On a Subfactor Construction of a Factor Not Anti–Isomorphic to Itself

by Maria Grazia Viola ^{∗} Mathematical Sciences Research Institute ^{†} Berkeley, CA mgviola@math.uiowa.edu

September 28, 2001

Abstract

Let α be a ℤ_{3}–kernel of 𝔏 (F_{\frac{11}{3}} ) with obstruction ϵ = e^{\frac{2πi}{3}} to lifting and β a ℤ_{3}–kernel of the hyperfinite factor R with obstruction ϵ̄ to lifting. Define an action γ of ℤ_{3} on 𝔏 (F_{\frac{11}{3}} ) ⊗ R by α ⊗ β, up to an inner automorphism. The aim of this paper is to show that the factor M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R)⋊_{γ} ℤ_{3}, which is similar to Connes’ example of a factor which is not anti–isomorphic to itself, is the enveloping algebra of an inclusion of II_{1} factors A ⊂ B. Here A is a free group factor and B is isomorphic to the crossed product A ⋊_{θ} ℤ_{9}, where θ is a ℤ_{3}–kernel of A with non-trivial obstruction to lifting. By using a Connes’ argument involving the invariant χ(M), we will verify that M is not anti–isomorphic to itself. Furthermore, we will prove that for one of the generator of χ(M), which we will denote by σ, the Jones invariant κ(σ) is equal to e^{\frac{2πi}{9}} .

1 Introduction

In this paper we are going to show that there exists an inclusion A ⊂ B of free group factors, which by iteration of the Jones basic construction give us a II_{1} factor which is not anti–isomorphic to itself. The II_{1} factor we obtain is essentially the same example given by Connes in [Co3] and [Co4] of a II_{1} factor not anti–isomorphic to itself.

An anti–isomorphism of a von Neumann algebra M is a vector space isomorphism Φ : M → M with Φ(x^{∗}) = Φ(x)^{∗} and Φ(xy) = Φ(y)Φ(x). Note that a von Neumann algebra M is anti–isomorphic to itself if and only if it is isomorphic to its conjugate algebra M^{c} (cf. Section 6).

∗Research at MSRI is supported by the Graduate College and the Department of Mathematics of the University of Iowa.

†Address from September 2001: Department of Mathematics, University of Iowa, MacLean Hall, Iowa City, IA 52242.

In Section 3 we will first give a slightly different description of the Connes’ type II_{1} factor which is not anti–isomorphic to itself (see [Co3] or [Co4] for more details on the Connes factor). In our approach we will use the recently developed theory of free group factors. The II_{1} factor M we are going to study is constructed from the tensor product of a free group factor and the hyperfinite II_{1} factor R. We will use two ℤ_{3}–kernels, α ∈ Aut (𝔏 (F_{\frac{11}{3}} )) and β ∈ Aut(R), which have conjugate obstructions to lifting, to generate an action of ℤ_{3} on 𝔏 (F_{\frac{11}{3}} ) ⊗ R. The action is given, up to an inner automorphism, by α ⊗ β, and

the factor M is equal to the crossed product (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3} (cf. Section 4).

The main result of this paper is Theorem 4.3, where we show that M is the enveloping algebra of an inclusion A ⊂ B of free group factors, with A isomorphic to 𝔏 (F_{\frac{35}{27}} ) and B = A⋊_{θ} ℤ_{9}, for a ℤ_{3}–kernel θ of A with obstruction

e^{\frac{2πi}{3}} to lifting.

Though extensive work has been done both by V. Jones and A. Connes on examples of II_{1} and III_{λ} factors non anti-isomorphic to itself, it does not seem that the possibility of realizing any of these factors through a subfactor construction has been really studied, and there is little literature on the subject. The result we will present here is a generalization of a similar result obtained by F. Rădulescu in the case of ℤ_{2}–kernels. The proof is mainly based on Voiculescu’s random matrix model introduced in [Vo].

In Section 5 we will use Connes’ exact sequence (see [Co4]) to show that for the factor M the Connes invariant χ(M) is equal to ℤ_{9}, a result due to Connes ([Co3] and [Co4]). Furthermore, if σ denotes the generator of χ(M) described in Remark 5.4, then the invariant κ(σ) defined by Jones in [Jo2] is equal to e^{\frac{2πi}{9}} .

In the last section, we will use an argument of Connes (see [Co3]) to show that M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3} is not anti-isomorphic to itself. The main ingredients of this argument are the uniqueness (up to inner automorphism) of the decomposition of γ into the product of an approximately inner automorphism and a centrally trivial automorphism and the fact that the only subgroup of order 3 in χ(M) is generated by the dual action γ̂ on M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3}. This will provide us with a canonical way to associate to a von Neumann algebra a complex number, which is invariant under isomorphism, and distinguishes M from its conjugate algebra M^{c}.

2 Definitions

Let M be a finite factor with trace τ. For a unitary u in M we denote by Ad_{M} u the inner automorphism of M defined by Ad_{M} u(x) = uxu^{∗}. We let \overline{Int(M)} be the closure of the group Int(M) of inner automorphisms of M in the topology of pointwise weak convergence.

We say that an automorphism α ∈ Aut(M) is centrally trivial if for any central sequence x_{n} ∈ M (i.e. a bounded sequence in M with the property lim_{n→∞} ∥[x_{n}, y]∥_{2} = 0 for every y ∈ M) we have lim_{n→∞} ∥α(x_{n}) − x_{n}∥_{2} = 0. We denote by Ct(M) the set of centrally trivial automorphisms of M.

The Connes invariant is defined as follows:

χ(M) = \frac{Ct(M) ∩ \overline{Int(M)}}{Int(M)}.

A central sequence (x_{n}) is said to be hypercentral if lim_{n→∞} ∥[x_{n}, y_{n}]∥_{2} = 0 for every central sequence (y_{n}) ∈ M. Given a free ultrafilter ω we denote by C_{ω} the algebra of bounded sequence (x_{n}) in M with lim_{n→ω} ∥[x_{n}, y]∥_{2} = 0 for all y ∈ M, and by ℑ_{ω} the subalgebra of C_{ω} consisting of central sequences which tend ∗-strongly to zero. For x = (x_{n}) ∈ C_{ω}, we set τ^{ω}(x) = lim_{n→ω} τ(x_{n}). Then M_{ω} = C_{ω}/ℑ_{ω} is a von Neumann algebra with trace τ^{ω}.

Remark 2.1 By [McD] the existence of non-trivial hypercentral sequences is equivalent to the non-triviality of the center of M_{ω} for some (and then for all) free ultrafilter ω.

Definition 2.2 A ℤ_{3}–kernel on a von Neumann algebra M is an automorphism α ∈ Aut(M) such that there exists a unitary U in M with the property α^{3} = Ad_{M} U.

Note that if α is a ℤ_{3}–kernel then α(U) = λU, for λ a cube root of unity. If λ ≠ 1 we say that α has obstruction λ to lifting, meaning that such an automorphism cannot be lifted to an action of ℤ_{3} on M.

We conclude this section by defining an invariant κ(σ) for any element σ in χ(M). Define Out(M) = \frac{Aut(M)}{Int(M)}, and let ξ : Aut(M) → Out(M) be the usual quotient map.

