aa r X i v : . [ m a t h . OA ] A ug Maximal amenability and disjointness for the radial masa
Chenxu Wen
Abstract
We prove that the radial masa C in the free group factor is disjoint from othermaximal amenable subalgebras in the following sense: any distinct maximal amenablesubalgebra cannot have diffuse intersection with C . Introduction
Amenability is one of the most central notions in the study of von Neumann algebras.Amenable von Neumann algebras are very well understood: Connes’ work on the charac-terization of amenability [6] is a milestone for the entire theory. Thus, in order to studynon-amenable von Neumann algebras, it is natural to consider their amenable subalgebras.Fuglede and Kadison [7] showed that for any II factor, there always exists a maximalhyperfinite subfactor, thus answered a question of Murray and von Neumann about thedouble relative commutant. Later on, during a conference at Baton Rouge in 1967, Kadisonasked a series of famous questions about von Neumann algebras (see for example [8]).Among them is the following: Question.
Is every self-adjoint element in a II factor contained in a hyperfinite subfactor? Popa answered this question in the negative, by showing that the generator masa in thefree group factor is maximal amenable, [22].If (
M, τ ) is a finite von Neumann algebra with a faithful normal tracial state τ and ω is a free ultrafilter, we’ll write M ω as the ultraproduct of ( M, τ ). The key insight of Popa[22] is that the inclusion A ⊂ M , where M = L ( F n ) with n ≥ A the generator masa,satisfies the asymptotic orthogonality property , which we define below: Definition.
Let A ⊂ M be an inclusion of finite von Neumann algebras. We say thatthe inclusion satisfies the asymptotic orthogonality property (AOP for short), if for any freeultrafilter ω on N , ( x n ) n ∈ A ′ ∩ M ω ⊖ A ω and y , y ∈ M ⊖ A , we have that y ( x n ) n ⊥ ( x n ) n y . Since Popa, there are many results considering maximal amenable subalgebras. Ge[9, Theorem 4.5] showed that any diffuse amenable finite von Neumann algebra can be1ealized as a maximal amenable subalgebra of the free group factor. Shen [28] showed thatthe N n ∈ N A is maximal amenable inside N n ∈ N M , where A is the generator masa in thefree group factor M , thus gave an example of a maximal masa in a McDuff-II factor.Cameron, Fang, Ravichandran and White [4] proved that the radial masa in the free groupfactor is maximal amenable. Brothier [3] gave an example in the setting of planar algebras.Boutonnet and Carderi [1] showed that the subalgebra coming from a maximal amenablesubgroup in a hyperbolic group, is maximal amenable. Houdayer [11] showed that thefactors coming from free Bogoljubov actions contains concrete maximal amenable masa’s,see also [13]. All these results use Popa’s AOP approach.Very recently a new method via the study of centralizers of states, is developed by Bou-tonnet and Carderi [2]. In particular, they are able to show that the subalgebra coming fromthe upper-triangular matrix subgroup of SL (3 , Z ), is maximal amenable inside L ( SL (3 , Z )).See Ozawa’s remark [18] for an application of this new approach.Another central theme in the theory is the study of the free group factors ([16], [30],[10], [17], [19]). One of the motivating questions of this paper is a conjecture by J. Peterson(see the end of [21]): Conjecture.
For the free group factor, any diffuse amenable subalgebra is contained in aunique maximal amenable subalgebra.
Houdayer’s result on Gamma stability of free products [13, Theorem 4.1] implies thatthe generator masa satisfies Peterson’s conjecture. The proof again is relying on the AOP.See also Ozawa’s proof [18] via the centralizer approach.One subalgebra of the free group factor under intense study is the radial masa. So let M = L ( F N ) with 2 ≤ N < ∞ be the free group factor with finitely many generators anddenote by C the von Neumann subalgebra of M generated by ω := P g ∈ F N , | g | =1 u g . Notethat ω is only well-defined for free groups with finitely many generators. It was proved byPytlik [26, Theorem 4.3] that C is a masa in M , called the radial masa or the Laplacianmasa . Moreover, R˘adulescu [27, Theorem 7] showed that C is singular and Cameron, Fang,Ravichandra and White [4, Corollary 6.3] proved that it is maximal amenable in M .Recall that a result of Popa [23, Corollary 4.3] shows that generator masa’s coming fromdifferent generators cannot be unitarily conjugate inside M . This implies that the radialmasa C cannot be unitarily conjugate with the generator masa A inside M . However,whether they are conjugate via some automorphism, is still unknown.The main result of this paper is the following: Theorem.
