Maximal amenable MASAs of radial type in the free group factors

TL;DR

This paper constructs a broad family of maximal amenable MASAs in free group factors by analyzing von Neumann algebras generated by weighted radial elements formed from semicircular elements in free products.

Contribution

It introduces a new class of maximal amenable MASAs in free group factors using weighted radial elements with at least two non-zero weights, expanding known examples.

Findings

Maximal amenability of the constructed MASAs is proven.

Unitary conjugacy of these MASAs corresponds to proportional weights.

Provides a large family of maximal amenable MASAs in free group factors.

Abstract

We prove that if $\{(M_j, \tau_j)\}_{j\in J}$ are tracial von Neumann algebras, $s_j \in M_j$ are selfadjoint semicircular elements and $t=(t_j)_j$ is a square summable $J$-tuple of real numbers with at least two non-zero entries, then the von Neumann algebra $A(t)$ generated by the ``weighted radial element'' $\sum_j t_j s_j\in M:=*_{j\in J} M_j$ is maximal amenable in $M$, with $A(t)$, $A(t')$ unitary conjugate in $M$ iff $t, t'$ are proportional. Letting $M_j$ be diffuse amenable, $\forall j$, this provides a large family of maximal amenable MASAs in the free group factor $L\mathbb F_n$.

Equations

$$\theta (v)A(\xi )\theta (v{)}^{\ast }=\theta (vA(\xi )v^{\ast })=\theta (A(\eta )v{v}^{\ast })=A({\eta }^{\prime })\theta (v{v}^{\ast }),$$

$$(B\ast Q\subset P\ast Q)\simeq (B\ast Q\subset P{\ast }_B(B\ast Q)).$$

Full text

RB Supported by ANR grant AODynG 19-CE40-0008

SP Supported by NSF Grant DMS-1955812 and the Takesaki Endowed Chair at UCLA

1. Introduction

While the isomorphism problem for the free group factors $L\mathbb{F}_{n}$, $2\leq n\leq\infty$, remains unsolved, a series of absorption, in-decomposability and randomness phenomena in $L\mathbb{F}_{n}$, and more generally in amalgamated free product II1 factors $M=M_{1}*_{B}M_{2}$, unravelled over the years.

The simplest type of absorption result shows that whenever a subalgebra $Q$ of $M_{1}$ “stays away” from $B$, $Q\nprec_{M_{1}}B$, any element in $M=M_{1}*_{B}M_{2}$ that intertwines $Q$ into $M_{1}$ must be contained in $M_{1}$ ([Po81] in the case $B=\mathbb{C}$, [IPP05] in general). While elementary, this method did for instance allow to deduce in-decomposability, such as primeness and absence of Cartan subalgebras, in free group factors $L\mathbb{F}_{J}$, with uncountable set $J$ of generators.

Another type of absorption, triggered by the discovery in [Po82] that a diffuse amenable subalgebra $Q\subset M$ that’s freely complemented in $M$ is maximal amenable in $M$, has been much studied over the years. A general such amenable absorption result in [BH16] shows that if $Q\subset M_{1}$ is maximal amenable in $M_{1}$ and $Q\nprec_{M_{1}}B$, then $Q$ is maximal amenable in $M=M_{1}*_{B}M_{2}$.

A random matrix approach to in-decomposability was initiated in [Vo96], based on the discovery that “thinness over MASA” assumptions on $L\mathbb{F}_{n}$ come in contradiction with randomness, quantified by the free entropy dimension. This allowed proving that $L\mathbb{F}_{n}$ have no Cartan subalgebras [Vo96], primeness [Ge96], and more generally absence of “thinness” around diffuse AFD-subalgebras in $L\mathbb{F}_{n}$ (cf. Section 4 in [GP96]; see also [S07]).

Deformation-rigidity and boundary methods have been used to establish even stronger absorption results in $L\mathbb{F}_{n}$, more generally in amalgamated free product factors, [Po01, O03, IPP05, Po07, OP07, PV11, Io13, DKP22, Dr22], showing for instance that the normalizer of any diffuse amenable subalgebra of $L\mathbb{F}_{n}$ generate an amenable von Neumann algebra [OP07].