Definition 2.3 Let M be a II_{1} factor without non-trivial hypercentral sequences and σ ∈ χ(M). Let ϕ be any automorphism on M with ξ(ϕ) = σ and u_{n} a sequence of unitaries with the properties

ϕ = lim_{n→∞} Ad u_{n} and lim_{n→∞} ∥ϕ(u_{n}) − λ_{ϕ}u_{n}∥_{2} = 0

for some λ_{ϕ} of modulus one (such sequence always exists since ϕ ∈ \overline{Int(M)}). Define κ(σ) = λ_{ϕ}.

Jones proved in [Jo2] that this definition makes sense (i.e. κ(σ) does not depend on the choice of ϕ or u_{n}) and that κ is a conjugacy invariant, meaning that if α, β ∈ Ct(M)∩\overline{Int(M)} and there exists ψ ∈ Aut(M) such that ψαψ^{−1} = β, then κ(ξ(α)) = κ(ξ(β)).

3 Preliminaries

A von Neumann algebra M is said to be full if \overline{Int(M)} = Int(M). Obviously all type I factors are full, while the hyperfinite factor R provides an example of a II_{1} factor which is not full since \overline{Int(R)} = Aut(R) ≠ Int(R).

Remark 3.1 For an arbitrary factor, being full is equivalent to having no nontrivial central sequence (see [Co1]).

The following result is an easy consequence of Lemma 4.3.3 in [Sa].

Lemma 3.2 Let G be a discrete group containing a non-abelian free group and let τ be the usual trace on 𝔏(G). Then any central sequence in 𝔏(G) satisfies lim_{n→∞} ∥x_{n} − τ(x_{n})∥ = 0.

Proof Set E = G − {e}, where e denotes the identity element in G. Let g_{1}, g_{2} be two generators of the free group and F = {g ∈ E | g = g_{1}g̃, g̃ ∈ E}. Take x ∈ 𝔏(G). Then x can be expressed as x = ∑_{g∈G} λ_{g}δ_{g} and the function f : G → ℂ defined by f(g) = λ_{g} belongs to l^{2}(G).

For such f we have that

∑_{g∈E} |f(g)|^{2} = ∥x − τ(x)∥_{2}^{2} and ∑_{g∈G} |f(g_{i}gg_{i}^{−1}) − f(g)|^{2} = ∥[x, δ_{g_{i}}]∥_{2}^{2}.

Now if (x_{n}) is a central sequence in 𝔏(G) then ∥[x_{n}, δ_{g_{i}}]∥ → 0 for n → ∞, hence we can apply Lemma 4.3.3 in [Sa] and conclude that ∥x_{n} − τ(x_{n})∥_{2} → 0 for n → ∞.

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By applying the previous lemma and Lemma 2 in [Jo1] (see also Lemma 15.42 in [EK] for a printed version) to the factor 𝔏 (F_{\frac{11}{3}} ) ≅ 𝔏(F_{3} ∗ ℤ_{3}), we obtain the following corollary

Corollary 3.3 𝔏 (F_{\frac{11}{3}} ) is full.

Proposition 3.4 𝔏 (F_{\frac{11}{3}} ) ⊗ R has no non-trivial hypercentral sequences.

Proof First we want to show that any central sequence in 𝔏 (F_{\frac{11}{3}} ) ⊗ R has the form (1 ⊗ x_{n}), for a central sequence (x_{n}) in R.

By the proof of Lemma 3.2, 𝔏 (F_{\frac{11}{3}} ) satisfies the hypothesis of Lemma 2.11

in [Co1] with Q_{1} = 𝔏 (F_{\frac{11}{3}} ), Q_{2} = R and b_{i} = δ_{g_{i}}, where g_{i} is a generator of F_{2}

for i = 1, 2 and 𝔏(F_{2}) ⊂ 𝔏 (F_{\frac{11}{3}} ). Therefore we can apply Lemma 2.11 to any

central sequence (X_{n}) in 𝔏 (F_{\frac{11}{3}} ) ⊗ R to obtain that ∥X_{n} − (τ ⊗ 1)(X_{n})∥_{2} → 0 for n → ∞. Since (τ ⊗ 1)(X_{n}) ∈ ℂ ⊗ R this implies that X_{n} = 1 ⊗ x_{n} for a central sequence (x_{n}) in R.

Now suppose (Y_{n}) is a hypercentral sequence in 𝔏 (F_{\frac{11}{3}} ) ⊗ R. Since (Y_{n}) is central it has the form Y_{n} = 1 ⊗ y_{n}, for a hypercentral sequence (y_{n}) in R. So we need only to prove that R has no non-trivial hypercentral sequences.

By Remark 3.1 it is enough to show that R_{ω} has trivial center. But this follows from Theorem 15.15 in [EK].

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Theorem 3.5 If α ∈ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R) then α = Ad z(ν ⊗ id) for some unitary z ∈ 𝔏 (F_{\frac{11}{3}} ) ⊗ R and an automorphism ν of 𝔏 (F_{\frac{11}{3}} ).

Proof Let (K_{n})_{n∈ℕ} be an increasing sequence of finite dimensional subfactors of R generating R and R_{n} = Kn^{′} ∩ R the relative commutant of K_{n} in R.

Set L_{n} = 1 ⊗ R_{n} ⊂ 𝔏 (F_{\frac{11}{3}} ) ⊗ R. Then there exists an n_{0} such that for all x ∈ L_{n_{0}}, ∥x∥ ≤ 1 one has ∥α(x) − x∥_{2} ≤ \frac{1}{2}. In fact, otherwise it would exist a uniform bounded sequence (x_{n}), x_{n} ∈ L_{n} and ∥x_{n}∥ ≤ 1, such that ∥α(x_{n}) − x_{n}∥_{2} > \frac{1}{2}. But (x_{n}) is a central sequence in 𝔏 (F_{\frac{11}{3}} ) ⊗ R because for each m and n ≥ m, x_{n} commutes with 𝔏 (F_{\frac{11}{3}} )⊗K_{m}, so we get a contradiction.

By Lemma 3.3 in [Co2], up to inner automorphism, α is of the form α_{1}⊗1_{R_{n_{0}}} where α_{1} is an automorphism of 𝔏 (F_{\frac{11}{3}} ) ⊗ K_{n_{0}}. Set F = 1 ⊗ K_{n_{0}}. Then F is a type I subfactor of 𝔏 (F_{\frac{11}{3}} ) ⊗ K_{n_{0}}.

Applying [Lemma 3.11, [Co2]] to α_{1} we obtain that α_{1}|_{1⊗K_{n_{0}}} = Ad V |_{1⊗K_{n_{0}}} . This implies that α = Ad z(ν ⊗ 1) for some automorphism ν of 𝔏 (F_{\frac{11}{3}} ).

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4 The realization of the crossed product (𝔏 (F_{\frac{11}{3}}) ⊗ R) ⋊_{γ} ℤ_{3} as the enveloping algebra of an inclusion of free group subfactors

In this section we will show, using Voiculescu’s random matrix model for free group algebras (see [Vo]), that the crossed product (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3} can be realized as the enveloping algebra of an inclusion of free group factors. To that purpose we will first give an explicit construction of (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3}, by giving models for 𝔏 (F_{\frac{11}{3}} ) and R.