Let M = L ( F N ) with ≤ N < ∞ and let C ⊂ M be the radial masa. Thenevery amenable subalgebra of M having diffuse intersection with C , must be contained in C . This is the first example of such disjointness for an maximal amenable subalgebra whichis not known to be in a free position. 2he approach taken in this paper is to show a stronger version of Popa’s AOP. The mainidea is that, according to a computation by R˘adulescu [27], ℓ ( F N ) ⊖ L ( C ) admits a verynice decomposition as a direct sum of C - C -bimodules. Moreover, each bimodule contains aRiesz basis whose interaction with the left and right actions of C is very similar to that ofthe canonical basis of the free group. In other words, the radial masa C and L ( M ) ⊖ L ( C )behave almost freely. Acknowledgement
We would like to thank Jesse Peterson, for suggesting the question and for many valuablediscussions and consistent encouragement. We are very grateful to Cyril Houdayer, R´emiBoutonnet and Stuart White for a careful reading and numerous helpful suggestions on anearly version of the paper. We also thank Sorin Popa and Allan Sinclair for their comments.Last but not least, we would like to thank the anonymous referee for providing many helpfulcomments.
Proof of the Theorem
Recall that a von neumann algebra M is said to be solid , if the relative commutant of anydiffuse amenable subalgebra is amenable [17]. M is called strongly solid , if for any diffuseamenable subalgebra D of M , the normalizer of D generates an amenable subalgebra of M [19]. Clearly strong solidity implies solidity. It is shown in [19] that free group factors arestrongly solid.The following proposition is inspired by [22, Lemma 3.1, Theorem 3.2] and [4, Lemma2.2, Corollary 2.3]: Proposition 1.
Let M be a strongly solid II factor and A ⊂ M a singular masa in M .Assume in addition that for any diffuse von Neumann subalgebra B ⊂ A and any freeulltrafilter ω , the following holds:for any ( x k ) k ∈ B ′ T M ω ⊖ A ω and for any y , y ∈ M ⊖ A , we have that y ( x k ) k ⊥ ( x k ) k y .Then any amenable subalegbra of M containing B , must be contained in A .Proof. As shown by [4, Lemma 2.2, Corollary 2.3], AOP and singularity imply that A ismaximal amenable in M .Let B ⊂ Q ⊂ M be an amenable subalgebra. By solidity of M , A ⊂ B ′ T M is amenable.Since A is maximal amenable, we conclude Q ′ T Q ⊂ B ′ T M ⊂ A .Let z be the maximal central projection of Q such that Qz is type II . Now supposethat z = 0. 3ince Qz is amenable and of type II , Popa’s intertwining theorem ([24, Theorem A.1])easily implies that there is a unitary u ∈ ( Qz ) ′ T ( Qz ) ω , such that E A ω ( u ) = 0. For a proof,see [4, Lemma 2.2].Now let C be a masa in Qz which contains Bz . Again by solidity and maximal injectivity, C ⊂ Az . Since Qz is of type II , there exists two non-zero projections p , p ∈ C and apartial isometry v ∈ Qz , such that vv ∗ = p , v ∗ v = p , p p = 0. Then we have E A ( v ) = E A ( p vp ) = p E A ( v ) p = 0 so that vu ⊥ uv . However we also know that vu = uv , hence v = 0. This contradicts that p , p = 0.Thus, Q has to be of type I. Let C be a masa in Q containing B . Again C ⊂ A . ByKadison’s result [15], C is regular in Q . Both A and Q lie in the normalizer of C , so theytogether generate an amenable algebra containing A . By maximal amenability of A , itfollows that Q ⊂ A .Thus, in order to confirm Peterson’s conjecture for the radial masa, it suffices to provethe strong-AOP as in Proposition 1 for the radial masa. This section is mainly devoted toestablish this fact.Let Γ = F N +1 , N ∈ N . Write K := 2 N + 1 for later use. Denote by ω n = P g ∈ G, | g | = n u g ,for n = 1 , , , · · · and let ω = u e . Let M = L (Γ) be the free group factor and let C = { ω } ′′ ⊂ M be the radial masa. { ω n } n ≥ forms an orthogonal basis for L ( C ).Let K i be the finite-dimensional subspace of H := L ( M ) spanned by all words oflength i and we denote by Q i the orthogonal projection from H onto K i . For ξ ∈ K i and n, m ∈ N S { } , we define the following ξ n,m := Q i + m + n ( ω n ξω m ) K ( n + m ) / . R˘adulescu [27] discovered that there is a nice decomposition of
H ⊖ L ( C ) = L i ≥ H i into a direct sum of C - C -bimodules, each H i has a distinguished unit vector ξ i , which isfrom K l ( i ) , for some l ( i ) ∈ N , such that H i is generated by ξ i as a C - C -bimodule.Moreover, by [27, Lemma 3, Lemma 6], for those i with l ( i ) ≥
2, we have that { ξ in,m } n,m ≥ forms an orthonormal basis for H i . For those i with l ( i ) = 1 (there arefinitely many such i ’s), { ξ in,m } n,m ≥ is no longer an orthonormal basis for H i , however forany i, j ≥
1, the linear mapping T i,j : H i → H j , given by T i,j ( ξ in,m ) = ξ jn,m , extends uniquely to an invertible bounded operator. Furthermore, there is a universalconstant C > k T ± i,j k ≤ C , ∀ i, j ≥ . Remark 2.
