Property Γ and the algebra generated by a factor and its commutant

Written by Claude Opus 5.5 (Anthropic), September 2026, revised October 2026. The September text was spot-checked by Claude Opus 5.5 in a separate session; the October revisions are self-checked by the writing AI. Public domain (CC0).

Introduction

Let $N$ be a II₁ factor with trace $\tau$, acting on $H=L^2(N,\tau)$ by left multiplication. Its commutant $N'$ is the algebra of right multiplications. Neither $N$ nor $N'$ contains a nonzero compact operator. The C*-algebra $C^*(N,N')$ generated by the two of them, however, may contain compact operators, and then, being irreducible, it contains all of them. This lesson proves that this happens exactly when $N$ does not have Murray and von Neumann's property Γ, that is, when $N$ has no approximately central unitaries of trace zero.

The main result, Theorem 2.5, gives four equivalent conditions on $N$:

(a) $N$ has property Γ;
(b) every finite set of inner automorphisms leaves some state of $N$ invariant that is not normal;
(c) no finite set of unitaries of $N$ has a spectral gap on $L^2(N)$: there are always unit vectors orthogonal to $\hat1$ that almost commute with the given unitaries;
(d) $C^*(N,N')$ contains no nonzero compact operator.

The theorem turns property Γ, a statement about approximately central elements inside $N$, into a statement about finitely many unitary operators on $L^2(N)$ (condition (c)) and into a statement about a concrete C*-algebra (condition (d)). The route around the cycle of implications uses four quite different tools: an irreducibility argument for (a)⇒(d), spectral theory for (d)⇒(c), ultrapowers and weak* limits of states for (c)⇒(b), and a maximality argument with estimates on spectral projections for (b)⇒(a).

Section 6 gives two consequences. A semidiscrete factor of type II₁ has property Γ, because for such a factor $C^*(N,N')$ is a quotient of the minimal tensor product of two simple C*-algebras. And a tensor product of two factors of type II₁ has property Γ exactly when one of the two factors has it; with separable preduals, the tensor product is full exactly when both factors are full. The group von Neumann algebra of the free group on two generators gives an explicit example in which the rank-one projection onto $\mathbb C\hat1$ is written as a continuous function of a single self-adjoint operator in $C^*(N,N')$ (Example 3.3).

The later lessons "Approximately inner and centrally trivial automorphisms" and "Uniqueness of the injective II₁ factor" of this course use these results.

What is assumed. From Course 1 we use the lessons "Ultraproducts and the asymptotic centralizer" (the ultrapower $N^\omega$, its standard form on uniformly integrable sequences, and the completeness of the unit ball for the trace norm) and "Full factors" (the equivalence of fullness with the absence of property Γ, and a spectral gap estimate for the free group). From this course we use the lesson "Trace inequalities for finite von Neumann algebras", which is recalled in Section 1. We also use standard operator algebra, listed in "Results used from other lessons", each fact with the lesson that proves it.

Basic references are [Connes 1976], [Effros–Lance 1977], [Akemann–Ostrand 1975] and [Anantharaman–Popa, Chapter 15].

Results used from other lessons

Throughout, $N$ is a von Neumann algebra with a faithful normal tracial state $\tau$ (from Section 2 on, a factor of type II₁). We write $\|x\|_2=\tau(x^*x)^{1/2}$ and $\|x\|_1=\tau(|x|)$, and $\mathcal U(N)$ for the unitary group.

(B1) Trace norms. $\|\cdot\|_1$ and $\|\cdot\|_2$ are norms on $N$, and $\|x\|_1=\sup\{|\tau(xy)|:y\in N,\ \|y\|\le1\}$; in particular the functional $\tau(y\,\cdot\,)$ has norm at most $\|y\|_1$ in $N^*$. For $a,b,x,y\in N$: $\|axb\|_1\le\|a\|\|x\|_1\|b\|$, $\|axb\|_2\le\|a\|\|x\|_2\|b\|$, $\|x^*\|_2=\|x\|_2$, $|\tau(xy)|\le\|x\|_2\|y\|_2$ and $|\tau(x)|\le\|x\|_1\le\|x\|_2\le\|x\|$. If $0\le x\le y$ then $\tau(x)\le\tau(y)$. The estimates for $\|\cdot\|_1$, including $\|axb\|_1\le\|a\|\,\|x\|_1\|b\|$ and the triangle inequality, are proved in Traces on von Neumann algebras, Proposition 7.1, and the duality formula for $\|x\|_1$ in Integration for a trace, Theorem 1.1. The rest is short: $\|\cdot\|_2$ is the norm of the inner product $\tau(y^*x)$, which is definite because $\tau$ is faithful; $\|x^*\|_2=\|x\|_2$ because $\tau(xx^*)=\tau(x^*x)$; $|\tau(xy)|\le\|x^*\|_2\|y\|_2$ is the Cauchy–Schwarz inequality; $\tau(b^*x^*a^*axb)\le\|a\|^2\tau(xbb^*x^*)\le\|a\|^2\|b\|^2\tau(xx^*)$; for $x=u|x|$, $\|x\|_1=\tau(u^*x)\le\|x\|_2$; and $\tau(x^*x)\le\|x\|^2$. Monotonicity is positivity of $\tau$.

(B2) Square-integrable operators. $H=L^2(N,\tau)$ is realized as a space of closed densely defined operators affiliated with $N$ (the $\tau$-measurable operators $\xi$ with $\tau(|\xi|^2)<\infty$), containing $N$ as the dense subspace $\{\hat x:x\in N\}$, with $\langle\hat x,\hat y\rangle=\tau(y^*x)$ and $\hat1$ the trace vector. We write $\xi$ for elements of $H$ and often $x$ for $\hat x$.

(a) $N$ acts by left multiplication and also by right multiplication, $\|x\xi y\|\le\|x\|\|\xi\|\|y\|$. The modular conjugation is $J\xi=\xi^*$; thus $JxJ\xi=\xi x^*$, and $N'=JNJ$ is the algebra of right multiplications. For $x\in N$, $\langle\hat x,\hat1\rangle=\tau(x)$.
(b) Every $\xi\in H$ has a polar decomposition $\xi=v|\xi|$, with $v\in N$ a partial isometry whose initial projection is the support of $|\xi|$, and $|\xi|$ in the cone $H_+=L^2(N,\tau)_+$ of positive elements. For a bounded Borel function $g$ on $[0,\infty[$, $g(|\xi|)\in N$ and $\|\xi g(|\xi|)\|=\||\xi|g(|\xi|)\|$.
(c) For $h\in H_+$ and $a\ge0$ put $E_a(h)=\chi_{]a,\infty[}(h)\in N$. The measure $\nu_h(B)=\tau(\chi_B(h))$ on $[0,\infty[$ satisfies $\tau(E_a(h))=\nu_h(]a,\infty[)$, $\|h\|^2=\int t^2\,d\nu_h(t)$ and $\|hE_a(h)\|^2=\int_{]a,\infty[}t^2\,d\nu_h(t)$. In particular $a^2\tau(E_a(h))\le\|hE_a(h)\|^2\le\|h\|^2$. For $\xi\in H$ we write $E_a(|\xi|)$ accordingly.
(d) For $u\in\mathcal U(N)$ the unitary $uJuJ\colon\xi\mapsto u\xi u^*$ maps $H_+$ onto itself, and $E_a(uhu^*)=uE_a(h)u^*$ for $h\in H_+$.
(e) If $f\in N$ is a projection and $h\in H_+$ satisfies $fh=h$, then $E_a(h)\le f$ for every $a>0$.
(f) Every normal positive functional $\psi$ on $N$ is $\omega_h=\langle\,\cdot\,h,h\rangle$ for a unique $h\in H_+$. (Then $\psi=\tau(h^2\,\cdot\,)$.)
(g) The $\tau$-measurable operators $y$ with $\|y\|_1=\tau(|y|)<\infty$ form the space $L^1(N,\tau)$, which contains $\xi\eta$ for $\xi,\eta\in H$. The map $y\mapsto\tau(y\,\cdot\,)$ is an isometry of $L^1(N,\tau)$ onto the predual $N_*$: $\|\tau(y\,\cdot\,)\|=\|y\|_1$.

The Hilbert space $L^2(N,\tau)$, its two actions, $J$ and $N'=JNJ$ are proved in Integration for a trace, Sections 2–3 (Theorem 3.1). The $\tau$-measurable operators, their adjoints, sums, products, square roots and polar decompositions are constructed in Operators recovered from small trace defects, and $L^1(N,\tau)$, $L^2(N,\tau)$ as spaces of such operators, with the isometry of $L^1(N,\tau)$ onto $N_*$ in (g), in Trace densities and noncommutative integration, §§TI-06 and TI-15. The statement (f) is The positive cone of a standard representation, §SF-10 and §SF-07 for the trace, whose cone is $H_+$. Then (c)–(e) follow from the spectral theorem for the positive self-adjoint operator $h$: $\tau(\chi_B(h))=\|\chi_B(h)\hat1\|^2$ is a measure, $\|g(h)\|^2=\int|g(t)|^2\,d\nu_h(t)$, and $a^2\chi_{]a,\infty[}(t)\le t^2\chi_{]a,\infty[}(t)\le t^2$; the unitary $uJuJ$ is conjugation by $u$, which carries positive operators to positive operators and spectral projections to spectral projections; and if $fh=h$, then the range of $h$, hence of each spectral projection $E_a(h)$ with $a>0$, lies in $fH$.

(B3) Normal and singular functionals. A positive functional $\sigma$ on a von Neumann algebra $M$ is singular if the only normal positive functional $\psi$ with $\psi\le\sigma$ is $0$. Every positive functional $\varphi$ on $M$ decomposes uniquely as $\varphi=\varphi_n+\varphi_s$ with $\varphi_n$ normal positive and $\varphi_s$ singular positive. A positive functional $\sigma$ is singular if and only if every nonzero projection $p\in M$ majorizes a nonzero projection $q\in M$ with $\sigma(q)=0$. The decomposition is proved in The universal enveloping von Neumann algebra of a C*-algebra, and W*-algebras, Theorem 10.3 and Proposition 10.5, where a functional is called singular when it lies in $M_*^\perp$, and the projection criterion for that notion is Theorem 11.2 there. The two notions agree for positive $\sigma$: if $\sigma$ has no nonzero normal positive minorant, its normal part, which is such a minorant, vanishes; conversely, if $\sigma\in M_*^\perp$ and $0\le\psi\le\sigma$ with $\psi$ normal and nonzero, the criterion gives a nonzero $q\le s(\psi)$ with $\sigma(q)=0$, so $\psi(q)=0$, which is impossible because $\psi$ is faithful on its support.

(B4) Compact operators in irreducible algebras. If a C*-algebra $A\subset B(H)$ acts irreducibly (its commutant is $\mathbb C$) and $A\cap\mathcal K(H)\ne\{0\}$, then $\mathcal K(H)\subset A$. Proved in Representations and positive functionals: the GNS construction and the Gelfand–Naimark theorem, Corollary 9.3.