Let \{X_{1}, X_{2}, X_{3}\} be a free semicircular family and u = ∑_{j=1}^{3} e^{\frac{2πij}{3}} e_{j} a unitary whose spectral projections have trace \frac{1}{3}, such that alg\{e_{j}\}_{j=1}^{3} is free with respect to \{X_{1}, X_{2}, X_{3}\}′′. Then 𝔏 (F_{\frac{11}{3}} ) can be thought as the von Neumann algebra generated by {X_{1}, X_{2}, X_{3}, u}, as in [Vo] and [Ra1].

The model for R is outlined in the following lemma. It is analogous to the construction given in the case of ℤ_{2} by Rǎdulescu [Ra2] and some parts of the arguments are probably well-known in the literature on subfactors. We include them here for the sake of completeness.

Lemma 4.1 Let (U_{k})_{k∈ℤ} be a family of unitaries with nine point spectrum so that each spectral projection for U_{k} has trace \frac{1}{9} and let g = ∑_{j=1}^{3} e^{\frac{2πij}{3}} g_{j} be a unitary whose spectral projections \{g_{j}\}_{j=1}^{3} have trace \frac{1}{3}. We assume the following relations:

(i) U_{k}^{9} = 1 for all k ∈ ℤ,

(ii) U_{k}gU_{k}^{∗} = e^{\frac{−2πi}{3}} g if k = 0, −1, while U_{k}gU_{k}^{∗} = g if k ∈ ℤ/\{0, −1\},

(iii) U_{k}U_{k+1}U_{k}^{∗} = e^{\frac{2πi}{9}} U_{k+1}, for k ∈ ℤ,

(iv) U_{i}U_{j} = U_{j}U_{i} if |i − j| ≥ 2.

We define a trace on the algebra generated by these unitaries requiring that each non-trivial monomial in the U_{k}’s and g have zero trace. Let

R_{−1} = {gU_{0}^{3}, U_{1}, U_{2}, …}′′ ⊂ {g, U_{0}, U_{1}, U_{2}, …}′′ = R_{0}.

Then R_{−1} ⊂ R_{0} is an inclusion of type II_{1} factors of index 9 and the relative commutant R′_{−1} ∩ R_{0} is generated by the unitary g.

If θ = Ad_{R_{−1}}(U_{0}) then θ is a outer automorphism of R_{−1} with θ^{9} = Id. Moreover, R_{0} is equal to the crossed product R_{−1} ⋊_{θ} ℤ_{9} and θ defines a ℤ_{3}– kernel on R_{−1} with obstruction e^{\frac{−2πi}{3}} to lifting.

Also, the k-th step in the iterated Jones’ basic construction for the inclusion R_{−1} ⊂ R_{0} is given by

R_{k} = {gU_{−1}^{3} … U_{−k}^{3}, U_{−k}, U_{−k+1}, …}′′, k ≥ 1.

Proof The properties of the family (U_{k})_{k∈ℤ} and of the unitary g imply immediately that the automorphism Ad U_{0} normalizes R_{−1}. Furthermore g commutes with R_{−1} (and generates the relative commutant of R_{−1} in R_{0}) so that Ad_{R_{−1}}(U_{0}^{3}) = Ad_{R_{−1}}(gU_{0}^{3}). Since h = gU_{0}^{3} is a unitary in R_{−1} (with spectral projections of trace \frac{1}{3}) and (Ad_{R_{−1}} U_{0})(h) = e^{\frac{−2πi}{3}} h, we obtain that θ = Ad_{R_{−1}}(U_{0}) is a ℤ_{3}–kernel with obstruction e^{\frac{−2πi}{3}} to lifting.

Therefore, at least algebraically, Ad_{R_{−1}}(U_{0}) implements the crossed product R_{−1} ⋊_{θ} ℤ_{9}. But according to our definition, the trace on the algebra generated by the U_{k}’s and g is compatible with the usual trace defined on the crossed product so (R_{−1} ∪ {U_{0}})′′ = R_{−1} ⋊_{θ} ℤ_{9}.

Since (Ad_{R_{0}} U_{−1})(U_{0}) = e^{\frac{2πi}{9}} U_{0} while Ad_{R_{0}}(U_{−1}) acts identically on R_{−1}, we conclude that Ad_{R_{0}}(U_{−1}) implements the dual action of ℤ_{9} on the crossed product R_{−1} ⋊_{θ} ℤ_{9}. Hence, the next step in the iterated basic construction of the inclusion R_{−1} ⊆ R_{0} is

R_{1} = {U_{−1}, g, U_{0}, U_{1}, …}′′ = {gU_{−1}^{3}, U_{−1}, U_{0}, U_{1}, …}′′.

The following steps in the basic constructions are realized by successively adjoining the unitaries U_{−2}, U_{−3}, etc.

With the above description for R_{1} the notation becomes homogeneous, in the sense that gU_{−1}^{3} is a unitary with the property that if we multiply it by U_{−1} or U_{−2} on the right or on the left, the two products differ by e^{\frac{2πi}{3}} . Therefore, adjoining the unitary U_{−2} to the algebra R_{1} we obtain the next step in the Jones basic construction.

To construct unitaries with the properties in the statement one has just to look for the unitaries implementing the crossed product in the Jones tower for an inclusion of the form R ⊂ R ⋊_{β} ℤ_{9}, where β is a ℤ_{3}–kernel (with obstruction e^{\frac{−2πi}{3}} to lifting, β^{3} = Ad w, and w is a unitary with spectral projections of trace \frac{1}{3}). The unitary g of order 3 is chosen from the elements of the first relative commutant.

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The next step is to give a concrete realization of the crossed product (𝔏 (F_{\frac{11}{3}} ) ⊗ R) ⋊_{γ} ℤ_{3}, for a ℤ_{3}–action γ.

Using the model 𝔏 (F_{\frac{11}{3}} ) = \{X_{1}, X_{2}, X_{3}, u\}′′ we define the automorphism

α on 𝔏 (F_{\frac{11}{3}} ) by:

α(X_{i}) = X_{i+1} for i = 1, 2 and α(X_{3}) = uX_{1}u^{∗},
α(u) = e^{\frac{2πi}{3}} u.

Note that α^{3} = Ad u, so α is a ℤ_{3}–kernel with obstruction e^{\frac{2πi}{3}} to lifting.

For the automorphism β on the hyperfinite II_{1} factor we use the model described in Lemma 4.1: R ≅ R_{0} = {g, U_{0}, U_{1}, …}′′ and β = Ad_{R_{0}}(U_{−1}) with β^{3} = Ad_{R_{0}}(g) and β(g) = e^{\frac{−2πi}{3}} g.

The action of ℤ_{3} on 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} is defined by

γ = (Ad_{(𝔏(F_{\frac{11}{3}} )⊗R_{0})} W^{2}) (α ⊗ β),

where W is any cube root of u ⊗ g belonging to the fixed point algebra of α ⊗ β in 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}. For example, if δ = e^{\frac{2πi}{9}} , we can choose W equal to

W = δE_{1} + δ^{2}E_{2} + E_{3},

where E_{1} = e_{1}g_{3}+e_{2}g_{2}+e_{3}g_{1}, E_{2} = e_{1}g_{1}+e_{2}g_{3}+e_{3}g_{2} and E_{3} = e_{1}g_{2}+e_{2}g_{1}+e_{3}g_{3}.