Recall that in a separable Hilbert space, a sequence of vectors { v n } formsa Riesz basis (for the basics of Riesz basis, see, e.g. [5]), if { v i } is the image of some4rthonormal basis under some bounded invertible operator. It is also equivalent to the factthat there exists some A, B > c n ) ∈ ℓ , A P | c n | ≤ k P c n v n k ≤ B P | c n | . In this case, every vector x in the Hilbert space has a unique decomposition x = P c n v n , for some ( c n ) ∈ ℓ . It follows that (cid:8) ξ in,m (cid:9) i ≥ ,n,m ≥ forms a Riesz basis for L ( M ) ⊖ L ( C ). Consequently, for any x ∈ L ( M ) ⊖ L ( C ), there is a unique decomposition x = P i ≥ ,n,m ≥ a in,m ξ in,m for some (cid:0) a in,m (cid:1) n,m,i ∈ ℓ . We call { ξ in,m } i ≥ ,n,m ≥ the R˘adulescubasis for L ( M ) ⊖ L ( C ).Sometimes it will be convenient to use the following convention: we write ξ in,m for all n, m ∈ Z , where we define ξ in,m = 0 whenever n < m < { ω n ξ i ω m } , after proper normalization, is the more natural basis to work with.We collect some relations between ω n ξ i ω m and ξ in,m ’s, due to R˘adulescu, in the followinglemma: Lemma 3 (Lemma 2, 6 in [27]) . The following statements hold for all n, m ≥ :(1) If l ( i ) ≥ , then ω n ξ i ω m = K n + m ξ in,m − K n + m − (cid:0) ξ in,m − + ξ in − ,m (cid:1) + K n + m − ξ in − ,m − ;(2) If l ( i ) = 1 , then there is some σ = σ ( i ) ∈ {− , } such that ω n ξ i ω m = K n + m ξ in,m − K n + m − (cid:0) ξ in,m − + ξ in − ,m + σξ in − ,m − (cid:1) + X k ≥ ( − σ ) k K n + m − k (cid:0) σξ in − k − ,m − k +1 + σξ in − k +1 ,m − k − + 2 ξ in − k,m − k (cid:1) ; (3) For all i, j ≥ , the linear mapping S i,j : H i → H j given by S i,j (cid:0) ω n ξ i ω m (cid:1) = ω n ξ j ω m , ∀ n, m ≥ , is well-defined and extends to an invertible bounded operator between the two subspaces, with sup i,j k S ± i,j k ≤ C , for some uniform constants < C < ∞ . Lemma 4. (cid:26) η in,m := ω n ξ i ω m K ( n + m ) / (cid:27) i ≥ ,n,m ≥ forms a Riesz basis for L ( M ) ⊖ L ( C ) .Therefore, for any x ∈ L ( M ) ⊖ L ( C ) , there is a unique decomposition x = P i ≥ ,n,m ≥ b in,m η in,m where (cid:0) b in,m (cid:1) i ≥ ,n,m ≥ ∈ ℓ .Proof. By (3) of the previous lemma, it suffices to prove the conclusion for some fixed i ≥ l ( i ) ≥
2. 5ix i ≥ l ( i ) ≥ a n,m ) n,m ∈ ℓ . We will omit the superscript i , since noconfusion will appear. Using part (1) of the previous lemma, we have X n,m ≥ a n,m η n,m = X n,m ≥ a n,m (cid:18) ξ n,m − ξ n,m − K − ξ n − ,m K + ξ n − ,m − K (cid:19) = X n,m (cid:16) a n,m − a n,m +2 K − a n +2 ,m K + a n +2 ,m +2 K (cid:17) ξ n,m = X n,m (cid:18)(cid:16) a n,m − a n,m +2 K (cid:17) − K (cid:16) a n +2 ,m − a n +2 ,m +2 K (cid:17)(cid:19) ξ n,m , hence by repeated use of the triangle inequality, we have (cid:13)(cid:13)(cid:13)X a n,m η n,m (cid:13)(cid:13)(cid:13) = X n,m ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:18)(cid:16) a n,m − a n,m +2 K (cid:17) − K (cid:16) a n +2 ,m − a n +2 ,m +2 K (cid:17)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) / ≥ X n,m ≥ (cid:12)(cid:12)(cid:12) a n,m − a n,m +2 K (cid:12)(cid:12)(cid:12) / − K X n,m ≥ (cid:12)(cid:12)(cid:12) a n +2 ,m − a n +2 ,m +2 K (cid:12)(cid:12)(cid:12) / ≥ (cid:18) − K (cid:19) X n,m ≥ (cid:12)(cid:12)(cid:12) a n,m − a n,m +2 K (cid:12)(cid:12)(cid:12) / ≥ (cid:18) − K (cid:19) X n,m ≥ | a n,m | / . The other side of the inequality is easy, since each a n,m only appears at most four times.Thus there is a B >
0, such that (cid:13)(cid:13)(cid:13)X a n,m η n,m (cid:13)(cid:13)(cid:13) ≤ B X | a n,m | , So we are done.