(B5) Minimal tensor products. For C*-algebras $A\subset B(H)$, $C\subset B(K)$, the minimal (spatial) tensor product $A\otimes_{\min}C$ is the closure of the algebraic tensor product $A\odot C$ in $B(H\otimes K)$. (i) If $\alpha\colon A\to B$ is a completely positive contraction between C*-algebras, then $\alpha\otimes\mathrm{id}_C$ extends to a contraction $A\otimes_{\min}C\to B\otimes_{\min}C$. (ii) A $*$-homomorphism between C*-algebras has closed range. (iii) Takesaki's theorem: if $A$ and $C$ are simple, so is $A\otimes_{\min}C$. Statement (i) is proved in Completely positive finite models, Lemma 4.1, (ii) in C*-algebras: continuous functional calculus, automatic continuity, positive cones, approximate identities and quotients, Corollary 4.6 applied to the injective map induced on the quotient by the kernel, and (iii) in Tensor products of C*-algebras and the minimal norm, Theorem 5.2. See also [Takesaki 1964].

(B6) Factors of type II₁. Let $N$ be a II₁ factor. Its normal tracial state $\tau$ is unique and faithful. For each projection $q$ and $s\in[0,\tau(q)]$ there is a projection $q'\le q$ with $\tau(q')=s$; if $\tau(p)\le\tau(q)$, then $p$ is equivalent to a subprojection of $q$; in particular projections with the same trace are equivalent. For projections $p,q$ one has $\tau(p\vee q)\le\tau(p)+\tau(q)$ (by Kaplansky's formula $p\vee q-q\sim p-p\wedge q$, valid in any von Neumann algebra). For a nonzero projection $f$, $fNf$ is a II₁ factor with trace $\tau(\cdot)/\tau(f)$. On $L^2(N,\tau)$ the commutant $N'$ is a factor of type II₁. If $N_1,N_2$ are factors of type II₁, the von Neumann tensor product $N_1\bar\otimes N_2$ is a factor of type II₁ whose trace is $\tau_1\otimes\tau_2$. Dixmier approximation theorem: for $x\in N$ the norm closed convex hull of $\{uxu^*:u\in\mathcal U(N)\}$ contains $\tau(x)1$. Uniqueness and faithfulness of $\tau$: Integration for a trace, Corollary 7.4 and Traces on von Neumann algebras, Proposition 3.2. Subprojections of every trace: Measurable operators for a trace, Lemma 4.3. Comparison through the trace: Traces on von Neumann algebras, Corollary 5.4. Kaplansky's formula: Lemma 1.3 there; it gives $\tau(p\vee q)=\tau(q)+\tau(p-p\wedge q)\le\tau(p)+\tau(q)$. Corners: Lemma 1.5. The commutant: $N'=JNJ$ (Integration for a trace, Theorem 3.1), and $x\mapsto Jx^*J$ is a $*$-isomorphism of the opposite algebra of $N$ onto $N'$, so $N'$ is a II₁ factor, with tracial state $Jx^*J\mapsto\tau(x)$. Tensor products: Integration for a trace, Proposition 6.1 and Spatial tensor products of von Neumann algebras, Corollary 11.5(3). The Dixmier approximation theorem, used only in an exercise, is Traces on von Neumann algebras, Theorem 8.2 and Corollary 8.3; its σ-weakly closed version is Lemma 5.1 there.

(B7) Semidiscreteness and the product map. If a von Neumann algebra $N\subset B(H)$ is semidiscrete in the sense that the identity map of $N$ is a pointwise $\sigma$-weak limit of normal completely positive maps of finite rank, then $\|\sum_ka_kb_k\|\le\|\sum_ka_k\otimes b_k\|_{\min}$ for all finite families $a_k\in N$, $b_k\in N'$. [Effros–Lance 1977]. This form is used only as an alternative to Definition 6.2; for approximation through matrix algebras with normal first maps the inequality is Finite models of a von Neumann algebra, Theorem 2.1.

1. What is used from the trace inequalities

The following three statements are proved in the lesson "Trace inequalities for finite von Neumann algebras" of this course, there for a faithful normal semifinite trace; we give that lesson's numbering with each. We restate them for a von Neumann algebra $N$ with a faithful normal tracial state $\tau$, in the notation of (B2).

Theorem 1.1 (Powers–Størmer inequality). For $h,k\in H_+$, $\|h-k\|_2^2\le\|h^2-k^2\|_1=\|\omega_h-\omega_k\| ,$ where $\omega_h=\langle\,\cdot\,h,h\rangle$ and the norm on the right is that of $N_*$.

This is Theorem 3.1 of "Trace inequalities for finite von Neumann algebras". The equality on the right is (B2)(f) and (B2)(g): $\omega_h-\omega_k=\tau((h^2-k^2)\,\cdot\,)$.

Theorem 1.2 (spectral projections, integrated over the level). For $h,k\in H_+$, $\int_0^\infty\|E_{\sqrt s}(h)-E_{\sqrt s}(k)\|_2^2\,ds\le\|h-k\|_2\,\|h+k\|_2 .\tag{1.1}$ (The integrand is a Borel function of $s$; see the proof of Lemma 1.4.)

This is Proposition 4.3(b) of "Trace inequalities for finite von Neumann algebras".

Theorem 1.3 (conjugating close projections). Let $M$ be a von Neumann algebra and $p,q\in M$ equivalent finite projections. There is a unitary $W\in M$ with $WpW^*=q$, $W$ commuting with $|p-q|$, and $|W-1|\le\sqrt2\,|p-q|$. If $\tau$ is a faithful normal finite trace on $M$, then $\|W-1\|_1\le\sqrt2\,\|p-q\|_1$.

This is Corollary 9.5 of "Trace inequalities for finite von Neumann algebras" (with Theorem 9.4 there); the trace-norm statement is its $L^p$ estimate for the exponent $1$. Only this trace-norm consequence is used below.

Theorem 1.2 lets us choose one level at which spectral projections of nearby positive vectors are close, relative to their size.

Lemma 1.4 (choice of a level). Let $h\in H_+$, $h\ne0$, let $h_1,\dots,h_n\in H_+$, and let $c>\frac{1}{\|h\|_2^2}\sum_{j=1}^n\|h_j-h\|_2\,\|h_j+h\|_2 .$ Then there is $s>0$ such that $p=E_{\sqrt s}(h)$ is nonzero and $\sum_{j=1}^n\|E_{\sqrt s}(h_j)-p\|_2^2<c\,\|p\|_2^2$.

Proof. Put $G(s)=\|E_{\sqrt s}(h)\|_2^2=\tau(E_{\sqrt s}(h))$ and $D(s)=\sum_j\|E_{\sqrt s}(h_j)-E_{\sqrt s}(h)\|_2^2$. Both are right continuous, hence Borel. Indeed, for $k\in H_+$, as $s$ decreases to $s_0$ the sets $]\sqrt s,\infty[$ increase to $]\sqrt{s_0},\infty[$, so $E_{\sqrt s}(k)$ increases strongly to $E_{\sqrt{s_0}}(k)$; products of bounded strongly convergent nets converge strongly; and $\|p-q\|_2^2=\tau(p)+\tau(q)-2\operatorname{Re}\tau(pq)$ for projections, with $\tau(x)=\langle x\hat1,\hat1\rangle$. By (B2)(c) and Tonelli's theorem, $\int_0^\infty G(s)\,ds=\int_0^\infty\nu_h(]\sqrt s,\infty[)\,ds=\int\Big(\int_0^\infty\chi_{\{s<t^2\}}\,ds\Big)d\nu_h(t) =\int t^2\,d\nu_h(t)=\|h\|_2^2 .$ By (1.1), $\int_0^\infty D(s)\,ds\le\sum_j\|h_j-h\|_2\|h_j+h\|_2<c\|h\|_2^2=\int_0^\infty cG(s)\,ds$. Since $\int cG<\infty$, we get $\int_0^\infty(D(s)-cG(s))\,ds<0$, so the set of $s$ with $D(s)<cG(s)$ has positive measure. For such $s$, $G(s)>0$ because $D(s)\ge0$; so $p=E_{\sqrt s}(h)\ne0$ and $D(s)<c\|p\|_2^2$. $\square$

We also collect some elementary estimates.

Lemma 1.5. Let $N$ have a faithful normal tracial state $\tau$.

(i) If $z\in N$ and $s$ is a projection with $z=szs$, then $\|z\|_1\le\|z\|_2\|s\|_2$.
(ii) If $N$ is a II₁ factor and $p,q$ are projections with $\tau(p)=\tau(q)$, then $\|p-q\|_1\le\sqrt{2\tau(p)}\,\|p-q\|_2$.
(iii) $\|x\|_2^2\le\|x\|\,\|x\|_1$ for $x\in N$; in particular $\|x\|_2^2\le2\|x\|_1$ if $\|x\|\le2$.
(iv) Let $N$ be a II₁ factor, $e\ne0$ a projection, $u\in\mathcal U(N)$, and $\psi_e=\tau(e\,\cdot\,)/\tau(e)$. Then $\|\psi_e\circ\operatorname{Ad}u-\psi_e\|\le\sqrt2\,\|[u,e]\|_2/\|e\|_2$.

Proof. (i) By (B1), for $\|y\|\le1$, $|\tau(zy)|=|\tau(szsy)|=|\tau(z\,sys)|\le\|z\|_2\|sys\|_2\le \|z\|_2\|s\|_2$. Take the supremum over $y$. (ii) Apply (i) with $z=p-q$ and $s=p\vee q$; by (B6), $\|s\|_2^2=\tau(p\vee q)\le2\tau(p)$. (iii) $|x|^2\le\|x\|\,|x|$, so $\tau(x^*x)\le\|x\|\tau(|x|)$. (iv) $\psi_e(uxu^*)=\tau(u^*eu\,x)/\tau(e)$, so $\psi_e\circ\operatorname{Ad}u-\psi_e=\tau((u^*eu-e)\,\cdot\,)/ \tau(e)$, whose norm is at most $\|u^*eu-e\|_1/\tau(e)$ by (B1). By (ii), $\|u^*eu-e\|_1\le\sqrt{2\tau(e)}\, \|u^*eu-e\|_2$, and $\|u^*eu-e\|_2=\|eu-ue\|_2$. Since $\tau(e)=\|e\|_2^2$, the bound follows. $\square$

Parts (i) and (ii) hold, with the same proof, in any von Neumann algebra with a faithful normal tracial state (Kaplansky's formula gives $\tau(p\vee q)\le\tau(p)+\tau(q)$ there too). For $q=vpv^*$ with $v$ unitary, (ii) reads $\|vpv^*-p\|_1\le\sqrt2\,\|[v,p]\|_2\|p\|_2$; Lemma 8.1 of "Trace inequalities for finite von Neumann algebras" gives the same bound with $2$ in place of $\sqrt2$, and either would do below.

2. Property Γ, spectral gaps and the main theorem

From now on $N$ is a II₁ factor with trace $\tau$, and $H=L^2(N,\tau)$ is its standard space, with trace vector $\hat1$ and modular conjugation $J$ (B2). For $\xi\in H$ and $u\in N$ we write $[u,\xi]=u\xi-\xi u=(u-Ju^*J)\xi ,\qquad U_u=uJuJ,\quad U_u\xi=u\xi u^* .$ For a unitary $u$, $U_u$ is a unitary fixing $\hat1$, and $\|U_u\xi-\xi\|=\|(u\xi-\xi u)u^*\|=\|[u,\xi]\| .\tag{2.1}$ We write $C^*(N,N')$ for the C*-algebra generated by $N\cup N'$ in $B(H)$, $\mathcal K(H)$ for the compact operators, and $P_0$ for the orthogonal projection onto $\mathbb C\hat1$. Since $N$ and $N'$ commute, the $*$-algebra generated by $N\cup N'$ is the linear span of the products $ab$, $a\in N$, $b\in N'$, and it is norm dense in $C^*(N,N')$.

