we obtain a well defined functor and thus a unitary representation of F that we extend to V via permutations of the tensors; see Section 1 for more details.

In Section 2, we construct a unitary representation of V having an almost invariant vector but no nonzero [F,F]-invariant vectors showing that any intermediate subgroup between [F,F] and V does not have Kazhdan’s property (T) [8]. Recall that T was proved to not have property (T) initially by Reznikoff [11]. Another proof consists in embedding T in the group of diffeomorphisms of the circle via the smoothing methods of Ghys and Sergiescu [6] or Thurston (described in [2]) and using results of Navas concerning diffeomorphism groups that imply that T does not have property (T) [10].

In Section 3, we give a one parameter family (πα, 0α1) of unitary representations of V interpolating the trivial and the left regular one. We define some positive definite maps φα from those representations for which we can explicitly compute their values on T. They actually coincide on T with the family of maps constructed from the proper cocycle of Farley (see Remark 1), but differ on the larger group V. Hence they provide a C0 positive definite approximation of the identity for T, thus proving Farley’s result that T has the Haagerup property (though Farley actually proved it for all of V) [1, 4]. Note that a geometric approach to Haagerup and Kazhdan properties for Thompson’s groups follows from the actions on CAT(0) cube complexes constructed by Farley [3, 5] (which is related to Chatterji, Drutu and Haglund’s characterization of these properties via median metric spaces).

1 Definitions and notations

We briefly recall the construction of actions of groups of fractions for the particular cases of F, T and V and refer to [7, 2] for more details. Let be the category of (binary planar) forests whose objects are the natural numbers 𝐍:={1,2,} and morphisms (n,m) the set of binary planar forests with n roots and m leaves. We think of them as planar diagrams in the plane 𝐑2 whose roots and leaves are distinct points in 𝐑×{0} and 𝐑×{1}, respectively, and are counted from left to right. We compose forests by stacking them vertically so that pq is the forest obtained by stacking on top of q the forest p, where the ith root of p is attached to the ith leaf of q. We obtain a diagram in the strip 𝐑×[0,2] that we rescale in 𝐑×[0,1]. For any n𝐍, 1in, we consider the forest fi,n, or simply fi if the context is clear, the forest with n roots and n+1 leaves, where the ith tree of fi,n has two leaves. For example,

Consider the set of pairs of trees(t,s) with the same number of leaves, that we quotient by the relation generated by (t,s)(pt,ps) for any forest p. We write ts for the equivalence class of (t,s). These form the group of fractions of the category with the multiplication tssr=tr and inverse (ts)-1=st. It is isomorphic to Thompson’s group F.

Now consider the category of symmetric forests 𝒮 with objects 𝐍 and morphisms 𝒮(n,m)=(n,m)×Sm, where Sm is the symmetric group of m elements. Graphically, we interpret a morphism (p,τ)𝒮(n,m) as the concatenation of two diagrams. On the bottom, we have the diagram explained above for the forest p in the strip 𝐑×[0,1]. The diagram of τ is the union of m segments [xi,xτ(i)+(0,1)], i=1,,m, in 𝐑×[1,2], where the xi are m distinct points in 𝐑×{1} such that xi is on the left of xi+1. The full diagram of (p,τ) is obtained by stacking the diagram of τ on top of the diagram of p such that xi is the ith leaf of p. Given symmetric forests (q,τ)𝒮(n,m), (p,σ)𝒮(m,l), let li be the number of leaves of the ith tree of p. Then we define the composition of morphisms by (p,σ)(q,τ):=(τ(p)q,σS(p,τ)), where τ(p) is the forest obtained from p by permuting its trees such that the ith tree of τ(p) is the τ(i)th tree of p and S(p,τ) is the permutation corresponding to the diagram obtained from τ, where the ith segment [xi,xτ(i)+(0,1)] is replaced by lτ(i) parallel segments. Thompson’s group V is isomorphic to the group of fractions of the category 𝒮. Hence any element of V is an equivalence class of a pair of symmetric trees. Consider g=(t,τ)(s,σ)V and the standard dyadic partitions (I1,,In) and (J1,,Jn) of [0,1] associated to the trees s and t, respectively. The element g acting on [0,1] is the unique piecewise linear function with positive constant slope on each Ik that maps Iσ-1(i) onto Jτ-1(i) for any 1in.