Remark 5.
Because both k T ± i,j k and k S ± i,j k are uniformly bounded, there is a C > k T ± i,j k ≤ C , k S ± i,j k ≤ C , and for any (cid:0) c in,m (cid:1) ∈ ℓ ,1 C X i,n,m (cid:12)(cid:12) c in,m (cid:12)(cid:12) ≤ k X n,m ≥ ,i ≥ c in,m ξ in,m k ≤ C X i,n,m (cid:12)(cid:12) c in,m (cid:12)(cid:12) , C X i,n,m (cid:12)(cid:12) c in,m (cid:12)(cid:12) ≤ k X n,m ≥ ,i ≥ c in,m η in,m k ≤ C X i,n,m (cid:12)(cid:12) c in,m (cid:12)(cid:12) k ∈ N , define L k , L ′ k : L ( M ) ⊖ L ( C ) → L ( M ) ⊖ L ( C ) as follows L k X i ≥ ,n,m ≥ a in,m ξ in,m := X i ≥ ,n ≤ k,m ≥ a in,m ξ in,m ,L ′ k X i ≥ ,n,m ≥ b in,m η in,m := X i ≥ ,n ≤ k,m ≥ b in,m η in,m , i.e. L k (resp. L ′ k ) is the left “projection” onto the span of (cid:8) ξ in,m (cid:9) i,n,m (resp. (cid:8) η in,m (cid:9) i,n,m )with the first subscript no larger than k . However one should be warned that they are merelyidempotents, instead of projections, due to the presence of those i ’s with l ( i ) = 1. We canalso define R k , R ′ k for the right “projections” in the similar fashion. All these idempotentsare bounded operators. Let L k ∨ R k := L k + R k − L k R k , L ′ k ∨ R ′ k := L ′ k + R ′ k − L ′ k R ′ k . Lemma 6. L ′ k is right C -modular, ∀ k ≥ . Proof.
Since { ω n } n ≥ forms an orthogonal basis for L ( C ) and { η in,m } i ≥ ,n,m ≥ is a Rieszbasis for L ( M ) ⊖ L ( C ), it is sufficient to show that L ′ k ( η in,m ω l ) = L ′ k ( η in,m ) ω l .The definition of the η in,m ’s clearly implies that η in,m ω l ∈ span { η in,k } k ≥ , that is, mul-tiplying ω l on the right does not change neither the upper nor left index of η in,m , thus L ′ k ( η in,m ω l ) = L ′ k ( η in,m ) ω l and the proof is complete.We will need the following result from [4]: Lemma 7 (Lemma 4.3, Theorem 6.2 in [4]) . Given ( x k ) k ∈ M ω ⊖ C ω , if for every k ∈ N , we have that lim k → ω k ( L k ∨ R k ) ( x k ) k = 0 , then for any y , y ∈ L ( M ) ⊖ L ( C ) , y ( x k ) k ⊥ ( x k ) k y . Now we state the key technical result of this paper.
Proposition 8.
Let
Γ = F N +1 be a non-abelian free group with finitely many generatorsand M = L (Γ) the corresponding group von Neumann algebra. Denote by C the radialmasa of M and suppose that B ⊂ C is a diffuse von Neumann subalgebra. Then for any ( x k ) k ∈ B ′ T M ω ⊖ C ω and y , y ∈ M ⊖ C , where ω is a free ultrafilter, we have that y ( x k ) k ⊥ ( x k ) k y in L ( M ω ) . We will break the proof into several lemmas.Let ( x k ) k ∈ B ′ T M ω ⊖ C ω and y , y ∈ M ⊖ C be given. For each k , we can assume x k ∈ M ⊖ C ⊂ L ( M ) ⊖ L ( C ) , || x k || ≤ (cid:8) ξ in,m (cid:9) i ≥ ,n,m ≥ and (cid:8) η in,m (cid:9) i ≥ ,n,m ≥ , respectively: x k = X i ≥ ,n,m ≥ a i,kn,m ξ in,m = X i ≥ ,n,m ≥ b i,kn,m η in,m , (cid:16) a i,kn,m (cid:17) i ≥ ,n,m ≥ and (cid:16) b i,kn,m (cid:17) i ≥ ,n,m ≥ are from ℓ .Since B is diffuse, we can choose a sequence { u k } k in the unitary group of B , whichconverges to 0 weakly. Recall that (cid:26) ω i || ω i || (cid:27) i ≥ is an orthonormal basis for L ( C ). More-over, for any fixed N ≥ ω n ω m will be supported on those ω i ’s with i > N , providedthat | m − n | > N . We first need to approximate each u k using finite linear combinationsof ω i ’s. Lemma 9.
