Prime type III factors

Abstract

It is shown that for each 0 < λ < 1, the free Araki–Woods factor of type III_λ cannot be written as a tensor product of two diffuse von Neumann algebras (i.e., is prime) and does not contain a Cartan subalgebra.

A von Neumann algebra M is called prime if it cannot be written as a tensor product of two diffuse von Neumann algebras. Using Voiculescu's free entropy theory (1), Ge (2) and later Stefan (3) gave examples of prime factors of type II₁ (and hence of type II_∞). An example of a separable prime factor of type III is given here: I show that for each 0 < λ < 1, the type III_λ free Araki–Woods factor T_λ introduced in ref. 4 is prime. The main idea of the proof is to interpret the decomposition T_λ = A ⊗ B as a condition on its core, which is of type II. I then use Stefan's result (3) showing that L(𝔽_∞) cannot be written as the closure of the linear span of N⋅C₁⋅C₂ where N is a II₁ factor, which is not prime, and C_i are abelian von Neumann algebras.

I also prove the existence of separable type III factors that do not have Cartan subalgebras by showing that T_λ, 0 < λ < 1 has no Cartan subalgebras. The key ingredient is Voiculescu's result on the absence of Cartan subalgebras in L(𝔽_∞) (1).

Although the proofs are based on a reduction to the case of type II algebras (for which free entropy methods are available), I believe that the results of this paper should be viewed as an indication that free entropy theory should have an extension to algebras of type III.

T_λ Is Prime

We use the following theorem, due to Connes (see sections 4.2 and 4.3 of ref. 5):

Theorem 2.1. Let M be a separable type III_λ factor with 0 < λ < 1. Then there exists a faithful normal state φ on M, for which:

1.

The centralizer M^φ = {m ∈ M ∶ φ(mn) = φ(nm) ∀n ∈ M} is a factor of type II₁;

2.

The modular group σ_t^φ of φ is periodic, of period exactly 2π/logλ;

3.

M is generated as a von Neumann algebra by M^φ and an isometry V, satisfying:

(a)  V*V = 1, V^k(V*)^k ∈ M^φ for all k;

(b)  σ_t^φ(V) = λ^−it(V); thus, φ(V^k(V*)^k) = λ^kφ((V*)^kV^k) = λ^kφ(1) = λ^k;

(c)  V normalizes M^φ: VmV* and V* mV are both in M^φ if m ∈ M^φ.

The weight φ ⊗ Tr(B(ℓ²)) is unique up to scalar multiples and up to conjugation by (inner) automorphisms of M ≅ M ⊗ B(ℓ²).

Moreover, 2 ⇒ 1 and 3; 1 ⇒ 2 and 3. In particular, if φ₁ and φ₂ satisfy either 1 or 2, the centralizers and are stably isomorphic: ⊗ B(ℓ²) ≅ ⊗ B(ℓ²).

The existence of such a state can be easily seen by writing M as the crossed product of a type II_∞ factor C by a trace-scaling action of ℤ: set φ̂ to be the crossed-product weight (where C is taken with its semifinite trace). Next, compress to a finite projection p ∈ C and set φ = φ̂(p⋅p). The isometry V is precisely the compression of the unitary U, implementing the trace-scaling action of ℤ.

Recall that a von Neumann algebra M is called full if its group of inner automorphisms is closed in the u-topology inside its group of all automorphisms (see ref. 6).

Lemma 2.2. Let M be a full type III_λ factor. Assume that M = A₁ ⊗ A₂, where A₁ and A₂ are von Neumann algebras. Then A₁ and A₂ are both full factors, and exactly one of the following must hold true:

1.

A₁ and A₂ are both of type III_λ1 and III_λ2, respectively, and λ₁, λ₂ satisfy:

(i)  0 < λ_i < 1, i = 1, 2, (ii) λ₁^ℤλ₂^ℤ = λ^ℤ;

2.

For some i ≠ j, A_i is of type III_λ and A_j is of type II;

3.

For some i ≠ j, A_iis of type III_λ and A_jis of type I.

In particular, if we require that A₁ and A₂must both be diffuse, only 1 and 2 can occur. Moreover, if 2 occurs, we may assume that one of the algebras A₁, A₂is of type II₁.