Consider the cyclic group 𝐙/m𝐙 a subgroup of the symmetric group Sm and the subcategory 𝒜𝒮 of affine forests, where 𝒜(n,m)=(n,m)×𝐙/m𝐙. The group of fractions of 𝒜 is isomorphic to Thompson’s group T. We will often identify and 𝒜 as subcategories of 𝒮 giving embeddings at the group level FTV.

We say that a pair of symmetric trees ((t,τ),(s,σ)) is reduced if there are no pairs ((t,τ),(s,σ)) such that t has strictly less leaves than t and such that (t,τ)(s,σ)=(t,τ)(s,σ).

Let Hilb be the category of complex Hilbert spaces with isometries for morphisms. Given an isometry R:, we construct a functor

Φ=ΦR:Hilbsuch thatΦ(n)=n,Φ(fi,n)=idi-1Ridn-i

for i=1,,n. Consider the quotient space

{(t,ξ):ttree,ξΦ(target(t))}/,

where the equivalence relation is generated by (t,ξ)(pt,Φ(p)ξ) for all p.

This quotient space has a pre-Hilbert structure given by (t,ξ),(t,η):=ξ,η that we complete into a Hilbert space . Note that is the inductive limit of the system of Hilbert spaces t:={(t,ξ):ξΦ(target(t))} for trees t such that the embedding tpt is given by Φ(p). We denote by (t,ξ) or tξ the equivalence class of (t,ξ) inside and identify and t as subspaces of . We have a unitary representation π:F𝒰() given by the formula π(ts)sξ:=tξ that we extend to the group V as follows:

π((t,τ)(s,σ))sξ:=tθ(τ-1σ)ξ,whereθ(κ)(η1ηn):=ηκ-1(1)ηκ-1(n).

Note that if ξ,η are in the small Hilbert space and g=(t,τ)(s,σ), then

π((t,τ)(s,σ))ξ,η=θ(σ)Φ(s)ξ,θ(τ)Φ(t)η.

Consider an orthonormal basis {ξi:iI} of the Hilbert space . The isometry R can be thought of as a possibly infinite matrix with three indices Rij,k such that Rξi=j,kRij,kξjξk. We can reinterpret Φ(f) for a forest f as a partition function. Given a forest f with n roots and m leaves, we define the set of states Ω(f) on f as maps ω from the edges of f to the set of indices I. A vertex of f is a trivalent vertex, and thus roots and leaves are not vertices. If ω is a state on f and v a vertex, then we put Rvω the scalar equal to Rω(e-)ω(el),ω(er), where e- is the edge with target v and el,er are the edges with source v which goes to the left and right, respectively. Consider some multi-indices i¯:=(i1,,in)In and j¯:=(j1,,jm)Im, and say that a state ωΩ(f) is compatible with (i¯,j¯) if ω(ak)=ik and ω(b)=j for all 1kn, 1m, where ak is the edge with source the kth root of f and b is the edge with target the th leaf of f.

Lemma 1.

Using the notations of above, we have

Φ(f)ξi1ξin,ξj1ξjm=ωΩ(f)compatible with(i¯,j¯)va vertex offRvω,

with the convention that a product (resp. a sum) over an empty set is equal to one (resp. zero).

The infinite sum converges since the scalars Rij,k are matrix coefficients of an isometry. This formula can be proved by induction on the number of leaves and using the fact that any forest is the composition of some elementary forests fi,n. As an illustration, we compute explicitly this inner product for the simple case f=f2,3. Fix some multi-indices i¯=(i1,i2,i3) and j¯=(j1,j2,j3,j4). Observe that if Ω(f) is nonempty, then necessarily i1=j1 and i3=j4. Moreover, a state ωΩ(f) that is compatible with (i¯,j¯) is completely determined by the couple of multi-indices (i¯,j¯), so there is at most one term in the sum of the last lemma. Moreover, the forest f has only one vertex (roots and leaves are not considered as vertices), and so there is only one term in the product that is Ri2j2,j3. We obtain Φ(f)ξi1ξi2ξi3,ξj1ξj2ξj3ξj4=δi1,j1δi3,j4Ri2j2,j3. If the trees have more than one vertex, then the product formula becomes more complicated. For example, if we have the state

that we represent with the spin of an edge (i.e., the image by ω of this edge) next to it, then

va vertex offRvω=Rjl,mRij,k.