For each fixed N , there exists a sequence { S k } k ≥ of non-empty, disjoint, finitesubsets of N ∪ { } and a sequence of strictly increasing natural numbers { n k } k ≥ , such thatin the decomposition with respect to { ω i } i ≥ , the supports of elements from { ω m ω n : m ∈ S i , n ≤ N } and the supports of elements from { ω m ω n : m ∈ S j , n ≤ N } are disjoint,whenever i, j ≥ , i = j . Moreover, there exists a sequence { v k } k in C , with v k ∈ span { ω i : i ∈ S k } such that || v k || ≤ and || v k − u n k || ≤ k .Moreover, one can construct { v k } , { S k } such that there is a sequence { F k } of strictlyincreasing natural numbers such that L ′ N ( v i x ) = L ′ N (cid:16) v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x ) (cid:17) , for all x ∈ L ( M ) ⊖ L ( C ) .Proof. Throughout this lemma, for any x ∈ C , we always consider the Fourier expansionof x with respect to { ω i } i ≥ . Moreover, if x = P i ≥ a i ω i and F ⊂ N ∪ { } , we will use thenotation P F ( x ) := P i ∈ F a i ω i .We construct { S k } , { n k } and { v k } inductively. Since span { ω n } n ≥ is a weakly dense*-subalgebra of C , Kaplansky Density Theorem implies that there exists a sequence { z k } k of elements in C , whose Fourier expansions are finitely supported, such that || z k || ≤ / || u k − z k || ≤ k . For each k , suppose that z k is supported on { ω i } i ∈ T k , where T k ⊂ N ∪{ } issome finite subset. Let n = 1, v = z and S = T . Then || v || ≤ || v − u n || ≤ / v ∈ span { ω i : i ∈ S } .Now suppose that S , · · · , S k and n , · · · , n k have already been chosen. Then there existsa finite subset F k +1 ⊂ N ∪{ } , such that S ≤ i ≤ k S i ⊂ F k +1 and for any S ⊂ N ∪{ }\ F k +1 , wealways have that in the decomposition with respect to { ω i } i ≥ , the supports of elements from { ω m ω n : m ∈ ∪ ≤ i ≤ k S i , n ≤ N } and the supports of elements from { ω m ω n : m ∈ S, n ≤ N } are disjoint (for example, one can let F k +1 = { , , · · · , max ∪ ≤ i ≤ k S i + 3 N } ). Now since u k → n k +1 > n k , such that with respect to the basis { ω i } i ≥ , the Fourier coefficient of z n k +1 has absolute value less than 14 k | F k +1 ||| ω i || , for each8 ≤ i ≤ F k +1 . Let S k +1 := T n k +1 \ F k +1 , v k +1 := P S k +1 ( z n k +1 ). Then k v k +1 − u n k +1 k ≤ (cid:13)(cid:13) v k +1 − z n k +1 (cid:13)(cid:13) + (cid:13)(cid:13) z n k +1 − u n k +1 (cid:13)(cid:13) = k P ( T nk +1 \ S k +1 ) ( z n k +1 ) k + (cid:13)(cid:13) z n k +1 − u n k +1 (cid:13)(cid:13) = (cid:13)(cid:13) P F k +1 ( z n k +1 ) (cid:13)(cid:13) + (cid:13)(cid:13) z n k +1 − u n k +1 (cid:13)(cid:13) ≤ (cid:18) | F k +1 | k | F k +1 | (cid:19) / + 14 k +1 ≤ k +1 , and an easy estimate of the ℓ -norm gives us k v k +1 k ≤ k z k +1 k + | F k +1 | k | F k +1 | ≤ . The last statement can be achieved by letting the supports of { v k } k mutually far away.For example, choose the gap between S i and S j to be greater than 3 N and let F k be thecollection of elements of N ∪ { } between min n ∈ S k | n | − N and max n ∈ S k | n | + N .Thus by taking a subsequence if necessary, we may assume that { v k } is a sequence in C ,such that v k ∈ span { ω i : i ∈ S k } for some finite subset S k ⊂ N , || v k || ≤ || v k − u k || ≤ k and v i ω k ⊥ v j ω k , for all i, j ≥ , i = j and all 0 ≤ k ≤ N and there is a sequence { F k } ofstrictly increasing natural numbers such that L ′ N ( v i x ) = L ′ N (cid:16) v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x ) (cid:17) , forall x ∈ L ( M ) ⊖ L ( C ). Lemma 10. lim k → ω (cid:13)(cid:13) L ′ N ( x k ) (cid:13)(cid:13) = 0 .Proof. Fix a small ǫ >
