# Unusual crossed product constructions being factors

Let $$A$$ be an abelian von Neumann algebra and $$G$$ a countable group acting on $$A$$. In the literature we meet usually two kinds of crossed product $$A \rtimes G$$ being a factor:

• if the action is (essentially) free then $$A \rtimes G$$ is a factor iff the action is ergodic,
• the group von Neumann algebra $$LG$$, which is $$A \rtimes G$$ with $$A = \mathbb{C}$$ (and trivial action), is a factor iff $$G$$ is ICC (note that here the action is not free).

In this post we are interested in unusual crossed product factors, i.e. not of the above kinds.

Question: Take $$A = \mathbb{C}^{\oplus n}$$, what is a necessary and sufficient condition on $$G$$ and its action on $$A$$ for $$A \rtimes G$$ being a factor? Is it known for $$n=2$$?

Example: if $$G$$ is finite, $$n = |G|$$ and $$A=\mathbb{C}^{\oplus n} = \ell^{\infty}(G)$$ on which $$G$$ acts by translation then $$A \rtimes G$$ is a factor, isomorphic to $$M_{n}(\mathbb{C})$$.

Bonus question (related to this post): Is there a $$\mathrm{II}_1$$ factor $$\mathbb{C}^{\oplus n} \rtimes G$$ with $$n>1$$ which is not isomorphic to a group von Neumann algebra?

We fixed $$A = \mathbb{C}^{\oplus n}$$ for the question not being too broad, but all the results providing unusual crossed product constructions $$A \rtimes G$$ being factors, for any other $$A$$ (in particular atomless) are also welcome.

The following provides a necessary and sufficient condition for an arbitrary crossed product von Neumann algebra $$L^\infty(X) \rtimes G$$ to be a factor. As a corollary, I include a simpler criterion for actions that preserve a probability measure.

First note that a criterion for arbitrary actions necessarily has to refer to induced actions. Given an action $$G \curvearrowright (X,\mu)$$ of a countable group $$G$$ by nonsingular transformations (i.e. preserving sets of measure zero) of a standard probability space $$(X,\mu)$$, it may happen that there exists a subgroup $$\Gamma < G$$ and a $$\Gamma$$-invariant Borel set $$Y \subset X$$ such that, up to measure zero, the sets $$(g \cdot Y)_{g \in G/\Gamma}$$ form a partition of $$X$$. Then the crossed products satisfy $$L^\infty(X) \rtimes G \cong B(\ell^2(G/\Gamma)) \overline{\otimes} (L^\infty(Y) \rtimes \Gamma) \; .$$ In particular, the crossed product for $$G \curvearrowright X$$ is a factor if and only if the crossed product for $$\Gamma \curvearrowright Y$$ is a factor.

Conversely, given any nonsingular action $$\Gamma \curvearrowright Y$$ and any embedding of $$\Gamma$$ into a larger group $$G$$, there is a natural action $$G \curvearrowright X = G/\Gamma \times Y$$ that is induced from $$\Gamma \curvearrowright Y$$.

These induced actions are characterized by the existence of a $$G$$-equivariant map $$X \rightarrow G/\Gamma$$. Note that if $$\mu$$ is an invariant probability measure, this forces $$\Gamma$$ to be a finite index subgroup of $$G$$; see the corollary below.

Theorem. Let $$G$$ be a countable group and $$G \curvearrowright (X,\mu)$$ any action by nonsingular transformations of a standard probability space $$(X,\mu)$$. Then the following two conditions are equivalent.

(1) The crossed product $$L^\infty(X) \rtimes G$$ is a factor.

(2) The action $$G \curvearrowright (X,\mu)$$ is ergodic and whenever $$G \curvearrowright X$$ is induced from $$\Gamma \curvearrowright Y$$, the following holds: if $$h \in \Gamma \setminus \{e\}$$ acts trivially on $$Y$$, then $$h$$ has an infinite $$\Gamma$$-conjugacy class.

Proof. Write $$A = L^\infty(X)$$ and $$M = A \rtimes G$$. View $$A \subset M$$ and denote by $$(u_g)_{g \in G}$$ the canonical unitary operators in $$M$$.