Motivated by some of these results, as well as by their work in $L^{2}$-cohomology of groups, Peterson-Thom made in [PT07] the far reaching conjecture that if $Q_{i}\subset L\mathbb{F}_{n}$, $2\leq n\leq\infty$, is a family of amenable von Neumann subalgebras with $\cap_{i}Q_{i}$ diffuse, then $\vee_{i}Q_{i}$ is amenable. Equivalently, any maximal amenable $Q\subset L\mathbb{F}_{n}$ “absorbs” any other amenable $Q_{0}\subset L\mathbb{F}_{n}$ that has diffuse intersection with $Q$, a phenomenon superseding all above absorption statements.

This conjecture, and even stronger ones formulated later (in [Ha20, Po18]), seems now settled in the affirmative due to the combination of two results: on the one hand, Hayes used 1-bounded entropy [Ju07] to prove that if one could establish the strong convergence of tuples of tensor products of gaussian unitary ensemble (GUE) random matrices, see [Ha20], then the PT-conjecture holds true; on the other hand, Belinschi-Capitaine posted recently a preprint aiming to prove this random matrix convergence result, [BC22].

Note that by [Dy93] if $Q$ is a diffuse amenable tracial von Neumann algebra and $N$ is a tracial von Neumann algebra that’s either a free group factor $L\mathbb{F}_{n}$, $2\leq n\leq\infty$, or a diffuse amenable tracial von Neumann algebra, then $M=Q*N$ is isomorphic to the free group factor $L\mathbb{F}_{m}$, where $m=2$ when $N$ is amenable, and $m=n+1$ when $N=L\mathbb{F}_{n}$. So by [Po82], in all these cases $Q$ follows maximal amenable in $M=Q*N$, with the proof actually showing that any amenable von Neumann subalgebra of $M$ that has diffuse intersection with $Q$ must be contained in $Q$. Thus, whenever an amenable diffuse von Neumann subalgebra $Q\subset M$ is freely complemented, it satisfies the PT-absorption conjecture by [Po82].

While there are by now other constructions of maximal amenable von Neumann subalgebras in $L\mathbb{F}_{n}$, and more generally in (amalgamated) free product II1 factors $M$, one could not establish whether they are freely complemented or not. This fact is amply discussed in [Po18, Section 5]. But a resolution of the PT-conjecture makes it now particularly compelling to answer the question: are there actually any maximal amenable von Neumann subalgebras of $L\mathbb{F}_{n}$ that are not freely complemented ? It would of course be a very striking structural property of $L\mathbb{F}_{n}$ if all of its maximal amenable von Neumann subalgebras are freely complemented. But this is a possibility that should not be ruled out…

The so-called radial MASA $L_{n}$ in $L\mathbb{F}_{n}$, $2\leq n<\infty$, defined as the abelian von Neumann algebra generated by the “radial element” $\sum_{i=1}^{n}(u_{i}+u_{i}^{*})$, where $u_{1},...,u_{n}$ are the free generators of $L\mathbb{F}_{n}$, introduced in [Py81] and shown in [CFRW09] to be maximal amenable in $L\mathbb{F}_{n}$, is a good candidate for an example of a non-freely complemented maximal amenable subalgebra in the case $n<\infty$ (see [CFRW09, Question 1.1] and [Po18, Question 5.5]).

Our work here is motivated by an effort to produce examples of non-freely complemented maximal amenable von Neumann subalgebras in $L\mathbb{F}_{n}$. While we could not solve this problem, in this short note we produce a large family of distinct maximal amenable MASAs in $M=L\mathbb{F}_{n}$, and more generally in free products of factors $M=*_{j}M_{j}$. Our examples are “radial-type” MASAs, but they actually do not recover the “classic” radial MASA $L_{n}\subset L\mathbb{F}_{n}$. However, various considerations make them good candidates for not being freely complemented.

To state our result we use Voiculescu’s notion of semicircular elements in tracial von Neumann algebras and his free Gaussian formalism, allowing $\ell^{2}$-summations of freely independent such elements (see [Vo88]).

