2013
DOI: 10.1112/s0010437x13007537
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Amalgamated free product type III factors with at most one Cartan subalgebra

Abstract: International audienceWe investigate Cartan subalgebras in nontracial amalgamated free product von Neumann algebras M1 * B M2 over an amenable von Neumann subalgebra B. First, we settle the problem of the absence of Cartan subalgebra in arbitrary free product von Neumann algebras. Namely, we show that any nonamenable free product von Neumann algebra (M1, ϕ1) * (M2, ϕ2) with respect to faithful normal states has no Cartan subalgebra. This generalizes the tracial case that was established in [Io12a]. Next, we pr… Show more

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Cited by 22 publications
(25 citation statements)
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“…On the other hand, by Lemma 3.4.3, for any i there is some j such that N i⊗ LRp = pC φ (N i )p C h (M ) C h (M j ) = M j⊗ LR for any Tr-finite projection p ∈ LR N . By [1,Proposition 2.10], this implies N i M M j . So by the uniqueness, we get σ(i) = j and σ is determined from…”
Section: Proof Of Theorem Amentioning
confidence: 95%
“…On the other hand, by Lemma 3.4.3, for any i there is some j such that N i⊗ LRp = pC φ (N i )p C h (M ) C h (M j ) = M j⊗ LR for any Tr-finite projection p ∈ LR N . By [1,Proposition 2.10], this implies N i M M j . So by the uniqueness, we get σ(i) = j and σ is determined from…”
Section: Proof Of Theorem Amentioning
confidence: 95%
“…Thus one may expect that the continuous core of any 'type III 1 free product factor' has no regular diffuse hyperfinite von Neumann subalgebra. This is unclear at the moment of this writing, but the next weaker assertion follows directly from [8], [1].…”
Section: Remarksmentioning
confidence: 91%
“…It is well-known that if A is a Cartan subalgebra in a von Neumann algebra N , then so is A ⋊ σ ψ R = A⊗ L(R) in the continuous core N = N ⋊ σ ψ R with a faithful normal state ψ = ψ • E on N , where E denotes the unique normal conditional expectation from N onto A. The work [1] actually shows only the absence of such special Cartan subalgebras in the continuous core M (even when both M 1 , M 2 are hyperfinite). Hence Theorem 1 is seemingly stronger than the original one [1, Theorem A].…”
Section: Y Uedamentioning
confidence: 99%
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