(1) $$\Rightarrow$$ (2). If $$F \in A$$ is $$G$$-invariant, then $$F$$ belongs to the center of $$M$$ and hence $$F$$ is constant. So $$G \curvearrowright (X,\mu)$$ is ergodic. Assume that $$G \curvearrowright (X,\mu)$$ is induced from $$\Gamma \curvearrowright Y$$. Denote by $$N \lhd \Gamma$$ the normal subgroup that acts trivially on $$Y$$ and take $$h \in N \setminus \{e\}$$. If the $$\Gamma$$-conjugacy class $$C$$ of $$h$$ is finite, then $$\sum_{g \in C} u_g$$ defines a non scalar central element of the crossed product $$L^\infty(Y) \rtimes \Gamma$$. Then also $$M$$ would fail to be a factor. So $$C$$ must be infinite.

$$\neg$$ (1) $$\Rightarrow$$ $$\neg$$ (2). Assume that $$M$$ is not a factor and that $$G \curvearrowright (X,\mu)$$ is ergodic. We have to prove that $$G \curvearrowright X$$ is induced from $$\Gamma \curvearrowright Y$$ in such a way that there exists an element $$h \in \Gamma \setminus \{e\}$$ that acts trivially on $$Y$$ and that has a finite $$\Gamma$$-conjugacy class.

Let $$T$$ be a non scalar element in the center of $$M$$. Denote by $$E : M \rightarrow A$$ the normal conditional expectation satisfying $$E(u_g) = 0$$ for all $$g \in G \setminus \{e\}$$. Then, $$E(T)$$ is $$G$$-invariant and thus scalar. Replacing $$T$$ by $$T - E(T)$$, we may thus assume that $$T$$ is a central element with $$E(T) = 0$$ and $$T \neq 0$$.

View $$M$$ as the algebra of bounded measurable functions $$F : X \rightarrow B(\ell^2(G))$$ satisfying $$T(g \cdot x) = \rho_g T(x) \rho_g^*$$ for all $$g \in G$$ and a.e. $$x \in X$$. Here $$\rho$$ is the right regular representation of $$G$$. We then associate to $$T$$ the function $$f : X \rightarrow \ell^2(G) : f(x) = T(x) \delta_e \; .$$ Since $$T(x) \delta_h = \rho_h f(h^{-1} \cdot x)$$, we get that $$f$$ is not a.e. zero. Since $$E(T) = 0$$, we have $$f(x) \in \ell^2(G \setminus \{e\})$$. The fact that $$T$$ is central translates to the following two properties for the function $$f$$.

• The fact that $$T$$ commutes with the unitary operators $$u_g$$ translates to: $$f(g \cdot x) = \text{Ad}(g) f(x)$$ for all $$g \in G$$ and a.e. $$x \in X$$.

• The fact that $$T$$ commutes with $$A$$ translates to: if $$f(x)(h) \neq 0$$, then $$h \cdot x = x$$.

Define the measurable function $$S : X \rightarrow \mathbb{R} : S(x) = \max \{ |f(x)(h)| \mid h \in G \} \; .$$ This is well defined because $$f(x) \in \ell^2(G)$$. Since $$S$$ is $$G$$-invariant, it follows that $$S$$ is constant a.e. Also, $$S$$ is not a.e. zero. So we find $$0 < s < \infty$$ such that $$S(x) =s$$ for a.e. $$x \in X$$. For every $$x \in X$$, denote by $$R(x) \subset G$$ the finite subset given by $$R(x) = \{h \in G \mid |f(x)(h)| = s\} \; .$$ We have $$R(g \cdot x) = g R(x) g^{-1}$$. Also, $$R(x) \neq \emptyset$$ and $$R(x) \subset G \setminus \{e\}$$ for a.e. $$x \in X$$. There are only countably many finite subsets of $$G$$. So we can fix a nonempty finite subset $$C \subset G \setminus \{e\}$$ such that the set $$Y = \{x \in X \mid R(x) = C\}$$ has positive measure. When $$g \in G$$, the set $$g \cdot Y \cap Y$$ has positive measure if and only if $$g C g^{-1} = C$$. Defining the subgroup $$\Gamma < G$$ by $$\Gamma = \{g \in G \mid g C g^{-1} = C\} \; ,$$ it follows that $$G \curvearrowright X$$ is induced from $$\Gamma \curvearrowright Y$$. When $$g \in C$$ and $$x \in Y$$, we have $$|f(x)(g)| = s > 0$$ so that $$g \cdot x = x$$. In particular, $$R(g \cdot x) = R(x)$$, so that $$gCg^{-1} = C$$ and $$g \in \Gamma$$. We conclude that $$C$$ is a finite subset of $$\Gamma \setminus \{e\}$$ that is invariant under $$\Gamma$$-conjugation and that all elements of $$C$$ act trivially on $$Y$$. So, $$\neg 2$$ holds.

Corollary. Let $$G$$ be a countable group and $$G \curvearrowright (X,\mu)$$ any action by $$\mu$$-preserving transformations of a standard probability space $$(X,\mu)$$. Then the following two conditions are equivalent.

(1) The crossed product $$L^\infty(X) \rtimes G$$ is a factor.