2 Kazhdan’s property (T)

Recall that a countable discrete group G has Kazhdan’s property (T) if any unitary representation having an almost invariant vector has in fact a nonzero invariant vector [8].

If u𝒰() is a unitary and ζ is a unit vector, then the map

R:,ξu(ξ)ζ

is an isometry which provides us a unitary representation π:V𝒰() as described in Section 1. We claim that if |ζ,uζ|1, then π has no nonzero [F,F]-invariant vectors. Consider the following four trees

a=f3f3f1f1,b=f4f3f2f1,c=f1f1,d=f2f1,

and let tn be the complete binary trees with 2n leaves. Put g=ab, h=cd and k:=ghg-1h-1. Note that

k=aqwithq=f2f3f1f1.

Define the element

kn:=(a)ntn(q)ntn,

where (a)n is the forest with 2n roots in which every tree is a copy of a. Similarly, define

gn:=(a)ntn(b)ntnandhn:=(c)ntn(d)ntn,

and observe that kn=gnhn(gn)-1(hn)-1. Therefore, kn is in the commutator subgroup [F,F]. Observe that

Φ(q)ξ,Φ(a)η=u2ξuζζuζζ,u2ηζu2ζζζ=ξ,ηuζ,ζ2ζ,u2ζfor allξ,η,

and thus Φ(a)*Φ(q)=CidB() with C=uζ,ζ2ζ,u2ζ.

If ξ:=i=12nξi, η:=i=12nηi are elementary tensors of tn, then

π(kn)ξ,η=Φ((q)n)ξ,Φ((a)n)η=i=12nΦ(q)ξi,Φ(a)ηi=C2nξ,η.

By linearity and density, we obtain that π(kn)ξ,η=C2nξ,η for any vectors ξ,ηtn. Assume that ξ is an [F,F]-invariant unit vector and that |ζ,uζ|<1. This implies |C|<1. By density, there exists n and a unit vector ξtn such that ξ-ξ<14. We obtain

|C|2n=|π(kn)ξ,ξ||π(kn)ξ,ξ|-|π(kn)(ξ-ξ),ξ|-|π(kn)ξ,(ξ-ξ)|12,

for n as large as we want, which implies a contradiction since |C|<1 and proves the claim.

Consider :=2(𝐙) the shift operator u𝒰(2(𝐙)) and ζm2(𝐙) the characteristic function of {1,2,,h(m)} divided by h(m), where h(m)=2m8m, m1. Let (πm,(m)) be the associated sequence of unitary representations of V, and put (π,):=(mπm,m(m)) as their direct sum. Note that we have 0<ζm,uζm<1 for any m1, and thus the claim implies that π does not have any nonzero [F,F]-invariant vectors. Let ηm be the elementary tensor of 2m, where each entry is equal to ζm that we identify with the fraction tmηmtm viewed as an element of (m). Define ξm to be the unit vector of :=n(n), which is equal to tmηm in the mth slot and zero elsewhere. We claim that the sequence (ξm,m1) is an almost V-invariant vector. Fix gV, and note that, for m big enough, there exists s(1,2m), κ,ρS2m such that g=(s,κ)(tm,ρ). By increasing m, we can also assume that there exists p(2m,22m) such that ps=t2m. We obtain

π(g)ξm,ξm=j1πj(g)(tmηmδm,j),i1tmηmδm,i=πm(g)tmηm,tmηm=πm((s,κ)(tm,ρ))tmηm,tmηm=sθ(κ-1ρ)(ηm),tmηm=sηm,tmηm=Φ(p)ηm,Φ(q)ηm,whereps=t2m=qtm,=ia leaf oft2m(Φ(p)ηm)i,(Φ(q)ηm)i,where(Φ(p)ηm)iis theith tensor.