Theorem 1.1**.**

Let $\{(M_{j},\tau_{j})\}_{j\in J}$ be a family of tracial von Neumann algebras, with a self-adjoint semi-circular element $s_{j}$ in each $M_{j}$, $j\in J$. Denote by $\ell^{2}_{*}$ the set of square summable families of real numbers having at least two non-zero entries. For each $t=(t_{j})_{j}\in\ell^{2}_{*}$ denote by $A(t)$ the abelian von Neumann generated in $M=*_{j\in J}M_{j}$ by $s(t):=\sum_{j}t_{j}s_{j}\in M$. Then $A(t)$ is maximal amenable in $M$, $\forall t\in\ell^{2}_{*}$, with $A(t)\prec_{M}A(t^{\prime})$ if and only if $t,t^{\prime}\in\ell^{2}_{*}$ are proportional.

Note that in case $J$ is (at most) countable, with $M_{j}$ either separable amenable or $\simeq L\mathbb{F}_{n(j)}$, $\forall j\in J$, the resulting $M=*_{j}M_{j}$ follows of the form $L\mathbb{F}_{n(J)}$, for some $2\leq n(J)\leq\infty$, by [Dy93], thus making $A(t)\subset M$, $t\in\ell^{2}_{*},$ a large class of examples of maximal amenable MASAs in free group factors. Taking $M_{j}$ “much larger” than the abelian algebra $A_{j}=\{s_{j}\}^{\prime\prime}$, generated by the semicircular element $s_{j}\in M_{j}$, seems to indicate that $A(t)$ cannot be freely complemented in $M$. For instance, taking $M_{j}=A_{j}\rtimes\Gamma_{j}$, for some trace preserving action of an amenable group $\Gamma_{j}\curvearrowright A_{j}$. This heuristic seems particularly pertinent when taking $M_{j}=A$, $\forall j$, with $A$ abelian non-separable (e.g., an ultrapower of $L^{\infty}[0,1]$), where the statement could perhaps be used towards an existence result in $L\mathbb{F}_{n}$, for $n$ finite (see the remarks at the end of this paper for more comments along these lines).

For the proof of the theorem we will use the absorption results in [IPP05, BH16] and a trick for “glueing” intertwiners in the spirit of (page 398 in [Po03]).

First of all, note that we may assume $J$ is at most countable. Indeed, by [Po81] any intertwiner between some $A(t),A(t^{\prime})$, $t,t^{\prime}\in\ell^{2}_{*}$, lies in the von Neumann algebra generated by $M_{j}$ with $j$ in the supports of the $J$-tuples $t=(t_{j})_{j},t^{\prime}=(t^{\prime}_{j})_{j}$, which are countable. Also, if $A(t)$ would be contained in some amenable $B\subset M$ and $b\in B$ is an arbitrary element, then there exists a countable $J_{0}\subset J$ that contains the support of $t$, with $b\in M_{J_{0}}:=*_{j\in J_{0}}M_{j}$. Thus, if $A(t)$ is maximal amenable in $M_{J_{0}}$, then $b\in A(t)$. This shows that if we can show that $A(t)$ is maximal amenable in any $M_{J_{0}}$ with $J_{0}\subset J$ countable, then the result follows for all $J$, thus reducing the theorem to the case $J$ at most countable.

Our argument makes crucial use of the von Neumann algebra $N\subset M$ generated by the semicircular elements, $N:=\{s_{j}\mid j\in J\}^{\prime\prime}$, which we describe via the free Gaussian functor [Vo88]. Thus, letting $H=H(J)$ denote the $|J|$-dimensional real Hilbert space, we alternatively view $N$ as the von Neumann algebra generated by $s(\xi)$, $\xi\in H$, on $\mathcal{F}(H)$, where $\mathcal{F}(H)$ is the full Fock space of $H$ and $s(\xi)\in N$ is the semi-circular operator associated to the unit vector $\xi$ in $H$.

Note that $N\simeq L\mathbb{F}_{|J|}$ and that $N$ contains all the abelian von Neumann algebras $A(t),t\in\ell^{2}_{*}$, with $\ell^{2}_{*}$ identifying naturally with the subset of vectors having at least two non-zero coordinates in the $J$-dimensional real Hilbert space $H=H(J)$. We view $A(t),t\in\ell^{2}_{*}$ as part of the larger family of abelian von Neumann subalgebras $A(\xi)=\{s(\xi)\}^{\prime\prime}\subset N$, $\xi\in H$.