Observation 4.2 Note that W belongs to the center of the fixed point algebra of α ⊗ β since for any element z in the fixed point algebra we have

z (u ⊗ g) = (α ⊗ β)^{3}(z (u ⊗ g)) = Ad (u ⊗ g)(z (u ⊗ g)) = (u ⊗ g) z.

In M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3}, we denote by v the unitary implementing the

crossed product. Thus the spectral projection of v in M have trace \frac{1}{3}. Using the relations [W, u] = 0 and [W, g] = 0 we conclude that

Ad v(g) = e^{\frac{−2πi}{3}} g,  (1)

and

Ad v(u) = e^{\frac{2πi}{3}} u.  (2)

In addition,

Ad v(x) = Ad W^{2}(α(x)), for all x ∈ {X_{1}, X_{2}, X_{3}}′′  (3)

Ad v(y) = Ad W^{2}(β(y)), for all y ∈ {U_{0}, U_{1}, U_{2}, …}′′.  (4)

We can now prove our main theorem:

Theorem 4.3 Let M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3}. Then there exist subalgebras A ⊂ B in M such that A is isomorphic to 𝔏(F_{\frac{35}{27}} ) and B is equal to the crossed product A ⋊_{θ} ℤ_{9}, where θ is a ℤ_{3}–kernel on A with non-trivial obstruction to lifting. Furthermore, M is the enveloping algebra for the inclusion A ⊂ B.

Proof Let A denote the von Neumann subalgebra of M generated by {X_{1}, X_{2}, X_{3}, u, g, v}. We want to describe how Ad U_{0} acts on the elements generating A.

Obviously

(Ad U_{0})(X_{i}) = X_{i} for i = 1, …, 3,

(Ad U_{0})(u) = u and (Ad U_{0})(g) = e^{\frac{−2πi}{3}} g.

Moreover, using (4) and the explicit expression for W^{2} = δ^{2}E_{1} + δ^{4}E_{2} + E_{3}, where δ = e^{\frac{2πi}{9}} , we obtain

(Ad U_{0})(v) = U_{0}(vU_{0}^{∗}v^{∗})v = U_{0} Ad W^{2}(β(U_{0}^{∗}))v = δ̄(Ad U_{0}(W^{2}))(W^{2})^{∗}v

= δ̄(E_{1} + δ^{2}E_{2} + δ^{4}E_{3})(W^{2})^{∗}v = e^{\frac{−2πi}{3}} (E_{1} + E_{2} + e^{\frac{−2πi}{3}} E_{3})v.

This implies that Ad U_{0} leaves A invariant. Furthermore, Ad_{A}(U_{0}^{3}) acts identically on \{X_{1}, X_{2}, X_{3}, u, g\}′′, while Ad_{A}(U_{0}^{3})(v) = e^{\frac{−2πi}{3}} v.

Set θ = Ad_{A}(U_{0}). Then, using (1) and the fact that Ad_{A}(g^{∗}) acts identically on X_{1}, X_{2}, X_{3}, u, g we conclude that

θ^{3} = Ad_{A}(g^{∗}) and θ(g^{∗}) = e^{\frac{2πi}{3}} g^{∗}.

Thus θ is a ℤ_{3}–kernel on A with obstruction e^{\frac{2πi}{3}} to lifting. Furthermore, at least algebraically, the algebra B = (A ∪ {U_{0}})′′ can be identified with the crossed product A ⋊_{θ} ℤ_{9} with U_{0} implementing the crossed product.

To complete the proof that B = A⋊_{θ} ℤ_{9} we need to check that any monomial in the variables {X_{1}, X_{2}, X_{3}, u, v, g, U_{0}} can be written using only one occurrence of U_{0} to some power. We also need to verify that any monomial containing U_{0} has zero trace.

Since the crossed product (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3} is implemented by the unitary v, any monomial in the variables {X_{1}, X_{2}, X_{3}, u, g, v, U_{0}} has the form v^{k}m, where m is an element in 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} and k = 0, 1, 2. But by the construction of R_{0}, U_{0} can appear at most once in m, so any monomial in B can be written using only one occurrence of U_{0}.

Furthermore, by the definition of the trace on the crossed product, the trace of any monomial of the form v^{k}m for k = 1, 2, is zero. So we have only to look at the trace of a monomial of the form m containing U_{0}. But looking at the way the trace is defined on R_{0} and on 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}, we can conclude immediately that m has zero trace. Therefore B = A ⋊_{θ} ℤ_{9}.

In addition, using an argument similar to the one used to build the model for R (Lemma 4.1), one can easily verify that the unitaries U_{1}, U_{2}, … implement the consecutive terms in the iterated basic construction of A ⊂ A⋊_{θ}ℤ_{9}. This implies that (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3} is the enveloping algebra of the above inclusion of subfactors.

The last step of the proof is to show that A is isomorphic to 𝔏 (F_{\frac{35}{27}} ). This will be the content of the next proposition.

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Proposition 4.4 With the same notation as in the previous theorem, define A = {X_{1}, X_{2}, X_{3}, u, g, v}′′ endowed with a trace with respect to which X_{1}, X_{2}, X_{3}, u are free, X_{1}, X_{2}, X_{3} are semicircular and u, g, v are unitaries with spectral projections of trace \frac{1}{3}.

In addition, the trace on the algebra A has the following properties

(i) Any monomial m in X_{1}, X_{2}, X_{3}, u, g, v which contains v has zero trace.

(ii) If m does not contain v then its trace in A is the same as its trace in {X_{1}, X_{2}, X_{3}, u}′′ ⊗ {g}′′.

Under these conditions A is a free group factor with fractional number of generators equal to \frac{35}{27}.

Proof This proof is based on Voiculescu’s random matrix model introduced in [Vo]. We intend to give an explicit random matrix model for the algebra A and then use it to show that A is a free group factor. This model will be a subalgebra of the algebra of 9 × 9 matrices with entries in a von Neumann algebra.

Let D be any von Neumann algebra with finite trace τ, containing free elements {a_{i}}_{i=1,...,18}, where a_{1}, a_{4}, a_{7}, a_{10}, a_{12}, a_{14}, a_{16}, a_{17}, a_{18} are semicircular, while a_{2}, a_{3}, a_{5}, a_{6}, a_{8}, a_{9}, a_{11}, a_{13}, a_{15} are circular. Denote by (e_{ij})_{i,j=1,...,9} the canonical system of matrix units in M_{9}(ℂ). Note that by [Jo3] the unitaries {u, g, v} generate a copy of M_{3}(ℂ) ⊕ M_{3}(ℂ) ⊕ M_{3}(ℂ) ⊆ M_{9}(ℂ), with minimal projections of trace \frac{1}{9}.