Proof: If one of A₁, A₂ fails to be a factor, then their tensor product would fail to be a factor, hence both A₁ and A₂ must be factors. Similarly, if at least one of A₁ and A₂ fails to be full, their tensor product would fail to be full.

If, say, A₁ is of type I or type II, then A₂ must be type III, because otherwise A₁ ⊗ A₂ would be of type II or type I. Hence if at least one of A₁ and A₂ is not type III, the situation described in 2 or 3 must occur.

If A₁ and A₂ are both type III, so that A is type III_λ1 and A₂ is type III_λ2, we must prove that λ^ℤ = λ₁^ℤλ₂^ℤ. Neither λ₁ nor λ₂ can be zero, because then at least one of A₁, A₂ would then fail to be full, and hence A₁ ⊗ A₂ would fail to be full.

Denote by T(M) the T invariant of Connes (see section 1.3 of ref. 5). Since

[ref. 5, Theorem 1.3.4(c)] and T(A_j) = (2πℤ/logλ_j), statement 1 must hold.

Proposition 2.3. Let M be a type III_λ factor, and assume that M = A₁ ⊗ A₂, where A₁ is a type III_λ1 factor, A is a type III_λ2 factor, and λ^ℤ = λ₁^ℤλ₂^ℤ. Let φ_ibe a normal faithful state on A_i as in Theorem 2.1, and let φ = φ₁ ⊗ φ₂be a normal faithful state on M.

Then the centralizer M^φ of φ in M is a factor, which can be written as a closure of the linear span of N⋅C₁⋅C₂, where N is a tensor product of two type II₁ factors, and C_i are abelian von Neumann algebras. In particular, M^φis not isomorphic to L(𝔽_∞).

Proof: Since the modular group of φ₁ ⊗ φ₂ is σ_t^φ₁ ⊗ σ_t^φ₂, it follows that has period exactly 2π/log λ. Hence the centralizer of φ₁ ⊗ φ₂ is a factor.

Choose now a decreasing sequence of projections p_k⁽¹⁾ ∈ A₁^φ₁, p_k⁽²⁾ ∈ A₂^φ₂, with φ_i(p_k⁽ⁱ⁾) = λ_i^k, and isometries V_i ∈ A_i, so that V^*_iV_i = 1, V_i^k(V^*_i)^k = p_k⁽ⁱ⁾, so that V_i normalizes A_i^φ_i, and A_i = W*(A_i^φ_i, V_i). Then A₁ ⊗ A₂ is densely spanned by elements of the form

with the convention that V_i⁻ⁿ = (V^*_i)ⁿ if n ≥ 0.

Using the fact that V^*_iaV_i, V_iaV^*_i ∈ A_i^φ_i whenever a ∈ A_i^φ_i, we can rewrite W as

Let now p_k = p_k⁽¹⁾ = V₁^k(V^*₁)^k ∈ A₁^φ₁ be as above. One can choose a diffuse commutative von Neumann algebra 𝒜, containing p_k, k ≥ 0, and so that 𝒜 ⊂ A₁^φ₁ and V₁𝒜V^*₁, V^*₁𝒜V₁ ⊂ 𝒜. Choose a projection 𝒜 ∋ q₀ ≤ p₁, so that q₀ ⊥ p₂ and φ₁(q) = Nφ₁(1 − p₁) = N(1 − λ₁) for some integer N. Choose projections 𝒜 ∋ q₁, … , q_n ≤ 1 − p₁, so that Σ_i=1^Nq_i = 1 − p₁ and φ₁(q_i) = φ₁(q₀) = (1/N)φ₁(1 − p₁). Choose matrix units {e_ij}_0≤i,j≤N ⊂ A₁^φ₁, so that e_ii = q_i, 0 ≤ i ≤ N. Let C be the von Neumann algebra generated by

By our choice of e_ij, C is hyperfinite (notice that V₁^ke_ii(V^*₁)^k ∈ 𝒜, and C is in fact the crossed product of 𝒜 ≅ L^∞(X) by a singly generated equivalence relation). Let R₁ = W*(C, V) ⊂ A₁. Then R₁ is also hyperfinite; in fact, it is the crossed product of C by the endomorphism x ↦ V₁xV^*₁. Notice that R₁ contains V₁. Furthermore, for all k ≥ 0, there is a d ≥ 0 and partial isometries r₁, … , r_d ∈ R₁ ∩ A₁^φ₁, so that