Since ps=t2m=qtm, any branch of p (resp. q) has length smaller than or equal to 2m (resp. m). This implies that any component of Φ(p)ηm and Φ(q)ηm is equal to uaζm for some 0a2m. Since the map a𝐍uaζm,ζm𝐑+ is decreasing, we obtain

(Φ(p)ηm)i,(Φ(q)ηm)iu2mζm,ζm=h(m)-2mh(m).

Therefore,

π(g)ξm,ξm(h(m)-2mh(m))22m=(1-8-m)4mm1,

and thus (ξm)m is an almost V-invariant vector. Since π does not have any nonzero [F,F]-invariant vectors, we obtain that any intermediate group between [F,F] and V does not have property (T).

3 Haagerup property

Recall that a countable discrete group G has the Haagerup property if there exists a sequence φn of positive definite functions which vanish at infinity on G and such that limnφn(g)=1 for all gG (see [1]).

Consider the free group 𝔽2 freely generated by a,b, and let {δg:g𝔽2} be the classical orthonormal basis of the Hilbert space 2(𝔽2). Identify 2(𝔽2)n with 2(𝔽2n) and δg1δgn with δg1,,gn. Set :=2(𝔽2), and define for 0α1 the isometry

Rα:2(𝔽2)2(𝔽2×𝔽2),δeαδe,e+1-α2δa,b,δgδag,bgfor allg𝔽2,ge.

This defines a functor Φα:Hilb and a unitary representation πα:V𝒰(α) as described in Section 1. The associated infinite matrix

(Rgh,k:=Rδg,δhδk)gh,k

is particularly simple since

Ree,e=α,Rea,b=1-α2,Rgag,bg=1for alleg𝔽2

and zero elsewhere.

Consider the case when α=0 and thus Rgag,bg=1 for any g𝔽2 and zero elsewhere. Let f be a forest with n roots and m leaves, and observe that

Φ0(f)δe,,e=δP(f),whereP(f)𝔽2m.

The ith component P(f)i is the word in a,b written from right to left corresponding to the path from a root of f to its ith leaf such that a left turn (resp. right turn) contributes in adding the letter “a” (resp. the letter “b”). For example, if f=f1,4f3,3f1,2, then

Φ0(f)δe,e=δaaδbaδbδaδb=δaa,ba,b,a,b.

The next lemma proves that if t is a tree, then the set of words in the tuple P(t) remembers completely t.

Lemma 2.

Consider two trees s,t with n leaves. Assume that there exists a permutation σSn acting on the leaves such that σ(P(s))=P(t). Then σ=id and s=t.

Proof.

We prove the lemma by induction on the number of leaves n1. The result is immediate for n=1 and is also clear for n=2 since there is only one tree with two leaves and thus P(s)=P(t)=(a,b). The permutation σ is necessarily trivial.

Suppose the result is true for any k between 1 and n, and consider s,t trees with n+1 leaves and a permutation σ such that σ(P(s))=P(t). Note that there exist trees s1,s2,t1,t2 such that s=(s1s2)f1 and t=(t1t2)f1, where s1s2 is the forest with two roots whose first tree is s1 and whose second is s2. Note that the word P(s)i finishes by the letter a (resp. the letter b) if and only if i is a leaf of s1 (resp. a leaf of s2). We have the same characterization for the leaves of t, and thus necessarily σ realizes a bijection from the leaves of sj onto the leaves of tj for j=1,2. Observe that P(s)i=P(s1)ia for any leaves of s1. This implies σ(P(s1))=P(t1). Similarly, we have σ(P(s2))=P(t2), and thus, by the induction hypothesis, we have s1=t1, s2=t2, and σ is the identity on the leaves of s1 and on the leaves of s2 implying σ=id and s=t. ∎

The lemma implies that π0 contains the left regular representation of V. Indeed, consider some symmetric trees (t,τ),(s,σ) in 𝒮 with the same number of leaves. We have

(3.1)π0((t,τ)(s,σ))δe,δe=θ(σ)Φ(s)δe,θ(τ)Φ(t)δe=δP(s),δσ-1τ(P(t)).