Definition 2.1 (property Γ). $N$ has property Γ if for every finite set $x_1,\dots,x_m\in N$ and every $\varepsilon>0$ there is a unitary $u\in N$ with $\tau(u)=0$ and $\|ux_k-x_ku\|_2\le\varepsilon$ for all $k$.

Reference: property Γ was introduced by Murray and von Neumann; see [Anantharaman–Popa, §15.2] and [Connes 1976, §3.9].

This is Definition 6.1 of "Full factors", where it is stated for factors with separable predual; here no separability is assumed.

It is enough to test property Γ on unitaries, and a projection of trace $\frac12$ gives a unitary of trace $0$.

Lemma 2.2. Suppose that for every finite set $u_1,\dots,u_n\in\mathcal U(N)$ and every $\delta>0$ there is a projection $E\in N$ with $\tau(E)=\frac12$ and $\|[u_j,E]\|_2\le\delta$ for all $j$. Then $N$ has property Γ.

Proof. Every $x\in N$ with $\|x\|\le1$ can be written $x=\frac12(w_1+w_1^*)+\frac i2(w_2+w_2^*)$ with unitaries $w_1,w_2$: write $x=a+ib$ with $a,b$ self-adjoint of norm at most $1$, and put $w_1=a+i(1-a^2)^{1/2}$, $w_2=b+i(1-b^2)^{1/2}$.

Let $x_1,\dots,x_m\in N$ and $\varepsilon>0$; put $C=1+\max_k\|x_k\|$. Write each $x_k/C$ in this way, and let $u_1,\dots,u_{2m}$ be the unitaries so obtained. Take a projection $E$ with $\tau(E)=\frac12$ and $\|[u_j,E]\|_2\le\varepsilon/(4C)$ for all $j$. Then $u=2E-1$ is a unitary with $\tau(u)=0$. For each $u_j$, $\|[u,u_j]\|_2=2\|[E,u_j]\|_2$, and $\|[u,u_j^*]\|_2=\|[u,u_j]\|_2$ because $[u,u_j^*]=-[u,u_j]^*$ when $u$ is self-adjoint. Hence, if $x_k/C=\frac12(w_1+w_1^*)+\frac i2(w_2+w_2^*)$, $\|[u,x_k]\|_2\le C\cdot\tfrac12\big(2\|[u,w_1]\|_2+2\|[u,w_2]\|_2\big)\le C\cdot2\cdot2\cdot\frac{\varepsilon}{4C} =\varepsilon . \qquad\square$

The spectral form of condition (c) below deserves a name.

Definition 2.3 (spectral gap). Unitaries $u_1,\dots,u_n\in N$ have a spectral gap if there is a constant $K\ge0$ with $\|\xi-\langle\xi,\hat1\rangle\hat1\|\le K\max_j\|[u_j,\xi]\|\qquad\text{for all }\xi\in H .\tag{2.2}$

Lemma 2.4. For unitaries $u_1,\dots,u_n\in N$ the following are equivalent.

(i) There is a sequence of unit vectors $\xi_k\in H$ with $\|[u_j,\xi_k]\|\to0$ for all $j$, such that $|\langle\xi_k,\hat1\rangle|$ does not converge to $1$.
(ii) There is a sequence of unit vectors $\xi_k\in H$ with $\langle\xi_k,\hat1\rangle=0$ and $\|[u_j,\xi_k]\|\to0$ for all $j$.
(iii) $u_1,\dots,u_n$ do not have a spectral gap.

Theorem 2.5. For a II₁ factor $N$ in standard form on $H=L^2(N,\tau)$, the four conditions below are equivalent.

(a) $N$ has property Γ (Definition 2.1).
(b) For every finite set $u_1,\dots,u_n\in\mathcal U(N)$ there is a state $\varphi$ on $N$ that is not normal and satisfies $\varphi(u_jxu_j^*)=\varphi(x)$ for all $x\in N$ and all $j$.
(c) For every finite set $u_1,\dots,u_n\in\mathcal U(N)$ there is a sequence of unit vectors $\xi_k\in H$ with $(u_j-Ju_j^*J)\xi_k\to0$ for all $j$, such that $|\langle\xi_k,\hat1\rangle|$ does not converge to $1$. Equivalently (Lemma 2.4), no finite set of unitaries of $N$ has a spectral gap.
(d) $C^*(N,N')\cap\mathcal K(H)=\{0\}$.

Reference: [Connes 1976, Theorem 3.9.1] states (a)⇔(d), without proof; Sections 4 and 5 give Connes's proof. See also [Anantharaman–Popa, §15.2].

In (b), invariance under $\operatorname{Ad}u_1,\dots,\operatorname{Ad}u_n$ is the same as invariance under the finitely generated group of inner automorphisms that they generate. Invariance under all inner automorphisms is too much to ask: by Exercise 3, the only state invariant under all of them is $\tau$, which is normal.

We prove (a)⇒(d)⇒(c) in Section 3, (c)⇒(b) in Section 4 and (b)⇒(a) in Section 5. No separability of the predual is needed anywhere.

3. Central sequences exclude compact operators; spectral gaps produce them

The first implication holds for any factor in any representation.

Proposition 3.1. Let $M\subset B(H)$ be a factor, $\xi_0\in H$ a unit vector and $\varepsilon\in\,]0,1]$. Suppose that for every finite set $a_1,\dots,a_m\in M$, every finite set $\eta_1,\dots,\eta_m\in H$ and every $\delta>0$ there is a unitary $v\in M$ with $|\langle v\xi_0,\xi_0\rangle|\le1-\varepsilon$ and $\|(a_lv-va_l)\eta_l\|\le\delta$ for all $l$. Then $C^*(M,M')\cap\mathcal K(H)=\{0\}$.

Proof. The commutant of $M\cup M'$ is $M'\cap M''=M'\cap M=\mathbb C$, so $C^*(M,M')$ acts irreducibly. Suppose it contains a nonzero compact operator. By (B4) it contains the rank-one projection $P$ onto $\mathbb C\xi_0$. By density there are $a_1,\dots,a_m\in M$ and $b_1,\dots,b_m\in M'$ with $\|S-P\|\le\varepsilon/3$, where $S=\sum_la_lb_l$. Let $v$ be as in the hypothesis for these $a_l$, the vectors $\eta_l=b_l\xi_0$, and $\delta<\varepsilon/(3m)$. Since $b_l$ commutes with $v$, $Sv\xi_0-vS\xi_0=\sum_l(a_lv-va_l)b_l\xi_0,\qquad\text{so}\qquad\|Sv\xi_0-vS\xi_0\|\le m\delta<\varepsilon/3 .$ On the other hand $\|vS\xi_0\|=\|S\xi_0\|\ge\|P\xi_0\|-\varepsilon/3=1-\varepsilon/3$, while $\|Sv\xi_0\|\le\|Pv\xi_0\|+\varepsilon/3=|\langle v\xi_0,\xi_0\rangle|+\varepsilon/3\le1-2\varepsilon/3$. Hence $\|Sv\xi_0-vS\xi_0\|\ge\varepsilon/3$, a contradiction. $\square$

Proof of (a)⇒(d). Let $N$ have property Γ; we check the hypothesis of Proposition 3.1 with $\xi_0=\hat1$ and $\varepsilon=1$. Let $a_1,\dots,a_m\in N$, $\eta_1,\dots,\eta_m\in H$ and $\delta>0$; put $A=1+\max_l\|a_l\|$. Since $\hat N$ is dense in $H$, choose $y_l\in N$ with $\|\eta_l-\hat y_l\|\le\delta/(4A)$, and put $Y=1+\max_l\|y_l\|$. By property Γ there is a unitary $v\in N$ with $\tau(v)=0$ and $\|a_lv-va_l\|_2\le\delta/(2Y)$ for all $l$. By (B1), $\|(a_lv-va_l)\eta_l\|\le\|(a_lv-va_l)y_l\|_2+2A\|\eta_l-\hat y_l\|\le\|a_lv-va_l\|_2\,Y+\delta/2\le\delta,$ and $\langle v\hat1,\hat1\rangle=\tau(v)=0$. By Proposition 3.1, $C^*(N,N')\cap\mathcal K(H)=\{0\}$. $\square$

The next implication is proved in contrapositive form: a spectral gap puts $P_0$ into the C*-algebra generated by finitely many unitaries of $N$ and $N'$.

Proposition 3.2. Let $u_1,\dots,u_n\in\mathcal U(N)$ have a spectral gap with constant $K$, and put $T=\frac1{2n}\sum_{j=1}^n\big(U_{u_j}+U_{u_j}^*\big),\qquad U_{u_j}=u_jJu_jJ .$ Then $T$ is self-adjoint, $T\hat1=\hat1$, and the spectrum of $T$ is contained in $[-1,1-c]\cup\{1\}$ with $c=1/(2nK^2)$. Consequently $P_0=f(T)$ for any continuous function $f$ on $[-1,1]$ with $f(1)=1$ and $f=0$ on $[-1,1-c]$, and $P_0$ lies in the C*-algebra generated by $u_1,\dots,u_n,Ju_1J,\dots,Ju_nJ$.

Proof. $T$ is self-adjoint with $\|T\|\le1$, and $T\hat1=\hat1$ since each $U_{u_j}$ fixes $\hat1$. Hence $T$ commutes with $P_0$ and leaves $\hat1^\perp$ invariant. For a unitary $U$ and a vector $\xi$, $\langle(2-U-U^*)\xi,\xi\rangle=2\|\xi\|^2-2\operatorname{Re}\langle U\xi,\xi\rangle=\|U\xi-\xi\|^2$. So, for $\xi\perp\hat1$, using (2.1) and (2.2), $\langle(1-T)\xi,\xi\rangle=\frac1{2n}\sum_j\|U_{u_j}\xi-\xi\|^2\ge\frac1{2n}\max_j\|[u_j,\xi]\|^2\ge \frac{\|\xi\|^2}{2nK^2}.$ Thus the restriction $T_0$ of $T$ to $\hat1^\perp$ satisfies $-1\le T_0\le1-c$, and $\operatorname{Sp}T\subset\operatorname{Sp}T_0\cup\{1\}\subset[-1,1-c]\cup\{1\}$. For $f$ as stated, $f(T)=f(1)P_0+f(T_0)(1-P_0)=P_0$. Finally $f(T)$ lies in the C*-algebra generated by $T$, which lies in the one generated by the $u_j$ and $Ju_jJ$. $\square$

Proof of (d)⇒(c). If (c) fails, some $u_1,\dots,u_n\in\mathcal U(N)$ fail condition (i) of Lemma 2.4, hence have a spectral gap by that lemma. By Proposition 3.2 the nonzero compact operator $P_0$ lies in $C^*(N,N')$, so (d) fails. $\square$