0. First choose some large N < N such that 2 P N i = N k v i − u i k ≤ ǫ and 4 (cid:13)(cid:13) L ′ N (cid:13)(cid:13) C + 1 N − N ≤ ǫ . Then we havelim k → ω N X i = N (cid:10) L ′ N ( v i x k ) , L ′ N ( v i x k ) (cid:11) ≥ lim k → ω N X i = N (cid:10) L ′ N ( u i x k ) , L ′ N ( u i x k ) (cid:11) − ǫ = lim k → ω N X i = N (cid:10) L ′ N ( x k u i ) , L ′ N ( x k u i ) (cid:11) − ǫ = lim k → ω N X i = N (cid:10) L ′ N ( x k ) u i , L ′ N ( x k ) u i (cid:11) − ǫ = ( N − N ) lim k → ω (cid:13)(cid:13) L ′ N ( x k ) (cid:13)(cid:13) − ǫ. x k ) k ∈ B ′ ∩ M ω and the third line uses the factthat L ′ N is a right- C modular map, i.e. L ′ N ( xa ) = L ′ N ( x ) a, ∀ x ∈ L ( M ) ⊖ L ( C ) , ∀ a ∈ C. Meanwhile, N X i = N (cid:10) L ′ N ( v i x k ) , L ′ N ( v i x k ) (cid:11) = N X i = N D L ′ N (cid:16) v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x k ) (cid:17) , L ′ N (cid:16) v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x k ) (cid:17)E ≤ (cid:13)(cid:13) L ′ N (cid:13)(cid:13) N X i = N D v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x k ) , v i (cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x k ) E ≤ N X i = N (cid:13)(cid:13) L ′ N (cid:13)(cid:13) k v i k (cid:13)(cid:13)(cid:13)(cid:16) L ′ F i +1 − L ′ F i (cid:17) ( x k ) (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13) L ′ N (cid:13)(cid:13) N X i = N C X j ≥ ,F i +1 ≤ n ≤ F i +1 ,m ≥ (cid:12)(cid:12)(cid:12) b j,kn,m (cid:12)(cid:12)(cid:12) ≤ (cid:13)(cid:13) L ′ N (cid:13)(cid:13) C X j ≥ , ≤ n ≤ F N ,m ≥ (cid:12)(cid:12)(cid:12) b j,kn,m (cid:12)(cid:12)(cid:12) ≤ (cid:13)(cid:13) L ′ N (cid:13)(cid:13) C k x k k ≤ (cid:13)(cid:13) L ′ N (cid:13)(cid:13) C . Therefore, we conclude that lim k → ω (cid:13)(cid:13) L ′ N ( x k ) (cid:13)(cid:13) ≤ || L ′ N || C + 1 N − N ≤ ǫ can be made arbi-trarily small. Thus the proof for Lemma 10 is complete. Lemma 11. lim k → ω k L N ( x k ) k = 0 .Proof. We use the relations between η in,m and ξ in,m , as stated above in Lemma 3 andLemma 4.First, since x k = P i ≥ ,l ( i ) ≥ ,n,m ≥ a i,kn,m ξ in,m ⊕ P i ≥ ,l ( i )=1 ,n,m ≥ a i,kn,m ξ in,m , it suffices toconsider separately the part with i ’s such that l ( i ) ≥ i ’s such that l ( i ) = 1.For the i ’s with l ( i ) ≥
2, recall that η in,m = ξ in,m − K − (cid:0) ξ in,m − + ξ in − ,m (cid:1) + K − ξ in − ,m − so that a in,m = b in,m − b in,m +2 K − b in +2 ,m K + b in +2 ,m +2 K . Therefore10 (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L N X i ≥ ,l ( i ) ≥ ,n,m ≥ a i,kn,m ξ i,kn,m (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = X i ≥ ,l ( i ) ≥ ,N ≥ n ≥ ,m ≥ (cid:12)(cid:12)(cid:12) a i,kn,m (cid:12)(cid:12)(cid:12) = X i ≥ ,l ( i ) ≥ ,N ≥ n ≥ ,m ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b i,kn,m − b i,kn,m +2 K − b i,kn +2 ,m K + b i,kn +2 ,m +2 K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X i ≥ ,l ( i ) ≥ ,N +2 ≥ n ≥ ,m ≥ (cid:12)(cid:12)(cid:12) b