This is nonzero (and then equal to one) if and only if P(s)=σ-1τ(P(t)). In that case, Lemma 2 implies s=t and σ-1τ=id, and thus (t,τ)(s,σ) is the trivial group element. We obtain that the cyclic representation generated by δe for α=0 is the left regular representation of V. If α=1, then the cyclic representation generated by δe is the trivial one. Indeed, if t is a tree with n leaves, then Φ(t)δe=δeδe with n tensors, and thus coefficient (3.1) is always equal to one for any choice of g=(t,τ)(s,σ)V. Our family πα of representations of V provides an interpolation between the trivial and the left regular representations.

Consider the family of positive definite maps

φα(g):=πα(g)δe,δewith 0α<1.

We will show that they vanish at infinity for gT and tend to the identity when α tends to one. Fix 0α<1, and write R instead of Rα. Section 1 tells us that, given a tree t with n leaves, we have

Φα(t)δe,δj¯=ωΩ(t)compatible with(e,j¯)va vertex oftRvω

for any multi-index j¯=(j1,,jn). Fix ωΩ(t), and assume that the ω coefficient of above is nonzero for a certain j¯. Then there exists a maximal subrooted tree zω of t such that the spin (i.e., the value of ω) at each of its edges is the trivial group element e𝔽2. Any (trivalent) vertex v of the tree zω satisfies Rvω=α. If fω is the unique forest satisfying t=fωzω, then any root w of fω that is not a leaf of fω satisfies Rwω=1-α2 since spins around it are necessarily e,a,b. Then any other vertex u of fω has its spins around it equal to g,ga,gb for some eg𝔽2, and thus Ruω=1. We obtain

va vertex oftRvω=αtarget(zω)-1(1-α2)m(t,zω)/2,

where m(t,zω) is the number of leaves of zω that are not leaves of t (i.e., the number of nontrivial trees of fω) and target(zω)-1 corresponds to the number of (trivalent) vertices of the tree zω. Indeed, a binary tree with n leaves has n-1 bifurcations (i.e., n-1 trivalent vertices). The spin of the edge with target the th leaf of t is the word in a,b corresponding to the path in the forest fω starting at the root connected to and finishing at the leaf . Write P(t,zω) for this word and P(t,zω) for the corresponding n-tuple, and note that, by definition, the multi-index j¯ is equal to P(t,zω). Therefore, given a multi-index j¯, there is at most one state ωΩ(f) compatible with (e,j¯) and having a nonzero coefficient vRvω. Moreover, j¯ has to be of the form P(t,z) for some subrooted tree zt. Conversely, any subrooted tree z of t provides a unique state ωzΩ(t) that is compatible with (e,P(t,z)) defined inductively by

ωz(c)={eifcis an edge ofz,a.ωz(d)ifcgoes to the left,its source is the target of an edgedandcz,b.ωz(d)ifcgoes to the right,its source is the target of an edgedandcz.

We obtain

Φα(t)δe=zEtαtarget(z)-1(1-α2)m(t,z)/2δP(t,z),

where Et is the set of subrooted trees (including the trivial subtree) of t.

Consider a pair of symmetric trees ((t,τ),(s,σ)) and g=(t,τ)(s,σ)V. We have

(3.2)φα(g)=zEt,rEsαtarget(z)+target(r)-2(1-α2)(m(t,z)+m(s,r))/2×θ(σ)δP(s,r),θ(τ)δP(t,z).

Fix a reduced pair of affine trees ((t,τ),(s,σ)) and the group element g=(t,τ)(s,σ) that is in Thompson’s group T since our trees are affine. We will show that all the terms in the sum (3.2) are equal to zero but one.

Lemma 3.

Consider some forests p,q both of them having m leaves, and let σZ/mZ<Sm be a cyclic rotation. If σ(P(p))=P(q), then p and q have the same number of roots n, and there exists c0 such that the jth tree of p is equal to the (j+c)th modulo n tree of q for any 1jn.

Proof.