Example 3.3 (the free group). Let $\mathbb F_2$ be free on $a,b$, let $\lambda,\rho$ be its left and right regular representations on $\ell^2(\mathbb F_2)$, and $N=\lambda(\mathbb F_2)''$, a factor of type II₁. The vector $\delta_1$ is a cyclic and separating trace vector, so $\ell^2(\mathbb F_2)=L^2(N,\tau)$ with $\hat1=\delta_1$, $J\delta_g=\delta_{g^{-1}}$ and $J\lambda(g)J=\rho(g)$, where $\rho(g)\delta_s=\delta_{sg^{-1}}$. Hence $U_{\lambda(g)}\delta_s=\delta_{gsg^{-1}}$: on $\ell^2(\mathbb F_2\setminus\{1\})=\hat1^\perp$, the unitaries $U_{\lambda(g)}$ form the representation of $\mathbb F_2$ acting by conjugation. By Lemmas 8.1 and 8.2 of "Full factors", every $f\in\ell^2(\mathbb F_2\setminus\{1\})$ satisfies $\|f\|\le14\max(\|U_{\lambda(a)}f-f\|,\|U_{\lambda(b)}f-f\|)$. For $\xi\in\ell^2(\mathbb F_2)$, apply this to $f=\xi-\langle\xi,\delta_1\rangle\delta_1$, using $U_{\lambda(c)}f-f=U_{\lambda(c)}\xi-\xi$ and (2.1): the unitaries $\lambda(a),\lambda(b)$ have a spectral gap with $K=14$. By Proposition 3.2 with $n=2$, the operator $T=\tfrac14\big(\lambda(a)\rho(a)+\lambda(a)^*\rho(a)^*+\lambda(b)\rho(b)+\lambda(b)^*\rho(b)^*\big)$ has spectrum in $[-1,1-\frac1{784}]\cup\{1\}$, and the projection onto $\mathbb C\delta_1$ is a continuous function of $T$. So $C^*(N,N')\supset\mathcal K(\ell^2(\mathbb F_2))$, and by Theorem 2.5, $N$ does not have property Γ. That $C^*(N,N')$ contains compact operators for this $N$ was first observed in [Akemann–Ostrand 1975].

4. From almost commuting vectors to invariant states

This section and the next follow Connes's proof of [Connes 1976, Theorem 3.9.1]. In this section $N$ is a von Neumann algebra with a faithful normal tracial state $\tau$ unless it is said to be a factor. We fix a free ultrafilter $\omega$ on $\mathbb N$. For a bounded sequence of states $\psi_m$, the formula $\varphi(x)=\lim_{m\to\omega}\psi_m(x)$ defines a state, the weak* limit of $(\psi_m)$ along $\omega$.

The following lemma is how non-normal invariant states are produced.

Lemma 4.1. Let $u_1,\dots,u_n\in\mathcal U(N)$, $c>0$, and for each $m\in\mathbb N$ let $\psi_m$ be a state and $e_m$ a projection of $N$ such that $\sum_m\tau(e_m)<\infty,\qquad\psi_m(e_m)\ge c,\qquad\lim_{m\to\infty}\|\psi_m\circ\operatorname{Ad}u_j-\psi_m\|=0\quad (j=1,\dots,n).$ Then the weak* limit $\varphi$ of $(\psi_m)$ along $\omega$ is not normal and satisfies $\varphi\circ\operatorname{Ad}u_j=\varphi$ for all $j$.

Proof. For $x\in N$, $|\varphi(u_jxu_j^*)-\varphi(x)|=\lim_\omega|\psi_m(u_jxu_j^*)-\psi_m(x)|\le\|x\|\lim_\omega\|\psi_m\circ \operatorname{Ad}u_j-\psi_m\|=0$. For $M\in\mathbb N$ let $f_M=\bigvee_{m\ge M}e_m$. Kaplansky's formula (B6) holds in any von Neumann algebra and gives $\tau(p\vee q)\le\tau(p)+\tau(q)$ for a tracial state; by induction and normality of $\tau$, $\tau(f_M)\le\sum_{m\ge M}\tau(e_m)$, which tends to $0$. The projections $f_M$ decrease to a projection $f_\infty$ with $\tau(f_\infty)=0$, so $f_\infty=0$. For $m\ge M$, $\psi_m(f_M)\ge\psi_m(e_m)\ge c$; since $\{m:m\ge M\}\in\omega$, $\varphi(f_M)\ge c$ for every $M$. If $\varphi$ were normal, $\varphi(f_M)$ would tend to $\varphi(f_\infty)=0$. $\square$

The next lemma says that a non-normal invariant state can be replaced by an invariant state that lives on projections of arbitrarily small trace.

Lemma 4.2. Let $\mathcal G$ be a group of automorphisms of $N$ and $\varphi$ a $\mathcal G$-invariant state that is not normal. Then there is a $\mathcal G$-invariant singular state $\sigma$ on $N$, and for every such $\sigma$ and every $\varepsilon>0$ there is a projection $f\in N$ with $\tau(f)\le\varepsilon$ and $\sigma(f)=1$.

Proof. Write $\varphi=\varphi_n+\varphi_s$ as in (B3); $\varphi_s\ne0$ because $\varphi$ is not normal. For $\alpha\in\mathcal G$, $\varphi=\varphi\circ\alpha=\varphi_n\circ\alpha+\varphi_s\circ\alpha$. Here $\varphi_n\circ\alpha$ is normal (automorphisms of von Neumann algebras are normal), and $\varphi_s\circ\alpha$ is singular: if $0\le\psi\le\varphi_s\circ\alpha$ with $\psi$ normal, then $0\le\psi\circ\alpha^{-1}\le\varphi_s$, so $\psi=0$. By uniqueness, $\varphi_s\circ\alpha=\varphi_s$. So $\sigma=\varphi_s/\varphi_s(1)$ is a $\mathcal G$-invariant singular state.

Now let $\sigma$ be any singular state. By Zorn's lemma choose a maximal family $(p_i)_{i\in I}$ of mutually orthogonal nonzero projections with $\sigma(p_i)=0$. If $1-\sum_ip_i\ne0$, it majorizes a nonzero projection $q$ with $\sigma(q)=0$ (B3), contradicting maximality; so $\sum_ip_i=1$. By normality $\sum_i\tau(p_i)=1$, so for $\varepsilon>0$ there is a finite $F\subset I$ with $\sum_{i\in F}\tau(p_i)\ge1-\varepsilon$. The projection $f=1-\sum_{i\in F}p_i$ has $\tau(f)\le\varepsilon$ and $\sigma(f)=1-\sum_{i\in F}\sigma(p_i)=1$. $\square$

The third lemma identifies the vectors of $L^2$ fixed by a set of inner automorphisms.

Lemma 4.3. Let $P$ be a von Neumann algebra with a faithful normal tracial state $\tau$, $S\subset\mathcal U(P)$ a set of unitaries, and $Q=S'\cap P$ its relative commutant. On $L^2(P,\tau)$ let $U_u=uJuJ$ for $u\in S$. Then the set of vectors $\eta$ with $U_u\eta=\eta$ for all $u\in S$ is the closure of $\hat Q=\{\hat y:y\in Q\}$. In particular, if $Q=\mathbb C$, the only such vectors are the multiples of $\hat1$.

Proof. If $y\in Q$ then $U_u\hat y=(uyu^*)^\wedge=\hat y$, and the fixed vectors form a closed subspace; so it contains the closure of $\hat Q$. Let $G$ be the group generated by $S$. For $g\in G$, $U_g=gJgJ$ is a unitary, $U_gU_{g'}=U_{gg'}$ (because $JgJ\in P'$), and a vector fixed by all $U_u$, $u\in S$, is fixed by all $U_g$.

Fix $x\in P$ and let $C_x$ be the norm closure in $L^2(P)$ of the convex hull of $\{(gxg^*)^\wedge:g\in G\}$. The convex hull consists of vectors $\hat y$ with $y\in P$, $\|y\|\le\|x\|$. A norm limit of such vectors is again of this form: the corresponding elements form a $\|\cdot\|_2$-Cauchy sequence in the ball of radius $\|x\|$, which is complete for $\|\cdot\|_2$ by Lemma 1.4 of "Ultraproducts and the asymptotic centralizer" (for a trace, $\|y\|^\sharp_\tau=\sqrt2\|y\|_2$). So $C_x\subset\{\hat y:y\in P,\ \|y\|\le\|x\|\}$. Let $\hat y_0$ be the unique element of minimal norm in the closed convex set $C_x$. Each $U_g$ is a linear isometry mapping $C_x$ onto itself, so $U_g\hat y_0=\hat y_0$, that is $gy_0g^*=y_0$ for all $g\in G$ (as $\tau$ is faithful). Hence $y_0\in Q$. If $\eta$ is fixed by all $U_g$, then $\langle U_g\hat x,\eta\rangle=\langle\hat x,U_g^*\eta\rangle= \langle\hat x,\eta\rangle$, so by linearity and continuity $\langle z,\eta\rangle=\langle\hat x,\eta\rangle$ for all $z\in C_x$, and in particular $\langle\hat y_0,\eta\rangle=\langle\hat x,\eta\rangle$.

Now let $\eta$ be fixed and write $\eta=\eta_1+\eta_2$ with $\eta_1$ in the closure of $\hat Q$ and $\eta_2\perp\hat Q$. Then $\eta_1$ is fixed, hence so is $\eta_2$, and for every $x\in P$, $\langle\hat x,\eta_2\rangle=\langle\hat y_0,\eta_2\rangle=0$. As $\hat P$ is dense, $\eta_2=0$. $\square$

Lemma 4.4. Let $Q$ be a von Neumann algebra with a faithful normal tracial state $\tau$. If $Q$ is infinite-dimensional, then for every $\delta>0$ it contains a nonzero projection of trace at most $\delta$.

Proof. Suppose every nonzero projection of $Q$ has trace at least $\delta>0$; we show $\dim Q\le\delta^{-2}$.

Minimal projections. Let $p\ne0$ be a projection and $t$ the infimum of the traces of the nonzero projections $q\le p$. Choose such a $q$ with $\tau(q)<t+\delta$. If $0\ne q'\le q$ with $q'\ne q$, then $q-q'\ne0$ has trace at least $\delta$, so $\tau(q')\le\tau(q)-\delta<t$, which is impossible. So $q$ is a minimal projection of $Q$, and every nonzero projection majorizes one.

A finite partition of unity. Mutually orthogonal nonzero projections number at most $1/\delta$. Take a maximal family $p_1,\dots,p_r$ of mutually orthogonal minimal projections. Then $\sum_ip_i=1$, otherwise $1-\sum_ip_i$ would majorize a further minimal projection.

Corners. Let $y\in p_iQp_i$ be self-adjoint. The von Neumann algebra generated by $y$ in $p_iQp_i$ (with unit $p_i$) is abelian, and its projections are subprojections of $p_i$ in $Q$, hence $0$ or $p_i$. An abelian von Neumann algebra is the norm closed span of its projections, so $y\in\mathbb Cp_i$. Thus $p_iQp_i=\mathbb Cp_i$. Next let $x,x'\in p_iQp_l$ with $x\ne0$. Then $x^*x'\in p_lQp_l$, say $x^*x'=d\,p_l$, and $xx^*\in p_iQp_i$, so $xx^*=\|x\|^2p_i$. Hence $\|x\|^2x'=xx^*x'=d\,xp_l=d\,x$, and $x'\in\mathbb Cx$. So $\dim p_iQp_l\le1$, and $\dim Q=\dim\sum_{i,l}p_iQp_l\le r^2\le\delta^{-2}$. $\square$

The next lemma lets us enlarge a finite set of unitaries so that its relative commutant in the ultrapower becomes trivial, unless that relative commutant is infinite-dimensional. We use the ultrapower $N^\omega$ of Definition 3.4 of "Ultraproducts and the asymptotic centralizer": bounded sequences in $N$ modulo those tending to $0$ in $\|\cdot\|_2$ along $\omega$, with the trace $\tau_\omega(x)=\lim_\omega\tau(x_k)$, and with $N\subset N^\omega$ as the constant sequences. The inclusion preserves the trace, hence $\|\cdot\|_2$.

