Finite outer actions and Bernoulli shifts

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. Self-checked by the writing AI. New original text: public domain (CC0).

A finite group can act on the hyperfinite finite factor while every nonidentity element acts outerly. Averaging then produces a fixed algebra with finite-dimensional approximants. To see that this algebra is again a factor, we compute a relative commutant using finite Fourier coefficients.

We use the tracial infinite product and the finite AFD uniqueness theorem. The foundational inputs are the finite trace GNS representation, $JMJ=M'$, the bicommutant theorem and the fact that an ultraweakly closed ideal in a von Neumann algebra is a central summand. No classification theorem is used.

An action satisfies $\alpha_g\alpha_h=\alpha_{gh}$. Throughout Sections 1–3, $M$ is a factor of type $\mathrm{II}_1$, $\tau$ is its normalized trace and $G$ is finite. An automorphism is outer if it has no unitary implementer in $M$.

1. Finite Fourier coefficients

On $L^2(M,\tau)\otimes\ell^2(G)$, put

Then $U_g\pi(x)U_g^*=\pi(\alpha_g(x))$. The crossed product $P=M\rtimes_\alpha G$ is generated by these operators.

Lemma 1.1. Every element of $P$ has a unique expression

The coefficient maps are normal and satisfy $\|x_g\|\le\|X\|$. The formula $\tau_P(X)=\tau(x_e)$ defines a faithful normal tracial state.

Proof. The $(t,s)$ matrix entry of (2) is

In particular $x_g=b_{e,g^{-1}}$. Thus the coefficients are unique normal matrix slices and satisfy the norm bound. Conversely, the matrix relations

are ultraweakly closed. Their solution space is exactly the finite sums (2); it is a unital star algebra by covariance. It therefore is the generated von Neumann algebra.

The normalized matrix trace on $M_{|G|}(M)$, restricted to this algebra, is $\tau(x_e)$, because $\tau\alpha_t=\tau$. It is faithful, normal and tracial. Alternatively, multiplying two finite sums gives

and trace invariance converts each summand to the corresponding term of $\tau_P(YX)$. In particular

This also proves faithfulness directly. $\square$

Lemma 1.2. If $\alpha_g$ is outer for $g\ne e$, then

Consequently $P$ is a factor.

Proof. Commutation of (2) with $\pi(y)$ says

If $a\ne0$ satisfies (7), its adjoint relation shows that $a^*a$ commutes with $\alpha_g(M)$, while $aa^*$ commutes with $M$. Both are scalar. Their nonzero scalar values agree: the polar part of $a$ has both initial and final support $1$, so it is a unitary and $a^*a=c1=aa^*$. Thus $a=\sqrt c\,v$ and $\alpha_g(y)=v^*yv$. This contradicts outerness for $g\ne e$. The identity coefficient is central in $M$, hence scalar. The center of $P$ is contained in (6). $\square$

Lemma 1.3. Under the same outerness hypothesis, any faithful normal covariant representation $(\rho,V)$ gives a normal isomorphism

Proof. Covariance makes (8) a unital star homomorphism. The finite coefficient bound gives

so this homomorphism is bounded, hence contractive by the $C^*$-identity. It is normal because there are finitely many normal coefficient maps. Its kernel is a central ideal in the factor $P$, and it cannot be all of $P$ because the map is unital. It is therefore zero. A faithful normal representation of a von Neumann algebra has ultraweakly closed range: its bounded balls are ultraweakly compact and their image is compact, and the inverse is normal. Thus the range is precisely the indicated generated algebra. $\square$

Finiteness is used in both the coefficient expansion and the norm bound. This proof does not supply an integrated isomorphism for arbitrary infinite groups.

2. The fixed algebra and its relative commutant

Let $N=M^G=\{x:\alpha_g(x)=x\text{ for every }g\}$. On $L^2(M,\tau)$, each action has the canonical unitary

This is unitary because the trace is invariant. It commutes with $J\widehat x=\widehat{x^*}$.

Theorem 2.1. If all nonidentity $\alpha_g$ are outer, then

and $N$ is a factor of type $\mathrm{II}_1$. Neither conclusion requires separable predual.

Proof. Write $M'=JMJ$. The action on $M'$ implemented by $W_g$ is outer: an implementer in $M'$, conjugated by $J$, would implement $\alpha_g$ in $M$. Lemma 1.3, applied to the opposite finite factor $M'$, identifies

Here the representation of $M'$ on $L^2(M)$ is faithful and normal. Its commutant is

so $Q=N'$. Lemma 1.2 for $M'$ now gives

The center of $N$ lies in this intersection, so $N$ is a factor. The restricted trace makes it finite. If it were a finite type I factor $M_d$, its matrix units would decompose $M$ as

by the exact matrix-coordinate calculation in the preceding lesson. This contradicts the type of $M$. Hence $N$ has type $\mathrm{II}_1$. $\square$

Theorem 2.2. Suppose, in addition, that an increasing sequence of unital finite-dimensional algebras $F_n$ generates $M$, and $\alpha_g(F_n)=F_n$ for all $g,n$. Then $N$ is AFD and is isomorphic to $R$.