Set ϵ = e^{\frac{2πi}{3}} . We let in M_{9}(ℂ) ⊆ D ⊗ M_{9}(ℂ):

u = ϵ (e_{11} + e_{44} + e_{77}) + ϵ^{2}(e_{22} + e_{55} + e_{88}) + (e_{33} + e_{66} + e_{99}),
g = ϵ (e_{11} + e_{55} + e_{99}) + ϵ^{2}(e_{33} + e_{44} + e_{88}) + (e_{22} + e_{66} + e_{77}),
v = (e_{12} + e_{23} + e_{31}) + (e_{45} + e_{56} + e_{64}) + (e_{78} + e_{89} + e_{97})

or, in matrix notation

u = (\begin{matrix}ϵ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & ϵ^{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ϵ & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & ϵ^{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & ϵ & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & ϵ^{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix})

g = (\begin{matrix}ϵ & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & ϵ^{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & ϵ^{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & ϵ & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & ϵ^{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & ϵ \end{matrix})

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

With this choice of u, g, v and with δ = e^{\frac{2πi}{9}} we obtain

W = δ^{2}Id ⊕ Id ⊕ δ Id,

where Id : 2001 − 031.tex, v1.12001/10/1517 : 45 : 33levyExplevy denote the identity of M_{3}(ℂ). In matrix notation

W = (\begin{matrix}δ^{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & δ^{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & δ^{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & δ & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & δ & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & δ \end{matrix})

Furthermore, this choice of u, g, v satisfies the required conditions

ug = gu,  vu = ϵ uv,  vg = \overline{ϵ} gv.

Denote by Re(a) = \frac{1}{2}(a + a^{∗}), for a ∈ D ⊗ M_{9}(ℂ). We set:

X_{1} =a_{1} ⊗ e_{11} + a_{4} ⊗ e_{22} + a_{7} ⊗ e_{33} + a_{10} ⊗ e_{44} + a_{12} ⊗ e_{55} + a_{14} ⊗ e_{66}

+ a_{16} ⊗ e_{77} + a_{17} ⊗ e_{88} + a_{18} ⊗ e_{99} + 2Re(a_{2} ⊗ e_{15}) + 2Re(a_{3} ⊗ e_{19}) + 2Re(a_{5} ⊗ e_{26}) + 2Re(a_{6} ⊗ e_{27}) + 2Re(a_{8} ⊗ e_{34}) + 2Re(a_{9} ⊗ e_{38})

+ 2Re(a_{11} ⊗ e_{48}) + 2Re(a_{13} ⊗ e_{59}) + 2Re(a_{15} ⊗ e_{67}).

To find X_{2} and X_{3}, use the relations:

X_{2} = Ad ((W^{2})^{∗}v)(X_{1}) and X_{3} = Ad ((W^{2})^{∗}v)(X_{2})

Thus, using matrix notation we have:

X_{1} = (\begin{matrix}a_{1} & 0 & 0 & 0 & a_{2} & 0 & 0 & 0 & a_{3} \\ 0 & a_{4} & 0 & 0 & 0 & a_{5} & a_{6} & 0 & 0 \\ 0 & 0 & a_{7} & a_{8} & 0 & 0 & 0 & a_{9} & 0 \\ 0 & 0 & a_{8}^{∗} & a_{10} & 0 & 0 & 0 & a_{11} & 0 \\ a_{2}^{∗} & 0 & 0 & 0 & a_{12} & 0 & 0 & 0 & a_{13} \\ 0 & a_{5}^{∗} & 0 & 0 & 0 & a_{14} & a_{15} & 0 & 0 \\ 0 & a_{6}^{∗} & 0 & 0 & 0 & a_{15}^{∗} & a_{16} & 0 & 0 \\ 0 & 0 & a_{9}^{∗} & a_{11}^{∗} & 0 & 0 & 0 & a_{17} & 0 \\ a_{3}^{∗} & 0 & 0 & 0 & a_{13}^{∗} & 0 & 0 & 0 & a_{18} \end{matrix})

X_{2} = (\begin{matrix}a_{4} & 0 & 0 & 0 & δ̄^{4}a_{5} & 0 & 0 & 0 & δ̄^{2}a_{6} \\ 0 & a_{7} & 0 & 0 & 0 & δ̄^{4}a_{8} & δ̄^{2}a_{9} & 0 & 0 \\ 0 & 0 & a_{1} & δ̄^{4}a_{2} & 0 & 0 & 0 & δ̄^{2}a_{3} & 0 \\ 0 & 0 & δ^{4}a_{2}^{∗} & a_{12} & 0 & 0 & 0 & δ^{2}a_{13} & 0 \\ δ^{4}a_{5}^{∗} & 0 & 0 & 0 & a_{14} & 0 & 0 & 0 & δ^{2}a_{15} \\ 0 & δ^{4}a_{8}^{∗} & 0 & 0 & 0 & a_{10} & δ^{2}a_{11} & 0 & 0 \\ 0 & δ^{2}a_{9}^{∗} & 0 & 0 & 0 & δ̄^{2}a_{11}^{∗} & a_{17} & 0 & 0 \\ 0 & 0 & δ^{2}a_{3}^{∗} & δ̄^{2}a_{13}^{∗} & 0 & 0 & 0 & a_{18} & 0 \\ δ^{2}a_{6}^{∗} & 0 & 0 & 0 & δ̄^{2}a_{15}^{∗} & 0 & 0 & 0 & a_{16} \end{matrix})

X_{3} = (\begin{matrix}a_{7} & 0 & 0 & 0 & δa_{8} & 0 & 0 & 0 & δ̄^{4}a_{9} \\ 0 & a_{1} & 0 & 0 & 0 & δa_{2} & δ̄^{4}a_{3} & 0 & 0 \\ 0 & 0 & a_{4} & δa_{5} & 0 & 0 & 0 & δ̄^{4}a_{6} & 0 \\ 0 & 0 & δ̄a_{5}^{∗} & a_{14} & 0 & 0 & 0 & δ^{4}a_{15} & 0 \\ δ̄a_{8}^{∗} & 0 & 0 & 0 & a_{10} & 0 & 0 & 0 & δ^{4}a_{11} \\ 0 & δ̄a_{2}^{∗} & 0 & 0 & 0 & a_{12} & δ^{4}a_{13} & 0 & 0 \\ 0 & δ^{4}a_{3}^{∗} & 0 & 0 & 0 & δ̄^{4}a_{13}^{∗} & a_{18} & 0 & 0 \\ 0 & 0 & δ^{4}a_{6}^{∗} & δ̄^{4}a_{15}^{∗} & 0 & 0 & 0 & a_{16} & 0 \\ δ^{4}a_{9}^{∗} & 0 & 0 & 0 & δ̄^{4}a_{11}^{∗} & 0 & 0 & 0 & a_{17} \end{matrix})

Obviously X_{1}, X_{2}, X_{3} commute with g. Moreover, using Voiculescu’s random matrix theory [Vo] and the assumption that the family {a_{i}}_{i=1,...,18} is free, it is easy to check that X_{1}, X_{2}, X_{3}, u are free with respect to the unique normalized trace on D ⊗ M_{9}(ℂ). In addition, X_{1}, X_{2}, X_{3}, v are independent (with respect to the trace) of g.