Construct in a similar way the algebra R₂ ⊂ A₂, in such a way that V₂ ∈ R₂ and for all k ≥ 0, there is a d ≥ 0 and partial isometries r₁, … , r_d ∈ R₂ ∩ A₂^φ₂, so that

Notice that R₁ ⊗ R₂ ⊂ A₁ ⊗ A₂ is globally fixed by the modular group of φ₁ ⊗ φ₂. In particular, this means that

Assume now that W ∈ (A₁ ⊗ A₂). Then (W) = W. Hence λ₁^m−l⋅λ₂^n−k = 1. It follows that W can be written in one of the following forms, using the fact that V^*_iA_i^φ_iV_i ⊂ A_i^φ_i:

where a⁽¹⁾ ∈ A₁^φ₁, a⁽²⁾ ∈ A₂^φ₂ and λ₁^m = λ₂^k, λ₂ⁿ = λ₁^l. In the first case, choose r₁, … , r_d ∈ R₁ ∩ A₁^φ₁ for which 1 − V₁^m(V^*₁)^m = Σ_i=1^dr_iV₁^m(V^*₁)^mr^*_i. Then, writing

we obtain

Reversing the roles of A₁ and A₂, we get that in general, span{(A₁^φ₁ ⊗ A₂^φ₂)⋅(R₁ ⊗ R₂)} is dense in (A₁ ⊗ A₂).

Since each R_i is hyperfinite, the algebra R₁ ⊗ R₂ is also hyperfinite; hence (R₁ ⊗ R₂) is hyperfinite. It follows that the centralizer of M = A₁ ⊗ A₂ can be written as the closure of the span of NR, where N is a tensor product of two type II₁ factors, and R is a hyperfinite algebra. Since every hyperfinite algebra can be written as a linear span of the product C₁⋅C₂, where C_i are abelian von Neumann algebras, it follows that the centralizer M^φ is the closure of the span of N⋅C₁⋅C₂, with N a tensor product of two type II₁ factors, and C₁, C₂ abelian von Neumann algebras. Hence by Stefan's result (3), we get that M^φ cannot be isomorphic to L(𝔽_∞).

Theorem 2.4. Let T_λbe the free Araki–Woods factor constructed in ref. 4. Then T_λ ≇ A₁ ⊗ A₂, where A₁ and A₂ are any diffuse von Neumann algebras.

Proof: Since T_λ is a full III_λ factor, we have by Lemma 2.2 that the only possible tensor product decompositions with A₁ and A₂ diffuse are ones where either exactly one of A₁ and A₂ is type II₁ and the other is of type III_λ, or each A_i is of type III_λi, with λ₁^ℤλ₂^ℤ = λ^ℤ.

Denote by ψ the free quasifree state on T_λ. It is known (see ref. 4, Corollary 6.8) that T_λ^ψ is a factor, isomorphic to L(𝔽_∞). Let φ be an arbitrary normal faithful state on T_λ, such that T_λ^φ is a factor. Then (see Theorem 2.1), T_λ^φ ⊗ B(ℓ²) ≅ T_λ^ψ ⊗ B(ℓ²) ≅ L(𝔽_∞) ⊗ B(ℓ²). Since L(𝔽_∞) has ℝ₊ as its fundamental group (see ref. 7), it follows that whenever φ is a state on T_λ, and T_λ^φ is a factor, then T_λ^φ ≅ L(𝔽_∞).

Assume now that one of A₁, A₂ is of type II₁; for definiteness, assume that it is A₁. Choose on A₂ a normal faithful state φ₂ for which A^φ₂ is a factor, and let τ be the unique trace on A₁. Let φ = τ ⊗ φ₂ on T_λ. Then T_λ^φ ≅ A₁ ⊗ A₂^φ₂, and hence cannot be isomorphic to L(𝔽_∞) by the results of Stephan (3) and Ge (2). This is a contradiction.