i,kn,m (cid:12)(cid:12)(cid:12) ≤ C (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ′ N +2 X i ≥ ,l ( i ) ≥ ,n,m ≥ b i,kn,m η i,kn,m (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) , and the last term goes to 0 as k → ω , by the previous lemma.Now consider the i ’s with l ( i ) = 1. As there are only finitely many such i ’s, we mayrestrict our attention to a single fixed i .For some σ ∈ { , − } , we have that X n,m ≥ b i,kn,m η i,kn,m = X n,m b i,kn,m ξ i,kn,m − ξ i,kn − ,m K − ξ i,kn,m − K + σ ξ i,kn − ,m − K + X l ≥ ( − σ ) l K l (cid:16) σξ i,kn − l − ,m − l +1 + σξ i,kn − l +1 ,m − l − + 2 ξ i,kn − l,m − l (cid:17) ! = X n,m b i,kn,m − b i,kn +2 ,m K − b i,kn,m +2 K + σb i,kn +1 ,m +1 K + X l ≥ ( − σ ) l K l (cid:16) σb i,kn + l +1 ,m + l − + σb i,kn + l − ,m + l +1 + 2 b i,kn + l,m + l (cid:17) ! ξ i,kn,m . Therefore, for any fixed ǫ > , N ≥
0, we find a large integer N ≫ N , to be specified11ater, and we let K = N − N . By the triangle inequality, X n ≤ N ,m ≥ | a i,kn,m | / ≤ X n ≤ N ,m ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b i,kn,m − b i,kn +2 ,m K − b i,kn,m +2 K + σb i,kn +1 ,m +1 K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / + X ≤ l ≤ K K l X n ≤ N ,m ≥ (cid:12)(cid:12)(cid:12) σb i,kn + l +1 ,m + l − + σb i,kn + k − ,m + k +1 + 2 b i,kn + k,m + k (cid:12)(cid:12)(cid:12) / + X l ≥ K +1 K l X n ≤ N ,m ≥ (cid:12)(cid:12)(cid:12) σb i,kn + l +1 ,m + l − + σb i,kn + l − ,m + l +1 + 2 b i,kn + l,m + l (cid:12)(cid:12)(cid:12) / . We estimate the third term in the above inequality first: X l ≥ K +1 K l X n ≤ N ,m ≥ (cid:12)(cid:12)(cid:12) σb i,kn + l +1 ,m + l − + σb i,kn + l − ,m + l +1 + 2 b i,kn + l,m + l (cid:12)(cid:12)(cid:12) / ≤ X l ≥ K +1 K l X n,m ≥ (cid:12)(cid:12)(cid:12) σb i,kn + l +1 ,m + l − + σb i,kn + l − ,m + l +1 + 2 b i,kn + l,m + l (cid:12)(cid:12)(cid:12) / ≤ X l ≥ K +1 K l X n,m ≥ (cid:12)(cid:12)(cid:12) b i,kn,m (cid:12)(cid:12)(cid:12) / ≤ X l ≥ K +1 K l C k x k k ≤ C K K ( K − , hence we can choose N large enough so that K is large, such that the third term is lessthan ǫ/
3, for any k .Now we estimate the first and the second terms. To this end, we choose a large k = k ( N , ǫ ), such that for any k ≥ k , we have that 4 C K (cid:18)P m ≥ ,n ≤ N +1 (cid:12)(cid:12)(cid:12) b i,kn,m (cid:12)(cid:12)(cid:12) (cid:19) / is lessthan ǫ/
3. Thus both the first and the second term can be bounded above by ǫ/
3. Combineall these pieces together, we conclude that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L N X n,m ≥ a i,kn,m ξ i,kn,m (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ C X n ≤ N ,m ≥ | a i,kn,m | / ≤ C ǫ, when k is close enough to ω . Since ǫ > roof of Proposition 8. The same proof for Lemma 11 shows that lim k → ω k R N ( x k ) k = 0.So Lemma 7 applies. Remark 12.
In fact, the same conclusion as in Proposition 8 holds, if we replace theassumption “ B ⊂ C diffuse” by “ B ⊂ C ω diffuse”. Theorem 13.
The radial masa satisfies Peterson’s conjecture.Proof.
It is shown in [19] that L ( F N ) , N ≥ Remark 14.
One can also use [12, Theorem 8.1] and Proposition 8 to conclude Theorem 13.We thank Boutonnet for pointing it out to us.In fact, one can state a more general structural result for the inclusion C ⊂ L ( F N +1 ). Theorem 15.