One may adjoin $1_M$ to nonunital finite-dimensional approximants: their dimension stays finite, invariance and inclusion are preserved, and the generated algebra is unchanged.

Proof. The averaging map

is normal, unital, completely positive, trace preserving and $N$-bimodular. It fixes $N$, and trace invariance shows that it is the orthogonal projection onto $L^2(N)$. In particular it contracts the $2$-norm. Also $E(F_n)=F_n^G$, an increasing finite-dimensional algebra.

For $x\in N$, bounded elements of $\bigcup_nF_n$ approximate $x$ in $2$-norm by Kaplansky density. Their averages are bounded elements of $\bigcup_nF_n^G$ and still approximate $x$. Thus the latter union generates $N$. This countable union makes $N$ separable, and Theorem 2.1 and finite AFD uniqueness identify it with $R$. $\square$

The same averaging argument works for an increasing invariant directed family of finite-dimensional algebras generating $M$. It proves local AFD without imposing separability; the conclusion of isomorphism with the separable factor $R$ uses the countable hypothesis.

3. An explicit outer finite action

For a finite group $H$, form the tracial product

Let $\beta_k$ move the algebra at $(h,n)$ to the algebra at $(kh,n)$. A permutation of the tensor coordinates preserves the product trace and extends to a normal automorphism. These maps satisfy $\beta_k\beta_l=\beta_{kl}$.

Put $Z=\operatorname{diag}(1,-1)$ and let $z_n$ be $Z$ at $(e,n)$. Then $\tau(z_n)=0$, $\|z_n\|=\|z_n\|_2=1$, and $z_n$ commutes eventually with every finite-coordinate element. For arbitrary $x\in R_H$, choose a finite-coordinate $a$ close in $2$-norm. Since

the sequence is central.

For $k\ne e$, the sites $(e,n)$ and $(k,n)$ are different, and independence gives

If $\beta_k=\operatorname{Ad}u$ with $u\in R_H$, the left side would tend to zero by (14) with $x=u$. Thus every nonidentity element acts outerly.

The algebras supported on $H\times\{1,\ldots,n\}$ are increasing invariant full matrix algebras. Theorems 2.1–2.2 show that $R_H^H$ is another copy of $R$ and that

For nontrivial $H$, the inclusion is proper, since $z_n$ is not fixed.

4. Countable Bernoulli actions and the inverse index

Let $\Gamma$ be an infinite countable discrete group and use the tracial product $R_\Gamma$ indexed by $\Gamma$. The left Bernoulli action is

where the entry on the right is placed at site $g$. Thus a local element at site $g$ moves to site $hg$.

Proposition 4.1. Formula (17) extends to a trace-preserving normal action, and $\beta_h$ is outer for every $h\ne e$.

Proof. Coordinate permutation is a star automorphism of the algebraic local union, preserves its trace and has inverse $\beta_{h^{-1}}$. Its trace GNS unitary extends it normally to the generated von Neumann algebra. On local tensors direct substitution gives $\beta_h\beta_k=\beta_{hk}$, and normality extends the equality.

Fix $h\ne e$. Choose sites $g_n$ so that both $g_n$ and $hg_n$ eventually avoid every fixed finite subset. This is possible by avoiding $F\cup h^{-1}F$ at each step of an exhaustion. The $Z$-operators at $g_n$ are central by the estimate (14), and their images are the $Z$-operators at the distinct sites $hg_n$. Their $2$-distance is $\sqrt2$. An inner automorphism would move them a distance tending to zero. $\square$

Placing $x_{hg}$ at site $g$ instead defines

Each map is an automorphism and is outer away from the identity, but the maps form an action of the opposite group. They give an action of $\Gamma$ itself only when the relevant group elements commute. Formula (17) supplies the inverse needed for a left action.

5. Exercises with complete solutions

Exercise 1. In the regular representation, locate the coefficient $x_g$ in a matrix entry, and locate the coefficient of the identity in $X^*X$.

Solution. Equation (3) gives $x_g=b_{e,g^{-1}}$. Covariance gives $X^*=\sum_g\pi(\alpha_{g^{-1}}(x_g^*))U_{g^{-1}}$. The identity coefficient of $X^*X$ is $\sum_g\alpha_{g^{-1}}(x_g^*x_g)$. Applying the invariant trace yields (5).

Exercise 2. Suppose a nonzero $a$ satisfies $a\theta(y)=ya$ for every $y$ in a factor. Explain why a partial-isometry implementer cannot have a proper nonzero support.