To show that the normalized trace of D ⊗ M_{9}(ℂ) has the properties claimed in the the statement let v^{k}m, k = 1, 2 be any monomial in X_{1}, X_{2}, X_{3}, u, g, v containing v. Since m commutes with g, it must be of the form

m = (\begin{matrix}∗ & 0 & 0 & 0 & ∗ & 0 & 0 & 0 & ∗ \\ 0 & ∗ & 0 & 0 & 0 & ∗ & ∗ & 0 & 0 \\ 0 & 0 & ∗ & ∗ & 0 & 0 & 0 & ∗ & 0 \\ 0 & 0 & ∗ & ∗ & 0 & 0 & 0 & ∗ & 0 \\ ∗ & 0 & 0 & 0 & ∗ & 0 & 0 & 0 & ∗ \\ 0 & ∗ & 0 & 0 & 0 & ∗ & ∗ & 0 & 0 \\ 0 & ∗ & 0 & 0 & 0 & ∗ & ∗ & 0 & 0 \\ 0 & 0 & ∗ & ∗ & 0 & 0 & 0 & ∗ & 0 \\ ∗ & 0 & 0 & 0 & ∗ & 0 & 0 & 0 & ∗ \end{matrix})

If we multiply m by v or v^{2} we obtain a matrix with zero on the diagonal and a few entries outside the diagonal. This implies that v^{k}m has zero trace for k = 1, 2. Thus, we have built a matrix model for A which satisfies the conditions of the statement.

In order to show that A is a free group factor we will reduce A by one of the spectral projections of g of trace \frac{1}{3}, and show that the new factor we obtain is a free group factor. Reduce A by g_{3} = e_{22} + e_{66} + e_{77}.

Then g_{3}Ag_{3} is generated by:

(i) a_{4} ⊗ e_{22} + 2Re(a_{5} ⊗ e_{26}) + 2Re(a_{6} ⊗ e_{27}) + a_{14} ⊗ e_{66} + a_{16} ⊗ e_{77} + 2Re(a_{15} ⊗ e_{67})

(ii) a_{7} ⊗ e_{22} + 2Re(δ̄^{4}a_{8} ⊗ e_{26}) + 2Re(δ̄^{2}a_{9} ⊗ e_{27}) + a_{10} ⊗ e_{66} + a_{17} ⊗ e_{77} + 2Re(δ^{2}a_{11} ⊗ e_{67})

(iii) a_{1} ⊗ e_{22} + 2Re(δa_{2} ⊗ e_{26}) + 2Re(δ̄^{4}a_{3} ⊗ e_{27}) + a_{12} ⊗ e_{66} + a_{18} ⊗ e_{77} + 2Re(δ^{4}a_{13} ⊗ e_{67})

(iv) ϵ^{2} ⊗ e_{22} + 1 ⊗ e_{66} + ϵ ⊗ e_{77},

which, by Voiculescu’s random matrix model [VDN], is isomorphic to the free group factor 𝔏(F_{3} ∗ ℤ_{3}) ≅ 𝔏 (F_{\frac{11}{3}} ). This implies that A is also a free group factor, and using the well-known formula (see [Dy] or [Ra1])

(𝔏(F_{s}))_{\frac{1}{N}} = 𝔏(F_{N^{2}(s−1)+1}),

we obtain that A = 𝔏 (F_{\frac{35}{27}} ).

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5 The Connes invariant for the crossed product (𝔏 (F_{\frac{11}{3}}) ⊗ R_{0}) ⋊_{γ} ℤ_{3}

The arguments in this section are the same ones used by Jones in [Jo2] to compute the Connes invariant for his example of a factor which is anti-isomorphic to itself but has no involutory antiautomorphism. For the sake of completeness we will present them here for the factor we are considering. The definition of K, K^{⊥} and L given below are due to Connes (see [Co4]).

Let N be a factor without non-trivial hypercentral sequences and G a finite subgroup of Aut(N) such that G ∩ \overline{Int(N)} = \{Id\}. Let K = G ∩ Ct(N) and K^{⊥} = {f : G → 𝕋 | f is a homomorphism and f|_{K} ≡ 1}.

Let ξ : Aut(N) → Out(N) be the usual quotient map and Fint be the subgroup \{Ad u | u ∈ N is fixed by G\} ⊆ Aut(N). Denote by \overline{Fint} its closure in Aut(N) and by G ∨ Ct(N) the subgroup of Aut(N) generated by {G ∪ Ct(N)}. Let L = ξ((G ∨ Ct(N)) ∩ \overline{Fint}) ⊆ Out(N).

Connes shows in [Co4] that there exists an exact sequence

0 → K^{⊥} →^{∂} χ(M) →^{Π} L → 0

where M = W^{∗}(N, G). The map ∂ is defined in the following way: write an element in the crossed product W^{∗}(N, G) as ∑_{g∈G} a_{g}u_{g}, with a_{g} ∈ N and Ad u_{g} = g on N. For each η ∈ K^{⊥}, η : G → 𝕋, define

Δ(η) (∑_{g∈G} a_{g}u_{g}) = ∑_{g∈G} η(g)a_{g}u_{g}.

Then one can show that Δ(η) ∈ Ct(M)∩\overline{Int(M)} and define ∂ as the composition ε ∘ Δ.

The map Π : χ(M) → L, is defined as follows. For σ ∈ χ(M) choose α ∈ Ct(M) ∩ \overline{Int(M)} such that ξ(α) = σ.

From the hypothesis G ∩ \overline{Int(N)} = \{Id\}, it follows (see Corollary 6 and Lemma 2 in [Jo1], or Lemma 15.42 in [EK]) that there exists a unitary z ∈ M and a sequence u_{n} of unitaries in N^{G} such that α = Ad z lim_{n→∞} Ad u_{n}. Let ψ_{σ} = (Ad z^{∗}α)|_{N}.

By construction ψ_{σ} ∈ \overline{Fint} and one can prove that ψ_{σ} ∈ G ∨ Ct(N). Moreover, the image of ψ_{σ} in Out(N) does not depend on the particular choice of α. Therefore we can define a map Π : χ(M) → L by Π(σ) = ξ(ψ_{σ}).

To show that Π is surjective, consider an element μ in L and let α_{μ} ∈ \overline{Fint} be such that ε(α_{μ}) = μ. α_{μ} commutes with G since it is the limit of automorphisms with this property, thus we may define an automorphism β_{μ} of M by

β_{μ} (∑_{g∈G} a_{g}u_{g}) = ∑_{g∈G} α_{μ}(a_{g})u_{g}.  (5)

Then β_{μ} ∈ Ct(M) ∩ \overline{Int(M)} and Π(ξ(β_{μ})) = μ.

Remark 5.1 Note that if u_{n} is a sequence of unitaries left invariant by G, with the property α_{μ} = lim_{n→∞} Ad u_{n} in Aut(N), then β_{μ} = lim_{n→∞} Ad u_{n} in Aut(M). It follows that

κ(ε(β_{μ})) = lim_{n→∞}(u_{n})^{∗}β_{μ}(u_{n}) = lim_{n→∞} (u_{n})^{∗}α_{μ}(u_{n}).

Our next goal is to show that if N = 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} and G is the subgroup

⟨γ⟩ ⊂ Aut (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) then χ(M) ≅ ℤ_{9}, as has been observed by Connes in [Co3].

For all the rest of this paper we will identify ⟨γ⟩ with ℤ_{3}. Note that by Proposition 3.4, 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} has no non-trivial hypercentral sequences, so in order to use the exactness of the Connes sequence described above we only need

to show that ℤ_{3} ∩ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})} = \{Id\}. Obviously it is enough to check

that γ ∉ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})}.