Let M = L ( F N +1 ) be a free group factor with ≤ N < ∞ and let C ⊂ M be the radial masa. If Q ⊂ M is a von Neumann subalgebra that has a diffuse intersectionwith C , then there exists a sequence of central projections e n ∈ Z ( Q ) , n ≥ such that • e Q ⊂ C ; • For all n ≥ , e n Q is a non-amenable II factor such that e n ( Q ′ ∩ M ω ) = e n ( Q ′ ∩ M ) is discrete and abelian (even contained in C ).Proof. Let e ∈ Z ( Q ) be the maximal projection such that e Q is amenable. Then Qe ⊕ C (1 − e ) is amenable and has a diffuse intersection with C so it is contained in C byTheorem 13. Moreover, Q (1 − e ) has a discrete center, by solidity of M . This gives asequence of central projections { e n } n ≥ such that for all n ≥ e n Q is a non-amenable II factor.Now fix n ≥
1. By [14, Lemma 2.7], one can find a central projection e ∈ Z (( e n Q ) ′ ∩ e n M e n )such that • e (( e n Q ) ′ ∩ e n M e n ) = e (( e n Q ) ′ ∩ ( e n M e n ) ω ) is discrete; • ( e n − e )(( e n Q ) ′ ∩ ( e n M e n ) ω ) is diffuse.By [20, Proof of Theorem 4.3], the fact that ( e n − e )(( e n Q ) ′ ∩ ( e n M e n ) ω ) is diffuse impliesthat ( e n − e ) Q is amenable. Since e n Q has no direct summand, this forces e = e n .Finally, ( Q ∩ C ) ′ ∩ M is amenable, again by solidity. As it contains C , it has to be equalto C . In particular Q ′ ∩ M ⊂ ( Q ∩ C ) ′ ∩ M ∩ C . So the last part of the theorem is true.13 emark 16. In [13, Theorem 3.1], Houdayer showed the general situation for free productsof σ -finite von Neumann algebras, which contains the strong-AOP for the generator masain a free group factor as a special case. Also, the strong-AOP as in Proposition 8 meansthat for any diffuse subalgebra B of the radial masa C , the inclusion C ⊂ M has the AOPrelative to B , in the sense of [12, Definition 5.1]. The unique maximal injective extensionfor any diffuse subalgebra of the generator masa is first shown by Houdayer [13, Theorem4.1]. A proof via the study of centralizers is obtained by Ozawa [18]. Remark 17.
Note that the disjointness result as in Theorem 13 is not true for arbitrarymaximal amenable masa of a II factor. For instance, if the inclusion A ⊂ M has some nicedecomposition, then A does not have the uniqueness property as the generator masa in theabove corollary. We give some such examples: • Let M = A ∗ A A be the amalgamated free product with A i amenable, and A diffuse, A = A i , i = 1 ,
2, then A can be contained in different maximal amenablesubalgebras. • Let M , M both be the free group factor and A i ⊂ M i the corresponding gener-ator masa, i = 1 ,
2. Then A = A ⊗ A is a maximal injective subalgebra inside M = M ⊗ M . However, many other injective subalgebras could contain the diffusesubalgebra A ⊗ • Let Λ < Γ be a singular subgroup in the sense of Boutonnet and Carderi ([2, Definition1.2]) and suppose Γ acts on a finite diffuse amenable von Neumann algebra Q . Then Q ⋊ Λ is maximal injective inside Q ⋊ Γ, by [2, Theorem 1.3]. However again there arelots of different injective subalgebras containing Q but are not contained in Q ⋊ Λ. Remark 18.
We would like to mention an example in the ultra-product setting. Let A ⊂ M be a singular masa inside a separable II factor. Then for any free ultrafilter ω , A := A ω is amaximal injective masa in M := M ω , a result due to Popa ([25, Theorem 5.2.1]). However,it is well known that any two separable abelian subalegebras in a ultraproduct of II factorsare unitarily conjugate ([23, Lemma 7.1]). In particular, A is both contained in a maximalinjective masa and a maximal hyperfinite subfactor of M . References [1] R´emi Boutonnet and Alessandro Carderi. Maximal amenable subalgebras of von neumannalgebras associated with hyperbolic groups. arXiv , (1310.5864), 2013.[2] R´emi Boutonnet and Alessandro Carderi. Maximal amenable von neumann subalgebras arisingfrom maximal amenable subgroups. arXiv , (1411.4093), 2014.[3] Arnaud Brothier. The cup subalgebra of a II factor given by a subfactor planar algebra ismaximal amenable. Pacific J. Math. , 269(1):19–29, 2014.
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Department of Mathematics, Vanderbilt University, 1326 Stevenson Center, Nashville,TN 37240, United States
E-mail address : [email protected]@vanderbilt.edu