Solution. Taking adjoints of the relation with $y^*$ gives $\theta(y)a^*=a^*y$. Hence $a^*a\in\theta(M)'\cap M=\mathbb C1$ and $aa^*\in M'\cap M=\mathbb C1$. Their support projections are nonzero central projections, hence $1$. The polar partial isometry is therefore unitary, and the intertwining relation implements $\theta$ by its adjoint.

Exercise 3. Find a nonfactor on which a nonidentity automorphism is outer, although its fixed algebra has a nontrivial relative commutant.

Solution. Take $M=\mathbb C\oplus\mathbb C^2$ and let the generator of $\mathbb Z/2$ exchange the last two coordinates and fix the first. Every inner automorphism of an abelian algebra is the identity, so this nonidentity automorphism is outer. Its fixed algebra is $\mathbb C\oplus\{(z,z):z\in\mathbb C\}$. Since $M$ is abelian, its relative commutant in $M$ is all of $M$. It also acts trivially on a nonzero central summand. Thus ordinary outerness is insufficient for the factor conclusion off factors.

Exercise 4. For the swap action on $M_2\otimes M_2$, determine the fixed algebra and explain why it need not be a full matrix algebra.

Solution. The swap of the two tensor legs is implemented by the flip unitary on $\mathbb C^2\otimes\mathbb C^2$. Its eigenspaces are the symmetric space of dimension $3$ and the antisymmetric space of dimension $1$. The fixed algebra is its commutant, $M_3\oplus\mathbb C$. In Theorem 2.2 the finite approximants may therefore have centers even though their generated fixed algebra is a factor.

Exercise 5. Show directly that the average (12) is $N$-bimodular and is self-adjoint as an operator on $L^2(M)$.

Solution. For $a,b\in N$, each $\alpha_g(axb)=a\alpha_g(x)b$, so $E(axb)=aE(x)b$. For $x,y\in M$,

The substitution $g\mapsto g^{-1}$ gives the second equality. Together with $E^2=E$, this identifies the orthogonal projection onto the closed fixed subspace $L^2(N)$.

Exercise 6. Let $H=\mathbb Z/3$ in (13). Give an explicit element outside the fixed algebra and compute its distance from its image under a generator.

Solution. Choose the trace-zero diagonal $Z$ at $(e,1)$. Its image is $Z$ at the different site $(k,1)$. Both have squared $2$-norm $1$, and their inner product is the product of their traces, hence $0$. Their squared distance is $2$. This element is not fixed; the tail copies give the same calculation and prove outerness.

Exercise 7. Let $\Gamma=\mathbb F(a,b)$. Exhibit the failure of the left action law for the printed placement $x_{hg}$.

Solution. Let $z$ be $Z$ at site $e$. Then $\gamma_a\gamma_b(z)$ is $Z$ at $a^{-1}b^{-1}$, whereas $\gamma_{ab}(z)$ is $Z$ at $b^{-1}a^{-1}$. These reduced words differ. Independence gives $2$-distance $\sqrt2$, so the maps are different. Replacing $hg$ by $h^{-1}g$ restores the action law.

Exercise 8. Does the outer-shift proof require the nonidentity $h$ to have infinite order? What changes if $\Gamma$ is finite?

Solution. No order assumption is used: left multiplication has no fixed site for $h\ne e$, and an infinite indexing set permits the escaping sites $g_n$. If $\Gamma$ is finite, the tensor product is a finite matrix algebra. Every automorphism of that matrix algebra is inner, and no escaping sequence exists. The repeated finite group indexing $H\times\mathbb N$ in Section 3 restores an infinite tail and gives the outer action.

Reading and attribution

Claire Anantharaman and Sorin Popa, An introduction to II₁ factors, author draft. Section 5.2, printed pp.68–70, supplies the tracial Fourier construction and properly outer relative-commutant calculation. Exercise 5.11, printed p.81, asks for the normal faithful realization by canonical action unitaries. Sections 1–2 above give every finite coefficient, normality, factor-kernel and opposite-algebra commutant step explicitly, including an arbitrary faithful normal covariant representation and no separability hypothesis. Invariant finite stages then supply the complete AFD fixed-factor argument.

The Bernoulli method is Example 5.2.4. With entries placed at the output index, its displayed $x_{gh}$ convention composes in reverse order; (17) uses the inverse for a left action. The draft explicitly defines “countable group” to mean countably infinite in footnote 13, printed p.17. Its escaping-site proof uses that infinite index and a nontrivial tensor algebra. The examples above supply an infinite repeated index for a finite group, and Exercise 8 gives the finite-index matrix counterexample. The index and tensor-base hypotheses are explicit in the constructions above. Finite trace GNS duality, the bicommutant theorem, closed ideals and the tracial product remain declared foundations.