This is equivalent to show that α ⊗ β ∉ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})}. By Corollary

3.3 in [Co5] α ⊗ β ∈ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})} if and only if α ∈ \overline{Int (𝔏 (F_{\frac{11}{3}} ))} and

β ∈ \overline{Int(R)}. But 𝔏 (F_{\frac{11}{3}} ) is full (Corollary 3.3) and α ∉ Int (𝔏 (F_{\frac{11}{3}} )), thus

we can conclude that ℤ_{3} ∩ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})} = \{Id\}.

Hence by Connes [Co4], the sequence

0 → K^{⊥} →^{∂} χ(M) →^{Π} L → 0

is exact. In order to compute χ(M) we intend to show that

K^{⊥} = ℤ_{3} and L = ℤ_{3},

and that the lifting to χ(M) of the generator of L produces an element of order 9.

Lemma 5.2 The group K = ℤ_{3} ∩ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) is trivial.

Proof Obviously it is enough to show that the automorphism γ = Ad W^{2}(α⊗β) is not in Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}). We have already shown in the proof of Proposition

3.4 that any central sequence in 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} has the form (1 ⊗ x_{n})_{n∈ℕ} for a

central sequence (x_{n})_{n∈ℕ} in R_{0}. It follows that γ ∈ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) if and

only if β ∈ Ct(R_{0}). But for ϵ = e^{\frac{2πi}{3}} , β is outer conjugate to the automorphism sϵ̄ ; 3 described by Connes in [Co6], and sϵ̄ ; 3 ∉ Ct(R_{0}) by Proposition 1.6 in the same paper. Therefore β ∉ Ct(R_{0}).

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Lemma 5.3 L ≅ ℤ_{3}, and a generator of L is given by

μ = ξ (Id ⊗ (Ad U_{0}^{∗} β))) ,

where U_{0} is the unitary in R_{0} defined in Lemma 4.1 with the property β(U_{0}) = e^{\frac{2πi}{9}} U_{0}.

Proof We want to show that

Id ⊗ (Ad U_{0}^{∗} β) ∈ (ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})) ∩ \overline{Fint}.

Let’s look first at ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}): multiplying Id ⊗ (Ad U_{0}^{∗} β) by the

automorphism γ^{−1} = (Ad W^{2}(α ⊗ β))^{−1} we obtain

Ad ((W^{2})^{∗}(1 ⊗ U_{0}^{∗})) (α^{−1} ⊗ Id),

which is in Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) since, by Remark 3.1, 𝔏 (F_{\frac{11}{3}} ) does not have

non-trivial central sequences. Thus Id ⊗ (Ad U_{0}^{∗} β) ∈ ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}).

To prove that Id ⊗ (Ad U_{0}^{∗} β) ∈ \overline{Fint} we have to show that there exists a sequence (ũ_{n}) of unitaries in 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} invariant respect to γ, such that Id ⊗ (Ad U_{0}^{∗} β) = lim_{n→∞} Ad ũ_{n}. First of all observe that the sequence (x_{n}) of unitaries in R_{0}

x_{n} = \begin{cases}U_{0}U_{1}^{∗}U_{2}U_{3}^{∗} … U_{n}^{∗}, & if n is odd, \\ U_{0}U_{1}^{∗}U_{2}U_{3}^{∗} … U_{n}, & if n is even \end{cases}

has the properties

β = lim_{n→∞} Ad x_{n} and β(x_{n}) = e^{\frac{2πi}{9}} x_{n}.

Define u_{n} = U_{0}^{∗} x_{n}. Obviously

(Ad U_{0}^{∗}) β = lim_{n→∞} Ad u_{n} and β(u_{n}) = u_{n}

so that

(α ⊗ β)(1 ⊗ u_{n}) = 1 ⊗ u_{n}.  (6)

Since by Observation 4.2 W belongs to the center of the fixed point algebra of α ⊗ β, using (6) we obtain

γ(1 ⊗ u_{n}) = Ad W^{2}(α ⊗ β)(1 ⊗ u_{n}) = 1 ⊗ u_{n}

and Id⊗(Ad U_{0}^{∗} β) = lim_{n→∞} Ad (1 ⊗ u_{n}). Therefore ũ_{n} = 1⊗u_{n} is the desired sequence and Id ⊗ (Ad U_{0}^{∗} β) ∈ (ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})) ∩ \overline{Fint}.

So far we know that μ = ξ (Id ⊗ (Ad U_{0}^{∗} β)) is an element of order 3 of L = ξ ((ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})) ∩ \overline{Fint}). We want to show that it generates

L. This means that if φ is any element in (ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})) ∩ \overline{Fint}, then it differs from a power of Id ⊗ (Ad U_{0}^{∗} β) by an inner automorphism.

By Theorem 3.5 any automorphism in Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) can be written

as Ad z(ν ⊗ id) for some unitary z ∈ 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} and an automorphism

ν ∈ Aut (𝔏 (F_{\frac{11}{3}} )). Thus, since φ ∈ ℤ_{3} ∨ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) there is an

n ∈ {0, 1, 2} such that φ Ad (W^{2})^{n}(α ⊗ β)^{n} is of the form Ad z(ν ⊗ Id). Solving for φ we obtain

φ = Ad x(να^{−n} ⊗ β^{−n}),

where x = z(ν ⊗ Id)(W^{2n})^{∗} ∈ 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}.

Since φ ∈ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})} and by Corollary 3.3 𝔏 (F_{\frac{11}{3}} ) is full, we may apply Corollary 3.3 in [Co5] to conclude that να^{−n} = Ad w for some unitary w. This implies that φ = Ad x′(Id ⊗ β^{−n}), where x′ = x (w ⊗ 1). Thus φ differs from Id ⊗ (Ad U_{0}^{∗} β) only by an inner automorphism. Hence Id ⊗ (Ad U_{0}^{∗} β) generates L.

■

Note that if u_{n} is the sequence of unitaries defined in the previous lemma

(Id ⊗ (Ad U_{0}^{∗} β))(1 ⊗ u_{n}) = 1 ⊗ U_{0}^{∗}u_{n}U_{0}.

But lim_{n→∞} Ad u_{n} = Ad U_{0}^{∗} β and β(U_{0}) = e^{\frac{2πi}{9}} U_{0}, so

lim_{n→∞} u_{n}U_{0}u_{n}^{∗} = Ad U_{0}^{∗}(β(U_{0})) = e^{\frac{2πi}{9}} U_{0}

and

lim_{n→∞} (1 ⊗ u_{n}^{∗})(Id ⊗ (Ad U_{0}^{∗} β))(1 ⊗ u_{n}) = e^{\frac{2πi}{9}} .