Lemma 4.5. Let $N$ be a II₁ factor and $u_1,\dots,u_n\in\mathcal U(N)$. Suppose the relative commutant $Q=\{u_1,\dots,u_n\}'\cap N^\omega$ is finite-dimensional. Then $Q\subset N$, and there are unitaries $u_{n+1},\dots,u_q\in N$ such that $\{u_1,\dots,u_q\}'\cap N^\omega=\mathbb C$.

Reference: a lemma in Connes's proof of [Connes 1976, Theorem 3.9.1].

Suppose some $x\in Q$ is not in $N$; scaling, $\|x\|<1$. Then $d=\inf\{\|x-y\|_2:y\in B\}>0$. By the definition of the quotient norm, $x$ has a representing sequence $(x_m)$ with $\|x_m\|\le1$. For every $y\in B$, $\lim_{m\to\omega}\|x_m-y\|_2=\|x-y\|_2\ge d$, and $\lim_{m\to\omega}\|[u_j,x_m]\|_2=\|[u_j,x]\|_2=0$.

We construct, by induction on $p\ge1$, sequences $(y^{(p)}_k)_{k\in\mathbb N}$ in $B$ such that $\|[u_j,y^{(p)}_k]\|_2<1/k$ for all $j,k$, and $\|y^{(p)}_k-y^{(p')}_k\|_2>d/2$ for all $k$ and $p'<p$. Given the sequences for $p'<p$ and given $k$, the set of $m$ with $\max_j\|[u_j,x_m]\|_2<1/k$ and $\|x_m-y^{(p')}_k\|_2>d/2$ for all $p'<p$ is a finite intersection of sets in $\omega$, hence nonempty; let $y^{(p)}_k=x_m$ for such an $m$.

Let $y^{(p)}\in N^\omega$ be the class of $(y^{(p)}_k)_k$. Then $\|y^{(p)}\|\le1$, $\|[u_j,y^{(p)}]\|_2=\lim_\omega\|[u_j,y^{(p)}_k]\|_2=0$, so $y^{(p)}\in Q$, and $\|y^{(p)}-y^{(p')}\|_2=\lim_\omega\|y^{(p)}_k-y^{(p')}_k\|_2\ge d/2$ for $p'<p$. So the unit ball of $Q$ contains an infinite $d/2$-separated set for the norm $\|\cdot\|_2$. But $Q$ is finite-dimensional, all norms on it are equivalent, and its unit ball is compact. This contradiction proves $Q\subset N$.

Step 2: killing $Q$. Put $Q_0=Q$. If $Q_i\ne\mathbb C$, pick $x\in Q_i\setminus\mathbb C$. Since $N$ is a factor and $x\in N$, $x$ does not commute with $N$, and since $N$ is spanned by its unitaries, there is $w_{i+1}\in\mathcal U(N)$ with $w_{i+1}x\ne xw_{i+1}$. Put $Q_{i+1}=Q_i\cap\{w_{i+1}\}'$, a proper subspace of $Q_i$. As $\dim Q<\infty$, after finitely many steps $Q_r=\mathbb C$. Then $\{u_1,\dots,u_n,w_1,\dots,w_r\}'\cap N^\omega=Q\cap\{w_1,\dots,w_r\}'=Q_r=\mathbb C$. $\square$

Proof of (c)⇒(b). Assume (c) and let $u_1,\dots,u_n\in\mathcal U(N)$. Let $Q=\{u_1,\dots,u_n\}'\cap N^\omega$, a von Neumann subalgebra of $N^\omega$ on which $\tau_\omega$ is a faithful normal tracial state.

Case 1: $Q$ is infinite-dimensional. Let $m\in\mathbb N$. By Lemma 4.4 there is a nonzero projection $g\in Q$ with $t=\tau_\omega(g)\le4^{-m}$. By the first step of the proof of Theorem 3.3 of "Ultraproducts and the asymptotic centralizer", $g$ is the class of a sequence $(f_k)$ of projections of $N$. Along $\omega$, $\tau(f_k)\to t$ and $\|[u_j,f_k]\|_2\to\|[u_j,g]\|_2=0$. So there is $k$ with $t/2\le\tau(f_k)\le2\cdot4^{-m}$ and $\|[u_j,f_k]\|_2\le\frac1m(t/2)^{1/2}\le\frac1m\|f_k\|_2$ for all $j$; put $e_m=f_k\ne0$ and $\psi_m=\tau(e_m\,\cdot\,)/\tau(e_m)$. Then $\psi_m(e_m)=1$, $\sum_m\tau(e_m)<\infty$, and by Lemma 1.5(iv), $\|\psi_m\circ\operatorname{Ad}u_j-\psi_m\|\le\sqrt2/m$. Lemma 4.1 gives a non-normal state invariant under $\operatorname{Ad}u_1,\dots,\operatorname{Ad}u_n$.

Case 2: $Q$ is finite-dimensional. By Lemma 4.5 there are $u_{n+1},\dots,u_q\in\mathcal U(N)$ with $\{u_1,\dots,u_q\}'\cap N^\omega=\mathbb C$. By (c) and Lemma 2.4, there are unit vectors $\xi_k\in H$ with $\langle\xi_k,\hat1\rangle=0$ and $\|[u_j,\xi_k]\|\to0$ for $j=1,\dots,q$.

We claim that $(\xi_k)$ does not satisfy the uniform integrability condition (4.1) of "Ultraproducts and the asymptotic centralizer". Suppose it does. By Theorem 4.2 of that lesson, the class $\xi$ of $(\xi_k)$ in the ultraproduct $H_\omega$ of Hilbert spaces lies in the subspace $\mathcal H$ on which $N^\omega$ acts in standard form, with trace vector $\mathbf1=(\hat1)_k$ and modular conjugation the restriction of $(\eta_k)\mapsto(J\eta_k)$. For $u\in N\subset N^\omega$, $u$ acts on $\mathcal H$ by $(\eta_k)\mapsto(u\eta_k)$, so $uJuJ\xi$ is the class of $(u\xi_ku^*)$. Hence for $j\le q$, $\|u_jJu_jJ\xi-\xi\|=\lim_\omega\|u_j\xi_ku_j^*-\xi_k\|=\lim_\omega\|[u_j,\xi_k]\|=0$ by (2.1). Also $\|\xi\|=1$ and $\langle\xi,\mathbf1\rangle=\lim_\omega\langle\xi_k,\hat1\rangle=0$. Transporting to $L^2(N^\omega,\tau_\omega)$ by the unitary equivalence of Theorem 4.2(iii) of that lesson, we get a unit vector orthogonal to $\hat1$ that is fixed by $U_{u_j}$ for $j\le q$. This contradicts Lemma 4.3, since $\{u_1,\dots,u_q\}'\cap N^\omega=\mathbb C$.

So (4.1) fails: there is $\varepsilon>0$ such that $\lim_\omega\|E_a(|\xi_k|)|\xi_k|\|_2\ge\varepsilon$ for every $a>0$. For $m\in\mathbb N$ the set of $k$ with $\|E_{2^m}(|\xi_k|)|\xi_k|\|_2>\varepsilon/2$ and $\max_{j\le q}\|[u_j,\xi_k]\|<1/m$ belongs to $\omega$; pick such a $k=k_m$, and put $\zeta_m=\xi_{k_m}$, $e_m=E_{2^m}(|\zeta_m|)$, and $\psi_m(x)=\langle\zeta_mx,\zeta_m\rangle=\langle Jx^*J\zeta_m,\zeta_m\rangle\qquad(x\in N),$ a normal state on $N$. We check the hypotheses of Lemma 4.1.

By (B2)(c), $\tau(e_m)\le4^{-m}\|\zeta_m\|^2=4^{-m}$.
Since $Je_mJ$ is a projection, $\psi_m(e_m)=\|\zeta_me_m\|^2=\||\zeta_m|e_m\|_2^2>\varepsilon^2/4$, by (B2)(b).
For $u\in\mathcal U(N)$ and $x\in N$, $\psi_m(u^*xu)=\langle(\zeta_mu^*)x,\zeta_mu^*\rangle$, because right multiplication by $u^*xu$ is right multiplication by $u^*$, then $x$, then $u$. Also $\psi_m(x)=\langle(u^*\zeta_m)x,u^*\zeta_m\rangle$, because left multiplication by $u^*$ is a unitary commuting with right multiplications. As $\|\zeta_mu^*-u^*\zeta_m\|=\|u^*(u\zeta_m-\zeta_mu)u^*\|=\|[u,\zeta_m]\|$, we get $|\psi_m(u^*xu)-\psi_m(x)|\le2\|x\|\|[u,\zeta_m]\|$. So $\|\psi_m\circ\operatorname{Ad}u_j^*-\psi_m\|\le2/m$ for $j\le q$, and since composition with $\operatorname{Ad}u_j$ is isometric, $\|\psi_m\circ\operatorname{Ad}u_j-\psi_m\|=\|\psi_m-\psi_m\circ\operatorname{Ad}u_j^*\|\le2/m$.

Lemma 4.1 gives a non-normal state invariant under $\operatorname{Ad}u_1,\dots,\operatorname{Ad}u_q$, in particular under $\operatorname{Ad}u_1,\dots,\operatorname{Ad}u_n$. $\square$

Remark 4.6. The two cases correspond to two ways in which almost commuting unit vectors can avoid $\hat1$. In the hyperfinite factor, with $e_k$ a central sequence of projections with $\tau(e_k)\to0$ (Exercise 4), the vectors $\hat e_k/\|e_k\|_2$ have their mass escaping to large values, as in Case 2. A central sequence of projections of trace $\frac12$ gives almost commuting vectors $(2e_k-1)^\wedge$ of norm $1$ that are uniformly bounded; these produce a nontrivial relative commutant in the ultrapower, as in Case 1.

5. From invariant states to property Γ

This section completes Connes's proof of [Connes 1976, Theorem 3.9.1]. Throughout, $N$ is a II₁ factor satisfying condition (b) of Theorem 2.5.

Lemma 5.1. Let $N$ satisfy (b), let $u_1,\dots,u_n\in\mathcal U(N)$ and $\varepsilon\in\,]0,1[$. There is a nonzero projection $e\in N$ with $\tau(e)\le\varepsilon$ and $\|[u_j,e]\|_2\le\varepsilon\|e\|_2$ for all $j$.

Reference: a lemma in Connes's proof of [Connes 1976, Theorem 3.9.1].

Proof. Let $\eta=\min\big(1,(\varepsilon^2/(6n))^8\big)$.

Step 1: an invariant state on a small projection. By (b) and Lemma 4.2 (with $\mathcal G$ the group generated by the $\operatorname{Ad}u_j$) there are an invariant singular state $\sigma$ and a projection $f$ with $\tau(f)\le\varepsilon$ and $\sigma(f)=1$.