Given a non-zero vector $\xi\in H$, if we denote by $H_{0}\subset H$ the orthogonal complement of $\xi$, and by $N_{0}$ the von Neumann algebra associated with $H_{0}$ by the free Gaussian function, we see that the inclusion $N$ splits as the free product $A(\xi)\subset A*N_{0}$. So $A(\xi)$ is freely complemented in $N$ and is thus maximal amenable inside $N$ by [Po82].

Lemma 1.2**.**

Given unit vectors $\xi_{1},\xi_{2}\in H$, we have $A(\xi_{1})\prec_{N}A(\xi_{2})$ iff $A(\xi_{1})=A(\xi_{2})$ iff $\xi_{1}=\pm\xi_{2}$.

Proof.

It is clearly sufficient to show that if $\xi_{1},\xi_{2}\in H$ are unit vectors with $\xi_{1}\neq\pm\xi_{2}$, then $A(\xi_{1})\nprec_{N}A(\xi_{2})$, as all other implications are trivial. Note that in proving this, we may assume $|J|\geq 3$. Indeed, because if $A(\xi_{1})\nprec_{\tilde{N}}A(\xi_{2})$ in the larger factor $L\mathbb{F}_{|J|+1}=\tilde{N}\supset N$, then $A(\xi_{1})\nprec_{N}A(\xi_{2})$ as well.

Consider the set $\mathcal{S}$ of pairs of unit vectors $\xi_{1},\xi_{2}\in H$ such that $A(\xi_{1})\prec_{M}A(\xi)$. Assume by contradiction that this set contains a pair of non-proportional vectors. We will deduce that $\mathcal{S}$ must also contain a pair of orthogonal vectors. But this conclusion is absurd because if $\xi_{1}\perp\xi_{2}$ then $A(\xi_{1}),A(\xi_{2})$ are in free position in $N$, so $A(\xi_{1})\not\prec_{N}A(\xi_{2})$ by [Po81].

By functoriality, $\mathcal{S}$ is invariant under orthogonal transformations: if $(\xi,\eta)\in\mathcal{S}$ and if $\theta\in\mathcal{O}(H)$ is an orthogonal transformation, then $(\theta(\xi),\theta(\eta))\in\mathcal{S}$.

Claim. $\mathcal{S}$ also satisfies the following property: given unit vectors $\xi,\eta,\eta^{\prime}\in H$ with $\langle\xi,\eta\rangle=\langle\xi,\eta^{\prime}\rangle$, if $(\xi,\eta)\in\mathcal{S}$, then $(\eta,\eta^{\prime})\in\mathcal{S})$.

Indeed, if $(\xi,\eta)\in\mathcal{S}$, then by [Po01], since $A(\xi)$ and $A(\eta)$ are MASAs in $N$, there exist a non-zero partial isometry $v\in N$ such that $v^{*}v\in A(\xi_{1})$, $vv^{*}\in A(\xi_{2})$ and $vA(\xi_{1})v^{*}=A(\xi_{2})vv^{*}$. Moreover, by our assumption, there exists $\theta\in\mathcal{O}(H)$ such that $\theta(\xi)=\xi$ while, $\theta(\eta)=\eta^{\prime}$. The automorphism $\alpha\in\operatorname{Aut}(N)$ associated with $\theta$ leaves all elements in $A(\xi)$ fixed and satisfies $\theta(A(\eta))=A(\eta^{\prime})$. Thus

[TABLE]

implying that $w=\theta(v)v^{*}\in N$ is a partial isometry with right support $w^{*}w=vv^{*}\in A(\eta)$, left support $ww^{*}=\theta(vv^{*})\in A(\eta^{\prime})$ and satisfying $wA(\eta)w^{*}=A(\eta^{\prime})ww^{*}$. So $(\eta,\eta^{\prime})\in\mathcal{S}$.

Starting with $(\xi,\eta)\in\mathcal{S}$, $\xi\neq\pm\eta$, we may find a small rotation $\theta$ fixing $\xi$, such that the angle between $\eta$ and $\eta^{\prime}:=\theta(\eta)$ is of the form $\pi/2^{k}$ for some large enough $k$. So by the claim, we have found a pair of vectors $(\eta,\eta^{\prime})\in\mathcal{S}$ forming an angle of the form $\pi/2^{k}$, for some $k\in\mathbb{N}$.