Thus, in view of Remark 5.1 we obtain the following: Remark 5.4 If μ = Id ⊗ (Ad U_{0}^{∗} β), β_{μ} ∈ Ct(M) ∪ \overline{Int(M)} is as described in equation [5] and σ = ε(β_{μ}), then κ(σ) = e^{\frac{2πi}{9}} . Theorem 5.5 Let M denote the crossed product (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})⋊_{γ} ℤ_{3}. Then χ(M) = ℤ_{9}. Proof. By Connes [Co4] the sequence

0 → ℤ_{3} → χ(M) → ℤ_{3} → 0

is exact and a lifting to χ(M) of a generator of L ≅ ℤ_{3} is defined on M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3} by σ = ξ(β_{μ}), where

β_{μ} (∑_{k=0}^{2} a_{k}v^{k}) = ∑_{k=0}^{2} (Id ⊗ (Ad U_{0}^{∗} β))(a_{k})v^{k}.

Since 3 is the only non-trivial divisor of 9, to prove that χ(M) ≅ ℤ_{9} it suffices to show that σ^{3} ≠ 1, i.e. β_{μ}^{3} ∉ Int(M).

From the relations

(α ⊗ β)(1 ⊗ (U_{0}^{∗})^{3}g) = e^{\frac{2πi}{3}} (1 ⊗ (U_{0}^{∗})^{3}) g

and

Ad W^{2}(1 ⊗ (U_{0}^{∗})^{3}g) = 1 ⊗ (U_{0}^{∗})^{3}g

we obtain that

γ(1 ⊗ (U_{0}^{∗})^{3}g) = e^{\frac{2πi}{3}} (1 ⊗ (U_{0}^{∗})^{3}g).

But Ad v = γ on 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}, so that Ad v(1 ⊗ (U_{0}^{∗})^{3}g) = e^{\frac{2πi}{3}} (1 ⊗ (U_{0}^{∗})^{3} g) implies

Ad (1 ⊗ (U_{0}^{∗})^{3} g)(v) = e^{\frac{−2πi}{3}} v.

Using the definition of β_{μ} and what we have just observed we obtain

β_{μ}^{3} (∑_{k=0}^{2} a_{k} v^{k}) = ∑_{k=0}^{2} (Id ⊗ (Ad (U_{0}^{∗})^{3} β^{3})) (a_{k})v^{k}

= ∑_{k=0}^{2} (Id ⊗ Ad(U_{0}^{∗})^{3} Ad g) (a_{k})v^{k} = ∑_{k=0}^{2} Ad(1 ⊗ (U_{0}^{∗})^{3}g)(a_{k})v^{k}

= Ad (1 ⊗ (U_{0}^{∗})^{3}g) (∑_{k=0}^{2} e^{\frac{2πik}{3}} a_{k} v^{k}) .

Thus, being the product of a dual action and an inner automorphism β_{μ}^{3} must be outer. Hence χ(M) ≅ ℤ_{9}.

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6 (𝔏 (F_{\frac{11}{3}}) ⊗ R_{0}) ⋊_{γ} ℤ_{3} is not anti–isomorphic to itself.

In this section we are going to show that M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3} is not

anti-isomorphic to itself by using the dual action ℤ̂_{3} → Aut(M), which gives rise to the only subgroup of order 3 in χ(M). This argument has been described by Connes in [Co3] and [Co4].

First of all note that the action γ can be decomposed as γ = Ad w γ_{1}γ_{2} where

γ_{1} ∈ Ct (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) and γ_{2} ∈ \overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})} and w is a unitary in

𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}.

In fact, γ = Ad W^{2}(α ⊗ Id)(Id ⊗ β) and γ_{1} = α ⊗ Id is centrally trivial, since any central sequence in 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} has the form (1 ⊗ X_{n}), for a central sequence (X_{n}) in R_{0}. Furthermore in the proof of Lemma 4.1 we showed that β = lim_{n→∞} Ad x_{n} so that γ_{2} = Id ⊗ β = lim_{n→∞} Ad (1 ⊗ x_{n}) belongs to

\overline{Int (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})}.

Note also that this decomposition of γ into an approximately inner automorphism and a centrally trivial automorphism is unique up to inner automorphisms since χ (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) = 1.

Now let M be an arbitrary von Neumann algebra. Define the conjugate M^{c} of M as the algebra whose underlying vector space is the conjugate of M (for λ ∈ ℂ, x ∈ M the product λ by x in M^{c} is equal to λ̄x) and whose ring structure is the same as in M. The opposite M^{o} of M is by definition the algebra whose underlying vector space is the same as for M while the product of x by y is equal to yx instead of xy. M^{c} and M^{o} are clearly isomorphic through the map x → x^{∗}. For ψ ∈ Aut(M) we denote by ψ^{c} the automorphism of M^{c} induced by ψ.

The proof of the following theorem is due to Connes ([Co3] and [Co4]).

Theorem 6.1 M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⋊_{γ} ℤ_{3} is not anti–isomorphic to itself.

Proof We proved in the previous section (Theorem 5.5) that χ(M) ≅ ℤ_{9} and that the dual action ℤ̂_{3} → Aut(M) produces the only subgroup of order 3 in χ(M), namely ⟨σ^{3}⟩ = ⟨ξ(β_{μ}^{3})⟩.

Since χ(M) is an invariant of the von Neumann algebra M, it follows that ℤ̂_{3} is an invariant of M. This implies that M ⋊γ̂ ℤ̂_{3} is an invariant of M as it is the dual action ℤ_{3} → Aut(M ⋊γ̂ ℤ̂_{3}) on M ⋊γ̂ ℤ̂_{3}.

By Takesaki’s duality theory [Theorem 4.5, [Ta]]

M ⋊γ̂ ℤ̂_{3} = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0}) ⊗ B(ℓ^{2}(ℤ_{3})).

Moreover, the dual action ˆ γ̂ of γ̂ corresponds to the action γ ⊗Ad (λ(·)^{∗}), where λ(·) is the regular representation of ℤ_{3} on ℓ^{2}(ℤ_{3}) defined by

λ(h)ε(k) = ε(k − h) for h, k ∈ ℤ_{3}, ε ∈ ℓ^{2}(ℤ_{3}).

Since ⟨ˆγ̂⟩ ≅ ℤ_{3} is an invariant of M, it follows that the action γ of ℤ_{3} on 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} is an invariant of M. Consequently the approximately inner automorphism and the centrally trivial automorphism γ_{1} and γ_{2} which appear in the decomposition of γ are invariants of M.

Note that

γ_{1}^{3} = Ad (u ⊗ 1) and γ_{1}(u ⊗ 1) = e^{\frac{2πi}{3}} (u ⊗ 1)

so that γ_{1} is a ℤ_{3}–kernel of 𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0} with obstruction e^{\frac{2πi}{3}} to lifting.

Observe also that in the above argument we have only used the abstract group ℤ̂_{3} and the dual action defined on M ⋊γ̂ ℤ̂_{3}. Hence we have found a canonical way to associate to the von Neumann algebra M a number (equal to e^{\frac{2πi}{3}} in our case) which is invariant under isomorphisms.

Now if M = (𝔏 (F_{\frac{11}{3}} ) ⊗ R_{0})⋊_{γ} ℤ_{3} was anti–isomorphic to itself then M and M^{c} would be isomorphic. But the obstruction associated to γ_{1}^{c}, and therefore to M^{c} is equal to e^{\frac{−2πi}{3}} .

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Acknowledgements

We thank our advisor, Professor F. Rădulescu for his help and support, Professor D. Bisch and Professor D. Shlyakhtenko for useful discussions, and Dr. M. Müeger for his suggestions regarding this manuscript.

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