Assume now that A_i is type III_λi, with 0 < λ_i < 1. Then by Proposition 2.3 there is a state φ on T_λ, for which T_λ^φ is a factor, but is not isomorphic to L(𝔽_∞); contradiction.

3. T_λ Has No Cartan Subalgebras

Recall that a von Neumann algebra M is said to contain a Cartan subalgebra A if:

1.

A ⊂ M is a MASA (maximal abelian subalgebra).

2.

There exists a faithful normal conditional expectation E ∶ M → A.

3.

M = W*(𝒩(A)), where 𝒩(A) = {u ∈ M ∶ uAu* = A, u*u = uu* = 1} is the normalizer of A.

For type II₁ factors M, condition 2 is automatically implied by condition 1.

Proposition 3.1. Let M be a factor of type III_λ, 0 < λ < 1. Then there exists a normal faithful state ψ on M, so that σ_2π/logλ^ψ = id, and that the centralizer M^ψ is a II₁ factor containing a Cartan subalgebra.

Proof: Let A ⊂ M be a Cartan subalgebra. Let E ∶ M → A be a normal faithful conditional expectation. Let φ be a normal faithful state on A ≅ L^∞[0, 1], and denote by θ the state φ○E on M. Then θ is a normal faithful state. Furthermore, M^θ ⊃ A, because E is θ-preserving and hence σ^θ|_A = σ^θ|_A = id. Since M is type III_λ, it follows that σ_t0^θ is inner if t₀ = 2π/log λ. Let u ∈ M be a unitary for which σ_t0^θ(m) = umu*, ∀m ∈ M. Then uxu* = x for all x ∈ A, since σ^θ|_A = id. It follows that u ∈ A′ ∩ M = A′, since A is a MASA. Choose d ∈ A positive so that d^it₀ = u. Note that d is in the centralizer of θ (which contains A). Set ψ(m) = θ(d⁻¹m) for all m ∈ M. Then the modular group of ψ at time t₀ is given by Ad_u* ○ σ_t0^θ = id. It follows that ψ is a normal faithful state on M, so that σ_t0^ψ = id. It furthermore follows from Theorem 2.1 that the centralizer of M^ψ is a factor of type II₁. By the choice of ψ, its modular group fixes A pointwise, hence A ⊂ M^ψ.

I claim that A is a Cartan subalgebra in N = M^ψ. First, A′ ∩ N ⊂ A′ ∩ M = A, hence A is a MASA. Since A is a Cartan subalgebra in M, M is densely linearly spanned by elements of the form f⋅u, where u ∈ 𝒩(A) is a unitary and f ∈ A. The map

is a normal faithful conditional expectation from M onto N. If u ∈ 𝒩(A) is a unitary, so that ufu* = α(f) for all f ∈ A and α ∈ Aut(A), then uf = α(f)u. Hence

It follows that N is densely linearly spanned by elements of the form E(f⋅u) = f⋅E(u) for f ∈ A and u ∈ 𝒩(A). Let w(u) be the polar part of E(u), and let p(u) = E(u)*E(u) be the positive part of E(u), so that E(u) = w(u)p(u) is the polar decomposition of E(u). Since

it follows that p(u) commutes with A and hence is in A. Moreover, we then have that

so that w(u) ∈ 𝒩(A) ∩ N. Thus N is densely linearly spanned by elements of the form f⋅u for f ∈ A and u ∈ 𝒩(A) ∩ N, hence A is a Cartan subalgebra of N.

Corollary 3.2. For each 0 < λ < 1 the III_λ free Araki–Woods factor T_λdoes not have a Cartan subalgebra.

Proof: If T_λ were to contain a Cartan subalgebra, it would follow that for a certain state ψ on T_λ, the centralizer of ψ is a factor containing a Cartan subalgebra. Let φ be the free quasifree state on T_λ. Then by Theorem 2.1, one has

Since (T_λ)^φ ≅ L(𝔽_∞) (see Corollary 6.8 of ref. 5), and because the fundamental group of L(𝔽_∞) is all of ℝ₊ (see ref. 7) we conclude that L(𝔽_∞) contains a Cartan subalgebra. But this is in contradiction to a result of Voiculescu that L(𝔽_∞) has no Cartan subalgebras (see ref. 1).