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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.

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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.

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    $\begingroup$ Really excellent answer. $\endgroup$ – Nik Weaver Jun 6 at 16:10
  • $\begingroup$ 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? $\endgroup$ – Sebastien Palcoux Jun 7 at 13:41
  • $\begingroup$ Thank you, Sébastien. I corrected the typo "finite -> infinite" and I added a few sentences with extra details for the argument. $\endgroup$ – Stefaan Vaes Jun 8 at 13:32
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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.

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  • $\begingroup$ Thanks! Should $(X,\mu)$ be nonatomic? Can $G$ be locally compact? Is your result in the literature? Where? $\endgroup$ – Sebastien Palcoux Nov 18 '19 at 8:35
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    $\begingroup$ 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. $\endgroup$ – Vadim Alekseev Nov 18 '19 at 9:56
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    $\begingroup$ I'm sorry to hear that. Which aspect of the answer do you exactly consider problematic? $\endgroup$ – Vadim Alekseev Nov 19 '19 at 21:48
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    $\begingroup$ Alright, thanks! $\endgroup$ – Vadim Alekseev Nov 20 '19 at 10:31
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    $\begingroup$ 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). $\endgroup$ – Vadim Alekseev Nov 20 '19 at 18:11

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