Step 2: an almost invariant normal state on the same projection. The normal states are weak* dense in the state space: otherwise, by the Hahn–Banach separation theorem in $N^*$ with its weak* topology, there would be a state $\sigma'$ and a self-adjoint $y\in N$ with $\sigma'(y)>\sup\{\psi(y):\psi\text{ a normal state}\}$; but the supremum over the vector states $\langle y\xi,\xi\rangle$, $\|\xi\|=1$, is $\max\operatorname{Sp}y\ge\sigma'(y)$. So some net of normal states $\psi_i$ converges weak* to $\sigma$. Then $\psi_i(f)\to1$, and $\psi_i\circ\operatorname{Ad}u_j\to\sigma\circ\operatorname{Ad}u_j=\sigma$ weak*. Let $V$ be the convex set of normal states $\psi$ with $\psi(f)\ge1-\eta$, and $W=\{(\psi-\psi\circ\operatorname{Ad}u_1,\dots,\psi-\psi\circ\operatorname{Ad}u_n):\psi\in V\}\subset(N_*)^n ,$ a convex set. Eventually $\psi_i\in V$, and the corresponding elements of $W$ tend to $0$ in the topology of pointwise convergence on $N^n$. Since the dual of the Banach space $(N_*)^n$ is $N^n$, this is the weak topology of $(N_*)^n$, and $0$ lies in the weak closure of $W$. For convex sets the weak and norm closures agree (Hahn–Banach), so there is $\psi\in V$ with $\|\psi\circ\operatorname{Ad}u_j-\psi\|\le\eta$ for all $j$.

Step 3: cutting down to $f$. Put $\psi'(x)=\psi(fxf)/\psi(f)$, a normal state. By the Cauchy–Schwarz inequality, $|\psi((1-f)x)|\le\psi(1-f)^{1/2}\psi(x^*x)^{1/2}\le\eta^{1/2}\|x\|$ and $|\psi(fx(1-f))|\le\psi(fxx^*f)^{1/2}\psi(1-f)^{1/2}\le\eta^{1/2}\|x\|$; since $\psi(x)-\psi(fxf)=\psi((1-f)x)+\psi(fx(1-f))$, we get $\|\psi-\psi(f\,\cdot\,f)\|\le2\eta^{1/2}$. Also $\|\psi(f\,\cdot\,f)-\psi'\|=(\psi(f)^{-1}-1)\psi(f)=1-\psi(f)\le\eta$. So $\|\psi-\psi'\|\le3\eta^{1/2}$, and $\|\psi'\circ\operatorname{Ad}u_j-\psi'\|\le2\|\psi'-\psi\|+\|\psi\circ\operatorname{Ad}u_j-\psi\|\le7\eta^{1/2}.$

Step 4: from the state to a vector. By (B2)(f), $\psi'=\omega_h$ for a unique $h\in H_+$; $\|h\|_2=1$, and $\|(1-f)h\|^2=\psi'(1-f)=0$, so $fh=h$. For a unitary $u$, put $uhu^*=uJuJh\in H_+$ (B2)(d). Then $\psi'(u^*xu)=\langle xuh,uh\rangle=\langle x\,uhu^*,uhu^*\rangle$, because $uh=Ju^*J(uhu^*)$ and $Ju^*J$ is a unitary in $N'$. So $\psi'\circ\operatorname{Ad}u^*=\omega_{uhu^*}$, and by Theorem 1.1, $\|u_jhu_j^*-h\|_2^2\le\|\psi'\circ\operatorname{Ad}u_j^*-\psi'\|=\|\psi'-\psi'\circ\operatorname{Ad}u_j\|\le 7\eta^{1/2}.$

Step 5: a spectral projection. Put $h_j=u_jhu_j^*$. Then $\sum_j\|h_j-h\|_2\|h_j+h\|_2\le n\cdot\sqrt7\,\eta^{1/4}\cdot2<6n\eta^{1/4}$. By Lemma 1.4 with $c=6n\eta^{1/4}$ there is $s>0$ such that $e=E_{\sqrt s}(h)\ne0$ and $\sum_j\|E_{\sqrt s}(h_j)-e\|_2^2<6n\eta^{1/4}\|e\|_2^2$. By (B2)(d), $E_{\sqrt s}(h_j)=u_jeu_j^*$, and $\|u_jeu_j^*-e\|_2=\|[u_j,e]\|_2$. Hence $\|[u_j,e]\|_2\le(6n)^{1/2}\eta^{1/8}\|e\|_2\le(6n)^{1/2}\frac{\varepsilon^2}{6n}\|e\|_2\le\varepsilon\|e\|_2 .$ Finally $e\le f$ by (B2)(e), so $\tau(e)\le\varepsilon$. $\square$

Lemma 5.2. If $N$ satisfies (b) and $f\in N$ is a nonzero projection, then the factor $fNf$ of type II₁ satisfies (b).

Reference: a lemma in Connes's proof of [Connes 1976, Theorem 3.9.1].

Proof. Let $u_1,\dots,u_n$ be unitaries of $fNf$ and $\bar u_j=u_j+(1-f)\in\mathcal U(N)$. Choose an integer $q\ge1/\tau(f)$. By (B6) there are mutually orthogonal projections $e_{11},\dots,e_{qq}$ of trace $1/q$ with sum $1$ and $e_{11}\le f$, and partial isometries $w_a$ with $w_a^*w_a=e_{11}$, $w_aw_a^*=e_{aa}$. The elements $e_{ab}=w_aw_b^*$ form a system of matrix units spanning a subalgebra $K\cong M_q(\mathbb C)$. Let $\zeta=e^{2\pi i/q}$, $D=\sum_a\zeta^ae_{aa}$ and $S=\sum_ae_{a+1,a}$ (indices modulo $q$): unitaries that generate $K$ as an algebra, since the $e_{aa}$ are polynomials in $D$ and $e_{a+1,a}=Se_{aa}$.

By (b) there is a non-normal state invariant under $\operatorname{Ad}\bar u_j$, $\operatorname{Ad}D$ and $\operatorname{Ad}S$, and by Lemma 4.2 a singular state $\sigma$ with the same invariance. On $K\cong M_q$, $\sigma$ has the form $\operatorname{Tr}(\rho\,\cdot\,)$ for a density matrix $\rho\in K$. Invariance under $\operatorname{Ad}D$ and $\operatorname{Ad}S$ gives $D^*\rho D=\rho$ and $S^*\rho S=\rho$, so $\rho$ commutes with $K$, hence is scalar: $\sigma(e_{11})=1/q$. Therefore $\sigma(f)\ge1/q>0$, and $\sigma_f(x)=\sigma(x)/\sigma(f)$, $x\in fNf$, is a state on $fNf$. For $x\in fNf$, $\bar u_jx\bar u_j^*=u_jxu_j^*$, so $\sigma_f$ is invariant under $\operatorname{Ad}u_j$. Every nonzero projection $p$ of $fNf$ is a projection of $N$, so by (B3) it majorizes a nonzero projection $r$ with $\sigma(r)=0$; $r\le p\le f$ lies in $fNf$. By (B3) applied in $fNf$, $\sigma_f$ is singular; being nonzero, it is not normal. $\square$

Proof of (b)⇒(a). Let $u_1,\dots,u_n\in\mathcal U(N)$ and $\delta\in\,]0,1]$. By Lemma 2.2 it suffices to find a projection $E$ with $\tau(E)=\frac12$ and $\|[u_j,E]\|_2\le2\delta^{1/2}$ for all $j$.

The ordered set. Let $\mathcal R$ be the set of tuples $r=(E,U_1,\dots,U_n)$ where $E\in N$ is a projection with $\tau(E)\le\frac12$, each $U_j\in\mathcal U(N)$ commutes with $E$, and $\|U_j-u_j\|_1\le\delta\tau(E)$ for all $j$. It contains $(0,u_1,\dots,u_n)$. Write $r\le r'$ if $E\le E'$ and $\|U_j'-U_j\|_1\le\delta\tau(E'-E)$ for all $j$. This is a partial order: transitivity follows from the triangle inequality and $\tau(E''-E')+\tau(E'-E)=\tau(E''-E)$; if $r\le r'\le r$, then $E=E'$ and hence $U_j=U_j'$. Moreover, if $r\le r'$ and $\tau(E)=\tau(E')$, then $E=E'$ by faithfulness, so $r=r'$.

Chains have upper bounds. Let $\mathcal C\subset\mathcal R$ be a nonempty chain and $t=\sup_{r\in\mathcal C}\tau(E_r)$. By the last remark, $r\mapsto\tau(E_r)$ is strictly increasing on $\mathcal C$. If $t$ is attained, the element attaining it is the largest element of $\mathcal C$. Otherwise choose $r_1\le r_2\le\cdots$ in $\mathcal C$ with $\tau(E_{r_k})\uparrow t$; every $r\in\mathcal C$ has $\tau(E_r)<\tau(E_{r_k})$ for some $k$, hence $r\le r_k$. So it suffices to bound an increasing sequence $r_k=(E_k,U_{k,1},\dots,U_{k,n})$.

The projections $E_k$ increase strongly to a projection $E$ with $\tau(E)=\lim\tau(E_k)\le\frac12$, so $\|E-E_k\|_2^2=\tau(E-E_k)\to0$. For $k\le m$, $\|U_{m,j}-U_{k,j}\|_1\le\delta\tau(E_m-E_k)$, and by Lemma 1.5(iii), $\|U_{m,j}-U_{k,j}\|_2^2\le2\delta\tau(E_m-E_k)$. So $(U_{k,j})_k$ is $\|\cdot\|_2$-Cauchy in the unit ball and converges in $\|\cdot\|_2$ to some $U_j$ in the ball (Lemma 1.4 of "Ultraproducts and the asymptotic centralizer"); $U_{k,j}^*\to U_j^*$ as well since $\|x^*\|_2=\|x\|_2$. By (B1), $\|U_j^*U_j-1\|_2\le\|U_j^*(U_j-U_{k,j})\|_2+\|(U_j^*-U_{k,j}^*)U_{k,j}\|_2\to0$, and likewise for $U_jU_j^*$; so $U_j$ is unitary. Similarly $\|[E,U_j]\|_2\le\|[E,U_j]-[E_k,U_{k,j}]\|_2\le2\|E-E_k\|_2+2\|U_j-U_{k,j}\|_2\to0$. Since $\|\cdot\|_1\le\|\cdot\|_2$, the $\|\cdot\|_1$-norms pass to the limit: $\|U_j-u_j\|_1=\lim_k\|U_{k,j}-u_j\|_1\le\delta\tau(E)$ and $\|U_j-U_{k,j}\|_1=\lim_m\|U_{m,j}-U_{k,j}\|_1\le\delta\tau(E-E_k)$. So $r=(E,U_1,\dots,U_n)\in\mathcal R$ and $r_k\le r$ for all $k$.

A maximal element has trace $\frac12$. By Zorn's lemma $\mathcal R$ has a maximal element $r=(E,U_1,\dots,U_n)$. Suppose $\tau(E)<\frac12$. Put $F=1-E$, $M=FNF$ with normalized trace $\tau'=\tau/\tau(F)$, and $v_j=U_jF=FU_j$, a unitary of $M$ (because $U_j$ commutes with $E$). Choose $\varepsilon>0$ with $\varepsilon\le\frac12-\tau(E)$ and $2\varepsilon\le\delta$. By Lemma 5.2, $M$ satisfies (b), and by Lemma 5.1 applied in $M$ there is a nonzero projection $e\in M$ with $\tau'(e)\le\varepsilon$ and $\|[v_j,e]\|_{2,\tau'}\le\varepsilon\|e\|_{2,\tau'}$. Both sides of the last inequality scale by $\tau(F)^{1/2}$ when $\tau'$ is replaced by $\tau$, so $\|[v_j,e]\|_2\le\varepsilon\|e\|_2$; and $\tau(e)=\tau(F)\tau'(e)\le\varepsilon$.