Applying again the claim, we see that if $(\xi,\eta)\in\mathcal{S}$ is a pair of vectors forming an angle $\alpha$ then there exists $\eta^{\prime}$ such that $(\eta,\eta^{\prime})\in\mathcal{S}$ and which forms an angle $2\alpha$ with $\eta$. So by iterating this procedure, we can find a pair $(\xi,\eta)\in\mathcal{S}$ with angle $\pi/2$, i.e., $\xi\perp\eta$, contradiction. ∎

With the above notations, we have $A(t)\subset N\subset M$ with the properties stated in the theorem being satisfied in $N$: the abelian von Neumann algebras $A(t),t\in\ell^{2}_{*}$, are maximal amenable in $N$ and the space of intertwiners in $N$ between any two of them is equal to [math], unless the corresponding $t\in\ell^{2}_{*}$ are proportional.

Thus, in order to prove the theorem, all we need is to “lift” these properties from $N$ to $M$. To prove that $A(t)$’s are non-intertwinable in $M$ as well, we’ll need an absorption result from [IPP05, Theorem 1.1].

Lemma 1.3**.**

Let $B\subset P_{1},B\subset P_{2}$ be inclusions of tracial von Neumann algebras, with ${\tau_{P_{1}}}_{|B}={\tau_{P_{2}}}_{|B}$, and let $P=P_{1}\ast_{B}P_{2}$ denote their amalgamated free product. Let $A_{1},A_{2}\subset P_{1}$ be von Neumann subalgebras and assume $A_{1}\nprec_{P_{1}}B$. Then $A_{1}\nprec_{P}A_{2}$ iff $A_{1}\nprec_{P_{1}}A_{2}$.

Proof.

The result in [IPP05] shows that any intertwiner from $A_{1}$ to $P_{1}$ inside $P$ must lie in $P_{1}$. That is, if $\xi\in L^{2}P$ is so that the Hilbert $A_{1}-P_{1}$ bimodule $\mathcal{H}:=\overline{\text{\rm sp}A_{1}\xi P_{1}}\subset L^{2}P$ is “finite over $P_{1}^{op}$” (i.e., $(P_{1}^{op})^{\prime}\cap\mathcal{B}(\mathcal{H})$ is a finite von Neumann algebra), then $\xi\in L^{2}P_{1}$. Since $A_{2}\subset P_{1}$, if $A_{1}\prec_{P}A_{2}$ then the corresponding non-zero $A_{1}-A_{2}$ intertwiner $\xi\in L^{2}P$ must lie in $P_{1}$, showing that $A_{1}\prec_{P_{1}}A_{2}$. The converse is of course trivial. ∎

Proof of Theorem 1.1.

We may assume $J$ is at most countable, with $|J|\geq 2$, i.e., either $J=\{1,2,...,n\}$ for some finite $n\geq 2$, or $J=\{1,2,...\}$. We already know that $A(\xi),A(\eta)$ are non-intertwinable in $N$ for any $\xi,\eta\in H$ that are not proportional (this includes the case $\eta=s_{j}$, for some $j\in J$).

Note that for any inclusion of tracial von Neumann algebras $B\subset P$ and $Q$ another tracial von Neumann algebra, we have an obvious identification

[TABLE]

We’ll use this for the inclusions $N_{i}\subset N_{i+1}$, $i\geq 0$, where $N_{0}=N=A(s_{1})*A(s_{2})*...$, $N_{1}=M_{1}*A(s_{2})*A(s_{3})*...$, and more generally $N_{i}=M_{1}*...*M_{i}*A(s_{i+1})*...*A(s_{n})$, $\forall i\geq 1$, to write each such inclusion as $(N_{i}\subset N_{i+1})=(N_{i}\subset N_{i}*_{A(s_{i+1})}M_{i+1})$.

We prove by induction on $i$ that $A(t)$ is maximal amenable inside $N_{i}$ and $A(t)\nprec_{N_{i}}A(s_{i})$ and $A(t)\nprec_{N_{i}}A(t^{\prime})$, for all $i\geq 0$, and all $t,t^{\prime}\in\ell^{2}_{*}$ non-proportional.

We already proved the case $i=0$. Let now $i\geq 0$ and assume that the result hold for $i$. By Lemma 1.3, this implies $A(t)\nprec_{N_{i+1}}A(t^{\prime})$, $A(t)\nprec_{N_{i+1}}A(s_{j})$, $\forall j$. By using this fact for $j=i+1$ we also obtain from [BH16] that $A(t)$ is maximal amenable in $N_{i+1}$.