(2) The action $$G \curvearrowright (X,\mu)$$ is ergodic and if $$g \in G \setminus \{e\}$$ is such that $$\{x \in X \mid g \cdot x = x\}$$ has positive measure, then $$g$$ has an infinite conjugacy class.

Proof. (2) $$\Rightarrow$$ (1). Assume that $$G \curvearrowright X$$ is induced from $$\Gamma \curvearrowright Y$$ and that $$g \in \Gamma \setminus \{e\}$$ acts trivially on $$Y$$. By the theorem above, it suffices to prove that $$g$$ has an infinite $$\Gamma$$-conjugacy class. Since $$\mu$$ is a $$G$$-invariant probability measure, $$\Gamma < G$$ has finite index. It thus suffices to prove that $$g$$ has an infinite $$G$$-conjugacy class. Since $$g$$ fixes all points of $$Y$$, this is indeed the case by the assumption in (2).

$$\neg$$ (2) $$\Rightarrow$$ $$\neg$$ (1). Assume that $$g_0 \in G \setminus \{e\}$$ fixes all points in a subset of $$X$$ of positive measure and that the $$G$$-conjugacy class $$C$$ of $$g_0$$ is finite. We prove that the second condition in the theorem above does not hold. Since $$C$$ is finite and since $$g_0 \in C$$ fixes all points in a set of positive measure, we can choose a maximal subset $$C_0 \subset C$$ containing $$g_0$$ and having the property that $$Y := \{x \in X \mid h \cdot x = x \;\;\text{for all h \in C_0}\;\}$$ has positive measure. When $$g \in G$$ and $$x \in g \cdot Y \cap Y$$, we have that $$h \cdot x = x$$ for all $$h \in C_0 \cup g C_0 g^{-1} \subset C$$. By the maximality of $$C_0$$ we conclude that $$g \cdot Y \cap Y$$ has positive measure if and only if $$gC_0 g^{-1} = C_0$$ if and only if $$g \cdot Y = Y$$. Defining $$\Gamma = \{g \in G \mid gC_0 g^{-1} = C_0\} \; ,$$ we find that $$G \curvearrowright X$$ is induced from $$\Gamma \curvearrowright Y$$. By construction, $$C_0 \subset \Gamma \setminus \{e\}$$ is a finite set of elements that act trivially on $$Y$$ and $$C_0$$ is invariant under conjugation by $$\Gamma$$. So the second condition in the theorem above does not hold and the corollary is proven.

• Really excellent answer. – Nik Weaver Jun 6 at 16:10
• Welcome to MathOverflow! Thanks for your great answer. Here are my first comments: at the end of the 2nd paragraph of the 1st proof, finite should be infinite. You removed the use of normal subgroup $N$ in the statement of the theorem but not from its proof (2nd and last paragraph) and from the corollary's proof (1st and last paragraph); it should also be removed from there (if I'm not mistaken). In the 3rd paragraph of the 1st proof you wrote << $E(T)$ is $G$-invariant and thus scalar >> so I guess you assume that the action is ergodic to then prove the negation of the 2nd part of (2), right? – Sebastien Palcoux Jun 7 at 13:41
• Thank you, Sébastien. I corrected the typo "finite -> infinite" and I added a few sentences with extra details for the argument. – Stefaan Vaes Jun 8 at 13:32

For a countable discrete group $$G$$, the crossed product $$L^\infty(X,\mu)\rtimes G$$ for a nonsingular action is a factor if

1. the action is ergodic and
2. almost all stabilisers $$G_x:=\mathop{\mathrm{Stab}}(x)$$ are i.c.c.

Proof. Suppose the action is ergodic and almost all $$G_x$$ are i.c.c., and let $$k$$ be a central element in $$L^\infty(X)\rtimes G$$. Then $$k$$ has a (unique) “Fourier series” $$k = \sum_{g\in G} k_g u_g$$. As $$k$$ commutes with every $$f\in L^\infty(X,\mu)$$, we have $$f\cdot k_g = \alpha_g(f)k_g$$ for all $$f$$, so $$\mathop{\mathrm{supp}} k_g \subset \{x\in X\mid g\in G_x\}$$, so $$k$$ is “supported in the bundle of stabilisers”.

Now, let $$h_n\colon X\to G$$ be measurable functions such that $$G_x = \{h_n(x)\}_{n\in\mathbb N}$$ for almost every $$x\in X$$ and let $$b_n\colon x\mapsto \chi_{\mathop{\mathrm{supp}}h_n}(x)\cdot h_n(x)$$. Now, $$b_n \in L^\infty(X)\rtimes G$$, and so have to commute with $$k$$. A little calculation then shows that for a conull subset $$Y_n\subset X$$ we then have that $$k_{h_n(x)gh_n(x)^{-1}}(x) = k_g(x)$$ for all $$g\in G$$. Intersecting $$Y_n$$'s, we get that on a conull subset $$k$$ is constant on conjugacy classes. Since $$G_x$$ is i.c.c., $$k_g = 0$$ for $$g\neq 1$$.