The projections $v_jev_j^*$ and $e$ are equivalent in $M$ (through the partial isometry $v_je$), and $M$ is finite. By Theorem 1.3 in $M$ (whose unit is $F$), with the faithful normal finite trace $\tau|_M$, there is a unitary $w_j\in M$ with $w_jv_jev_j^*w_j^*=e$ and $\|w_j-F\|_1\le\sqrt2\,\|v_jev_j^*-e\|_1$. The projections $v_jev_j^*$ and $e$ of $N$ have the same trace, and $\|v_jev_j^*-e\|_2=\|(v_je-ev_j)v_j^*\|_2=\|[v_j,e]\|_2$ (right multiplication by $v_j^*$ is isometric on $M$ for $\|\cdot\|_2$, as $v_j^*v_j=F$). By Lemma 1.5(ii) in $N$, $\|w_j-F\|_1\le\sqrt2\cdot\sqrt{2\tau(e)}\,\|[v_j,e]\|_2\le2\sqrt{\tau(e)}\,\varepsilon\|e\|_2=2\varepsilon\,\tau(e)\le \delta\tau(e).$ Put $E'=E+e$ and $U_j'=U_jE+w_jv_j$. Since $U_jE$ is a unitary of $ENE$ and $w_jv_j$ a unitary of $M$, $U_j'$ is unitary. The unitary $w_jv_j$ commutes with $e$ (it conjugates $e$ to itself), and $U_jE$ commutes with $E$ and is orthogonal to $e$; so $U_j'$ commutes with $E'$. Next $U_j=U_jE+v_j$, so $U_j'-U_j=(w_j-F)v_j$ and, by (B1), $\|U_j'-U_j\|_1\le\|w_j-F\|_1\le\delta\tau(e)=\delta\tau(E'-E)$. Hence $\|U_j'-u_j\|_1\le\delta\tau(E)+\delta\tau(e)=\delta\tau(E')$, and $\tau(E')\le\tau(E)+\varepsilon\le\frac12$. So $r'=(E',U_1',\dots,U_n')\in\mathcal R$, $r\le r'$ and $r'\ne r$ since $e\ne0$. This contradicts maximality. So $\tau(E)=\frac12$.

Conclusion. For the maximal $r$, $\|U_j-u_j\|_1\le\delta/2$, so by Lemma 1.5(iii), $\|U_j-u_j\|_2\le\delta^{1/2}$. As $[U_j,E]=0$, $\|[u_j,E]\|_2=\|[u_j-U_j,E]\|_2\le2\|u_j-U_j\|_2\le2\delta^{1/2}$. Since $\delta$ was arbitrary, Lemma 2.2 shows that $N$ has property Γ. $\square$

This completes the proof of Theorem 2.5. $\square$

6. Semidiscrete factors and tensor products

We first record what Theorem 2.5 says about factors without property Γ.

Corollary 6.1. For a factor $N$ of type II₁ on $H=L^2(N,\tau)$ the following are equivalent:

(i) $N$ does not have property Γ;
(ii) $C^*(N,N')\cap\mathcal K(H)\ne\{0\}$;
(iii) $\mathcal K(H)\subset C^*(N,N')$;
(iv) some finite set of unitaries $u_1,\dots,u_n\in N$ has a spectral gap;
(v) for some unitaries $u_1,\dots,u_n\in N$, the projection $P_0$ lies in the C*-algebra generated by $u_1,\dots,u_n,Ju_1J,\dots,Ju_nJ$.

If $N$ has separable predual, these are also equivalent to: (vi) $N$ is full.

Proof. (i)⇔(ii) and (i)⇔(iv) are Theorem 2.5 with Lemma 2.4. (iv)⇒(v) is Proposition 3.2, and (v)⇒(ii) is clear. (ii)⇔(iii) follows from (B4), since $C^*(N,N')$ acts irreducibly (proof of Proposition 3.1). (i)⇔(vi) is Theorem 6.5 of "Full factors". $\square$

6.1 Semidiscrete factors

Definition 6.2. A von Neumann algebra $N$ is semidiscrete if there are a net of integers $k(i)$ and nets of unital completely positive maps $\alpha_i\colon N\to M_{k(i)}(\mathbb C)$ and $\beta_i\colon M_{k(i)}(\mathbb C)\to N$ such that $\beta_i(\alpha_i(x))\to x$ $\sigma$-weakly for every $x\in N$.

Definition 6.2 does not ask the maps to be normal, and normality plays no role below. The form used in (B7), in which the identity is a pointwise $\sigma$-weak limit of normal completely positive maps of finite rank, is the hypothesis under which [Effros–Lance 1977] prove (B7). For that form we use (B7) instead of Proposition 6.4, and reach the same conclusion (Corollary 6.5).

Lemma 6.3. A factor $N$ of type II₁ is a simple C*-algebra: its only norm closed two-sided ideals are $\{0\}$ and $N$.

Proof. Let $I\ne\{0\}$ be a closed ideal and $0\ne x\in I$. Then $a=x^*x\in I$ is positive and nonzero. For $0<t<\|a\|$, the spectral projection $e=\chi_{]t,\infty[}(a)$ is nonzero, and $e=a\,g(a)$ with the bounded Borel function $g(s)=s^{-1}\chi_{]t,\infty[}(s)$; so $e\in I$. Choose an integer $m\ge1/\tau(e)$ and write $1$ as a sum of $m$ orthogonal projections $p_i$ of trace $1/m$ (B6). Each $p_i$ is equivalent to a subprojection of $e$: $p_i=v_i^*v_i$ with $v_iv_i^*\le e$, so $ev_i=v_i$ and $p_i=v_i^*ev_i\in I$. Hence $1\in I$. $\square$

Proposition 6.4. Let $N\subset B(H)$ be a semidiscrete von Neumann algebra. For $a_1,\dots,a_m\in N$ and $b_1,\dots,b_m\in N'$, $\Big\|\sum_ka_kb_k\Big\|\le\Big\|\sum_ka_k\otimes b_k\Big\|_{\min}.$ Consequently the product map $a\otimes b\mapsto ab$ extends to a $*$-homomorphism of $N\otimes_{\min}N'$ onto $C^*(N,N')$.

Proof. Put $x=\sum_ka_k\otimes b_k$ and let $\alpha\colon N\to M_d(\mathbb C)$, $\beta\colon M_d(\mathbb C)\to N$ be unital completely positive, with matrix units $\epsilon_{pq}$ of $M_d(\mathbb C)$. We first show $\Big\|\sum_k\beta(\alpha(a_k))\,b_k\Big\|\le\|x\|_{\min}.\tag{6.1}$ Write $\alpha(a_k)=\sum_{p,q}\alpha(a_k)_{pq}\epsilon_{pq}$ with scalars $\alpha(a_k)_{pq}$, and put $y_{pq}=\sum_k\alpha(a_k)_{pq}b_k\in N'$. The matrix $Y=[y_{pq}]\in M_d(N')\subset B(\mathbb C^d\otimes H)$ is $(\alpha\otimes\mathrm{id})(x)$, so $\|Y\|\le\|x\|_{\min}$ by (B5)(i). The matrix $[\epsilon_{pq}]_{p,q}\in M_d(M_d(\mathbb C))$ is positive (it is $d$ times a rank-one projection), so, $\beta$ being completely positive, $[\beta(\epsilon_{pq})]_{p,q}$ is a positive element of $M_d(N)$. Write it as $X^*X$ with $X=[x_{rp}]\in M_d(N)$: $\beta(\epsilon_{pq})=\sum_rx_{rp}^*x_{rq}$. Since the $x_{rp}\in N$ commute with the $y_{pq}\in N'$, $\sum_k\beta(\alpha(a_k))b_k=\sum_{p,q}\beta(\epsilon_{pq})y_{pq}=\sum_r\sum_{p,q}x_{rp}^*y_{pq}x_{rq}=Z^*(Y\oplus\cdots \oplus Y)Z,$ where $Z\colon H\to(\mathbb C^d\otimes H)^{\oplus d}$ has components $Z_r\xi=(x_{r1}\xi,\dots,x_{rd}\xi)$. Now $\|Z\|^2=\|\sum_rZ_r^*Z_r\|=\|\sum_{r,p}x_{rp}^*x_{rp}\|=\|\sum_p\beta(\epsilon_{pp})\|=\|\beta(1)\|=1$, which gives (6.1).

Now take the nets $\alpha_i,\beta_i$ of Definition 6.2. For unit vectors $\xi,\xi'\in H$, $\langle\sum_k\beta_i(\alpha_i(a_k))b_k\xi,\xi'\rangle\to\langle\sum_ka_kb_k\xi,\xi'\rangle$, since $\sigma$-weak convergence implies weak operator convergence. By (6.1), $|\langle\sum_ka_kb_k\xi,\xi'\rangle|\le\|x\|_{\min}$, which is the inequality. Since $N$ and $N'$ commute, the product map is a $*$-homomorphism on $N\odot N'$; by the inequality it extends to $N\otimes_{\min}N'$. Its range is closed (B5)(ii) and contains the dense subalgebra spanned by the products $ab$, so it is $C^*(N,N')$. $\square$

Corollary 6.5. A semidiscrete factor of type II₁, in the sense of Definition 6.2 or in the finite-rank sense of (B7), has property Γ. Moreover, on $L^2(N,\tau)$ the product map is an isomorphism $N\otimes_{\min}N'\cong C^*(N,N')$, and $C^*(N,N')$ is simple.

Reference: [Connes 1976, §4.2].

Proof. Let $N$ act on $H=L^2(N,\tau)$. By Proposition 6.4 (or by (B7) for the finite-rank sense, which gives the same inequality, hence the same $*$-homomorphism) there is a $*$-homomorphism $\Pi$ of $N\otimes_{\min}N'$ onto $C^*(N,N')$. The factors $N$ and $N'$ are of type II₁ (B6), hence simple (Lemma 6.3), and $N\otimes_{\min}N'$ is simple by Takesaki's theorem (B5)(iii). As $\Pi\ne0$, its kernel is $\{0\}$, and $C^*(N,N')\cong N\otimes_{\min}N'$ is simple. The closed ideal $C^*(N,N')\cap\mathcal K(H)$ is therefore $\{0\}$ or all of $C^*(N,N')$. The second is impossible: $1$ is not compact, since $H$ is infinite-dimensional. So (d) of Theorem 2.5 holds, and $N$ has property Γ. $\square$

Example 6.6 (the hyperfinite factor). Let $R=\bar\bigotimes_{n\ge1}(M_2(\mathbb C),\operatorname{tr})$ be the hyperfinite II₁ factor, and let $A_n\cong M_{2^n}(\mathbb C)$ be the tensor product of the first $n$ factors. Then $R=A_n\bar\otimes R_n$, with $R_n$ the tensor product of the remaining factors, and the slice map $E_n=\mathrm{id}\otimes\tau_{R_n}\colon R\to A_n$ is unital, completely positive and trace preserving. By the Kadison–Schwarz inequality $E_n(z)^*E_n(z)\le E_n(z^*z)$, so $\|E_n(z)\|_2\le\|z\|_2$. If $y$ lies in the union $\bigcup_mA_m$, then $E_n(y)=y$ for $n\ge m$; since this union is $\|\cdot\|_2$-dense in $R$, for every $x\in R$ we get $\|E_n(x)-x\|_2\le\|E_n(x-y)\|_2+\|y-x\|_2\le2\|x-y\|_2$ for large $n$, so $E_n(x)\to x$ in $\|\cdot\|_2$. On bounded sets this implies strong, hence $\sigma$-weak, convergence on $L^2(R)$ (as in the proof of (a)⇒(d), $\|z\hat y\|\le\|z\|_2\|y\|$). With $\alpha_n=E_n$ and $\beta_n$ the inclusion $A_n\subset R$, $R$ is semidiscrete. By Corollary 6.5, $R$ has property Γ and $C^*(R,R')\cong R\otimes_{\min}R'$ is simple. By Example 3.3 and Corollary 6.5, the group von Neumann algebra of $\mathbb F_2$ is not semidiscrete.