If $|J|<\infty$, the proof is complete. Otherwise, we need to derive the conclusion in $M$. Since $N_{n}\nearrow M$, if $A(t),A(t^{\prime})$ would be interwtinable in $M$, via some non-zero $\xi\in L^{2}M$, then its expectation onto $N_{n}$ for some large $n$ is non-zero and intertwines $A(t),A(t^{\prime})$ in $N_{n}$, a contradiction. This implies $A(t)\not\prec_{M}A(t^{\prime})$, whenever $t\in\ell^{2}_{*}$, $t^{\prime}\in H$ are not proportional (so including the case $s(t^{\prime})=s_{j},j\in J$). About the maximal amenability of $A(t)$ in case where $J$ is infinite, we may in fact use a gathering trick. Since $|J|\geq 2$ and $t$ has at least two non-zero entries, we can split $J$ as a disjoint union of two non-empty sets $J=J_{1}\sqcup J_{2}$ such that the support of $t$ intersects both $J_{1},J_{2}$. Thus, we can decompose $s(t)$ as $s(t)=(\sum_{j\in J_{1}}t_{j}s_{j})+(\sum_{j\in J_{2}}t_{j}s_{j})=c_{1}\tilde{s}_{1}+c_{2}\tilde{s}_{2}$, with $c_{1},c_{2}\in\mathbb{R}$ non-zero and $\tilde{s}_{i}\in\tilde{M}_{i}:=*_{j\in J_{i}}M_{j}$, $i=1,2$, self-adjoint semicircular elements. Thus, we can apply the case $|J|=2$ above to deduce that the von Neumann algebra generated by $s(t)$ in $\tilde{M}_{1}*\tilde{M}_{2}$, i.e., $A(t)\subset M$, is maximal amenable. ∎

Remark 1.4**.**

$1^{\circ}$ While we could not use Theorem 1.1 to prove that the free group factors $L\mathbb{F}_{n}$ contain maximal amenable MASAs that are not freely complemented, we believe the following particular cases of 1.1 deserve further investigation: $(a)$ the case $M_{j}=A(s_{j})\otimes R$, $\forall j\in J$, where $A(s_{j})=\{s_{j}\}^{\prime\prime}\subset M_{j}$ and $R$ is the hyperfinite II1 factor; $(b)$ the case $M_{j}=A(s_{j})\rtimes\Gamma_{j}$, where $\Gamma_{j}$ is an amenable group and $\Gamma_{j}\curvearrowright A_{j}$ is a trace preserving action, $\forall j\in J$; $(c)$ the case $M_{j}$ abelian non-separable, e.g., an ultrapower of $L^{\infty}[0,1]$, $\forall j$.

When studying these cases, it may be useful to take into account that if $s_{j}\in M_{j}$ are fixed, the map $t\mapsto s(t)$ is very smooth, a fact that entails pointwise continuity of $t\mapsto A_{t}$. It may be possible to use this continuity in combination with the fact that at all points $t\in\ell^{2}_{*}$ the algebra $A(t)$ is maximal amenable, while at the “singularity points $A(s_{j})$” this fails, once $M_{j}$ is “much larger” than the semicircular element $s_{j}\in M_{j}$.

$2^{\circ}$ As we mentioned before, a natural candidate for a non-freely complemented abelian subalgebra in $L\mathbb{F}_{n}$ is the radial MASA $L_{n}$ (shown to be a MASA in [Py81] and maximal amenable in [CFRW09]). Another natural class of candidates are the MASAs $A_{g}=\{u_{g}\}^{\prime\prime}\subset L\mathbb{F}_{n}$, where $g\in\mathbb{F}_{n}$ is so that $g^{\mathbb{Z}}$ is maximal abelian in $\mathbb{F}_{n}$, but not freely complemented in $\mathbb{F}_{n}$. A typical such example is the commutator element $g=aba^{-1}b^{-1}\in\mathbb{F}_{2}$. It was pointed out in ([Po82], Remark 3.5.1) that any such abelian subalgebra $A_{g}\subset L(\mathbb{F}_{n})$ is maximal amenable, so in particular a MASA. But it remains open whether such MASAs are freely complemented in $L(\mathbb{F}_{n})$ or not.