Observe that if f is a forest, then the word P(f)i is a power of a (resp. a power of b) if and only if i corresponds to the first leaf (resp. the last leaf) of a tree of f. Consider p,q,σ as above, and write pj and qk for the jth and kth trees of p and q, respectively. Fix j, and note that the observation implies that there exist some natural numbers c,d such that the first and the last leaves of pj are sent to the first leaf of qj+c and the last leaf of qj+d. If cd modulo the number of roots of q, then P(pj) would be equal to a tuple P(f) of a forest f having at least two trees and thus having at least two words that are powers of b, which is impossible. Therefore, σ realizes a bijection from the leaves of pj onto the leaves of qj+c for a certain c. Since σ is cyclic, the number c does not depend on j. We obtain P(pj)=P(qj+c) for any j, which, by Lemma 2, implies that the trees pj and qj+c are equal. ∎

Assume that the (z,r)-term of equality (3.2) is nonzero. Then

P(s,r)=σ-1τ(P(t,z)).

If f(s,r),f(t,z) are the forest satisfying s=f(s,r)r, t=f(t,z)z, then Lemma 3 implies that there exists a cyclic permutation ρ on the roots of f(t,z) such that the ith tree of f(s,r) is equal to the ρ(i)th tree of f(t,z), and thus s=ρ(f(t,z))r. This implies that g can be reduced as a fraction (z,τ~)(r,σ~) for some permutations τ~,σ~. Since the pair ((t,τ),(s,σ)) is already reduced, we obtain z=t and r=s, and thus all the terms in equality (3.2) are equal to zero except one. Therefore,

(3.3)φα((t,τ)(s,σ))=α2target(t)-2for any reduced pairof affine trees((t,τ),(s,σ)).

This implies

limgφα(g)=0insideTfor any 0α<1,
limα1φα(h)=1for anyhT.

Therefore, Thompson’s group T has the Haagerup property.

Remark 1.

Farley constructed a proper cocycle c:VH with values in a Hilbert space and showed in the proof of [4, Theorem 2.4] that if gV is described by a reduced pair of symmetric trees with n leaves, then c(g)2=2n-2. By Schoenberg’s theorem, this provides a family of positive definite functions

ϕβ(g):=exp(-βc(g)2)=exp(-β(2n-2)),β0.

Note that if gT, then formula (3.3) implies φα(g)=ϕ-ln(α)(g) for any α>0. However, this equality is no longer true for certain elements of V. Consider

g=(t,id)(t,(13))V,

where t=f3f1f1 is the full binary tree with four leaves. We have

Φα(t)δe=α3δe,e,e,e+α21-α2(δe,e,a,b+δa,b,e,e)+α(1-α2)δa,b,a,b+1-α2δaa,ba,ab,bb.

Then φα(g)=α6+α2(1-α2)2α6=ϕ-ln(α)(g).

We proved that T has the Haagerup property by using the net of maps φα, 0<α<1. One could hope to extend our proof using the same approximation of the identity for the larger group V. Unfortunately, the maps φα with 0<α<1 are no longer vanishing at infinity if we consider them as functions on V. Indeed, consider the sequence of trees xn such that x2=f1 is the tree with two leaves and xn+1=f1xn for n2. Note that xn is a tree with n leaves. Let sn:=(xnxn)f1 be the tree equal to the composition of f1 with a copy of xn attached to each leaf of f1. Define the permutation σnS2n that is an involution and such that σn(2i+1)=2i+1+n and σn(2j)=2j for any 12i+1, 2jn. Hence σn sends any odd leaf of the first copy of xn to the same leaf in the second copy of xn and lets invariant the others. We set gn:=(sn,σn)(sn,id) and note that this fraction is reduced. If we consider φα(g), we observe that the term corresponding to z=r=f1 in formula (3.2) is nonzero and is equal to α2(1-α2)2. Since all the terms of φα(gn) are positive, we obtain φα(gn)α2(1-α2)2>0 for any n2, and thus φα(g) does not tend to zero when g tends to infinity in V.


Communicated by Pierre-Emmanuel Caprace


Award Identifier / Grant number: 669240 QUEST

Award Identifier / Grant number: DP140100732

Funding statement: A. Brothier was supported by European Research Council Advanced Grant 669240 QUEST and is now supported by a UNSW Sydney starting grant. V. F. R. Jones is supported by the grant numbered DP140100732, Symmetries of subfactors.

Acknowledgements

We thank the generous support of the New Zealand Mathematics Research Institute and the warm hospitality we received in Raglan, which made this work possible.

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Received: 2018-05-08
Revised: 2019-03-14
Published Online: 2019-05-07
Published in Print: 2019-09-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

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