6.2 Tensor products

Corollary 6.7. Let $N_1$ and $N_2$ be II₁ factors. Then $N_1\bar\otimes N_2$ has property Γ exactly when at least one of $N_1$, $N_2$ has it. When $N_1$ and $N_2$ have separable preduals, $N_1\bar\otimes N_2$ is full exactly when both $N_1$ and $N_2$ are full.

Reference: [Connes 1976, §3.9], the corollary stated after Theorem 3.9.1; see also [Anantharaman–Popa, §15.3].

Proof. Let $N=N_1\bar\otimes N_2$, a factor of type II₁ with trace $\tau=\tau_1\otimes\tau_2$ (B6), acting on $H_1\otimes H_2$, $H_i=L^2(N_i,\tau_i)$. The vector $\hat1\otimes\hat1$ is cyclic for $N$ (the vectors $\hat a\otimes\hat b$ span a dense subspace) and defines the trace, so $H_1\otimes H_2$ is the standard space $L^2(N,\tau)$, with $\hat x$ corresponding to $x(\hat1\otimes\hat1)$.

Suppose $N_1$ has property Γ (the case of $N_2$ is symmetric). Let $x_1,\dots,x_m\in N$ and $\varepsilon>0$. Since the vectors $\hat a\otimes\hat b$ span a dense subspace, there are $y_k=\sum_la_{kl}\otimes b_{kl}$ in the algebraic tensor product with $\|x_k-y_k\|_2\le\varepsilon/4$. Let $B=1+\max_{k,l}\|b_{kl}\|_2$ and $L$ the largest number of terms. Choose a unitary $u\in N_1$ with $\tau_1(u)=0$ and $\|[u,a_{kl}]\|_2\le\varepsilon/(2LB)$. Then $u\otimes1$ is a unitary of $N$ with trace $0$, and $\|[u\otimes1,x_k]\|_2\le2\|x_k-y_k\|_2+\sum_l\|[u,a_{kl}]\otimes b_{kl}\|_2\le\frac\varepsilon2+\sum_l\|[u,a_{kl}]\|_2 \|b_{kl}\|_2\le\varepsilon .$ So $N$ has property Γ.

Conversely, suppose neither $N_1$ nor $N_2$ has property Γ. By Theorem 2.5 there are nonzero compact operators $K_i\in C^*(N_i,N_i')$ on $H_i$. The commutant of $N$ contains $N_1'\otimes1$ and $1\otimes N_2'$. Hence $C^*(N,N')$ contains $C^*(N_1,N_1')\otimes1$ and $1\otimes C^*(N_2,N_2')$, and therefore the nonzero compact operator $K_1\otimes K_2=(K_1\otimes1)(1\otimes K_2)$ (here $K_1\otimes1$ is the image of $K_1$ under the $*$-homomorphism $z\mapsto z\otimes1$, which maps $C^*(N_1,N_1')$ into $C^*(N,N')$ because it maps $N_1$ and $N_1'$ into $N$ and $N'$). By Theorem 2.5, $N$ does not have property Γ.

With separable preduals, $N$ also has separable predual, and fullness is the absence of property Γ (Theorem 6.5 of "Full factors"); so the second statement is the first one in contrapositive form. $\square$

Example 6.8. Let $L(\mathbb F_2)$ be the group von Neumann algebra of Example 3.3 and $R$ the hyperfinite factor. Then $L(\mathbb F_2)\bar\otimes L(\mathbb F_2)$ does not have property Γ, and is full; while $R\bar\otimes L(\mathbb F_2)$ has property Γ although one of its tensor factors is full. In the first case, with $T$ and $f$ as in Example 3.3, the operators $f(T)\otimes1=P_0\otimes1$ and $1\otimes f(T)=1\otimes P_0$ lie in $C^*(N,N')$ for $N=L(\mathbb F_2)\bar\otimes L(\mathbb F_2)$, and their product is the rank-one projection onto $\mathbb C(\delta_1\otimes\delta_1)$.

7. Exercises

Exercise 1. Let $N$ be a II₁ factor and $u_1,\dots,u_n\in\mathcal U(N)$. With $T$ as in Proposition 3.2, let $T_0$ be the restriction of $T$ to $\hat1^\perp$. Show that $u_1,\dots,u_n$ have a spectral gap if and only if $1\notin\operatorname{Sp}T_0$. Deduce that $N$ has property Γ if and only if $1\in\operatorname{Sp}T_0$ for every finite set of unitaries.

Solution. If there is a gap, Proposition 3.2 gives $\operatorname{Sp}T_0\subset[-1,1-c]$ with $c>0$, so $1\notin\operatorname{Sp}T_0$. Conversely, if $1\notin\operatorname{Sp}T_0$, then $\operatorname{Sp}T_0$ is a compact subset of $[-1,1[$ (recall $\|T\|\le1$), so $T_0\le1-c$ for some $c>0$. For $\xi\perp\hat1$, the computation in the proof of Proposition 3.2 gives $c\|\xi\|^2\le\langle(1-T)\xi,\xi\rangle=\frac1{2n}\sum_j\|[u_j,\xi]\|^2\le\frac12\max_j\|[u_j,\xi]\|^2$. So (2.2) holds for $\xi\perp\hat1$ with $K=(2c)^{-1/2}$; for general $\xi$ apply this to $\xi-\langle\xi,\hat1\rangle\hat1$, whose commutators with the $u_j$ are those of $\xi$. The last statement is Theorem 2.5, (a)⇔(c), in the form of Lemma 2.4(iii).

Exercise 2. Let $N$ be a II₁ factor and $f\in N$ a nonzero projection. Show that $N$ has property Γ if and only if $fNf$ has property Γ.

Solution. If $N$ has property Γ, it satisfies (b) of Theorem 2.5, so $fNf$ satisfies (b) by Lemma 5.2, and $fNf$ has property Γ by Theorem 2.5. Conversely, suppose $fNf$ has property Γ. Choose an integer $k$ with $1/k\le\tau(f)$ and $f'\le f$ with $\tau(f')=1/k$. Then $f'Nf'=f'(fNf)f'$ has property Γ by the first part applied to $fNf$. As in the proof of Lemma 5.2 there are matrix units $e_{ab}$, $1\le a,b\le k$, in $N$ with $e_{11}=f'$ and $\sum_ae_{aa}=1$, and $e_{ab}\otimes y\mapsto e_{a1}ye_{1b}$ extends to an isomorphism $M_k(\mathbb C)\otimes f'Nf'\cong N$. Given $x_1,\dots,x_m\in N$, write $x_l=\sum_{a,b}e_{a1}x_l^{ab}e_{1b}$ with $x_l^{ab}=e_{1a}x_le_{b1}\in f'Nf'$. If $u$ is a unitary of $f'Nf'$ with trace $0$ and small commutators with all $x_l^{ab}$, then $U=\sum_ae_{a1}ue_{1a}$ is a unitary of $N$ with $\tau(U)=k\tau(f')\tau'(u)=0$ (where $\tau'$ is the trace of $f'Nf'$), and $[U,x_l]=\sum_{a,b}e_{a1}[u,x_l^{ab}]e_{1b}$, so $\|[U,x_l]\|_2\le\sum_{a,b}\|[u,x_l^{ab}]\|_2$ is small. So $N$ has property Γ.

Exercise 3. Let $N$ be a II₁ factor. Show that the only state $\varphi$ on $N$ with $\varphi\circ\operatorname{Ad}u=\varphi$ for all $u\in\mathcal U(N)$ is $\tau$. Conclude that in condition (b) of Theorem 2.5 the finiteness of the set of unitaries cannot be dropped.

Solution. Let $x\in N$ and $\delta>0$. By the Dixmier approximation theorem (B6) there are unitaries $w_1,\dots,w_r$ and weights $t_i\ge0$, $\sum t_i=1$, with $\|\sum_it_iw_ixw_i^*-\tau(x)1\|\le\delta$. Applying $\varphi$, which is invariant and has norm $1$, gives $|\varphi(x)-\tau(x)|\le\delta$. So $\varphi=\tau$, which is normal. Thus no non-normal state is invariant under all inner automorphisms, while for property Γ factors (b) holds for every finite set.

Exercise 4. Take $R$ as in Example 6.6, and let $s_n=\operatorname{diag}(1,-1)$ placed in the $n$-th tensor factor. Put $e_n=\prod_{i=n+1}^{2n}\frac12(1+s_i)$. Show that $\tau(e_n)=2^{-n}$, that $(e_n)$ is a central sequence, and that the states $\psi_n=\tau(e_n\,\cdot\,)/\tau(e_n)$ have a weak* limit along any free ultrafilter which is a state invariant under all $\operatorname{Ad}u$, $u\in\bigcup_mA_m$, and not normal. Why does this not contradict Exercise 3?

Solution. The $\frac12(1+s_i)$ are commuting projections of trace $\frac12$ in different tensor factors, so $e_n$ is a projection with $\tau(e_n)=2^{-n}$. If $y\in A_m$, then $e_n$ commutes with $y$ for $n\ge m$. For $u\in\mathcal U(A_m)$ and $n\ge m$, $\psi_n\circ\operatorname{Ad}u=\psi_n$ exactly. Since $\sum_n\tau(e_n)<\infty$ and $\psi_n(e_n)=1$, the proof of Lemma 4.1 (with the unitaries of a fixed $A_m$, for which $\|\psi_n\circ\operatorname{Ad}u-\psi_n\|=0$ for large $n$) shows that the limit $\varphi$ is not normal and is invariant under every $\operatorname{Ad}u$, $u\in\bigcup_m\mathcal U(A_m)$. For a central sequence $(e_n)$ in general, $\|[e_n,y]\|_2\to0$ for all $y\in R$ by density of $\bigcup_mA_m$ in $\|\cdot\|_2$. There is no contradiction with Exercise 3: $\varphi$ is invariant only under the unitaries of the dense subalgebra $\bigcup_mA_m$, and invariance of a non-normal state does not pass to $\|\cdot\|_2$-limits of unitaries. Indeed, by Lemma 1.5(iv), $\|\psi_n\circ\operatorname{Ad}u-\psi_n\|\le\sqrt2\|[u,e_n]\|_2/\|e_n\|_2$, and while $\|[u,e_n]\|_2\to0$ for every unitary $u$ of $R$, the ratio need not.

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