In the original version of my answer I claimed that the converse is also true; however, my original answer contained a mistake. Still, one can say a lot:

For a countable discrete group $$G$$, the crossed product $$L^\infty(X,\mu)\rtimes G$$ for a nonsingular action being a factor implies that the action is ergodic. Moreover, if additionally one of the following is satisfied:

• the FC-centre $$FC(G_x)$$ is finite for a.e. $$x\in X$$ or
• $$X$$ is finite,

then almost all stabilisers $$G_x:=\mathop{\mathrm{Stab}}(x)$$ are i.c.c. (More precise conditions are discussed in the proof below.)

Proof.

If the action is not ergodic, $$\chi_B$$ for an invariant measurable subset $$B\subset X$$ is in the centre.

Suppose the action is ergodic but $$G_x$$ is not i.c.c. on a subset of positive measure; by ergodicity, $$G_x$$ is not i.c.c. for almost every $$x\in X$$.

Let $$Z_x:= \mathscr Z(L(G_x))$$ be the centre of the von Neumann algebra of $$G_x$$. By assumption, it is nontrivial for a.e. $$x\in X$$. Consider the direct integral $$Z:=\int^\oplus_X Z_x\, d\mu(x)\subseteq L^\infty(X)\rtimes G.$$ It is a $$G$$-invariant (w.r.t. conjugation action) abelian von Neumann subalgebra containing $$L^\infty(X)$$. The $$G$$-equivariant inclusion $$L^\infty(X)$$ corresponds to a factor map $$(\widehat Z,\nu) \to (X,\mu)$$, where $$\widehat Z$$ is the spectrum of $$Z$$. The fibres of this map are canonically identified with $$(\widehat{Z_x},\nu_x)$$, with the marginal measures $$\nu_x$$ canonically induced by the trace on $$L(G_x)$$.

By ergodicity the type of $$(\widehat{Z_x},\nu_x)$$ is the same a.e., and we get the following possibilities for the types of $$(Z_x,\nu_x)$$:

1. with an atomic and a nonatomic part
2. finite
3. countably infinite
4. atomless.

In the first case the element $$x\mapsto z_x$$, where $$z_x\in Z_x$$ is the projection onto the atomic part is a nontrivial central element. In the second case the FC-centre of $$G_x$$ is a.e. finite, so the element $$x\mapsto \sum_{g\in FC(G_x)} g$$ is nontrivial and central. In the third case, the element $$x\mapsto z_x$$, where $$z_x\in Z_x$$ is the projection onto the sum of the biggest atoms is central and non-trivial (as the subset of $$Z$$ consisting of the biggest atoms in each $$Z_x$$ is clearly invariant).

Finally, if $$X$$ is finite, then by ergodicity $$X$$ can be identified with $$G/S$$, and by assumption $$S$$ is not i.c.c. Let $$C\subset S$$ be a non-trivial finite $$S$$-conjugacy class; the element $$gx\mapsto \sum_{c\in C} gcg^{-1}$$ of the von Neumann algebra is well-defined, nontrivial and central.

P.S. It was pointed out by Stefaan Vaes that $$L^\infty(X)\rtimes G$$ can be a factor for an action with non-i.c.c. stabiliser; in fact, one cannot say anything about the factoriality of the crossed product by looking only at the structure of stabilisers.

• Thanks! Should $(X,\mu)$ be nonatomic? Can $G$ be locally compact? Is your result in the literature? Where? – Sebastien Palcoux Nov 18 '19 at 8:35
• Nonatomicity is not required. For locally compact groups the situation, I believe, is more difficult because even to determine whether $LG$ is a factor is not that easy (see, for instance, arxiv.org/pdf/1505.07793.pdf, Theorem E (7.2) and its proof). For the discrete p.m.p. case, the result can be found, for instance, in the thesis of Rahel Brugger (Lemma 2.1.10): ediss.uni-goettingen.de/bitstream/handle/11858/…' there it's written down and proven in the language of groupoids. – Vadim Alekseev Nov 18 '19 at 9:56
• I'm sorry to hear that. Which aspect of the answer do you exactly consider problematic? – Vadim Alekseev Nov 19 '19 at 21:48
• Alright, thanks! – Vadim Alekseev Nov 20 '19 at 10:31
• Your comment led me to check the argument and it indeed contained a mistake; thanks a lot for expressing your doubt. I could fill the gap in many cases, but one case is still missing (see the updated version). – Vadim Alekseev Nov 20 '19 at 18:11