5
$\begingroup$

Let $(G,\mu)$ be a measured groupoid and denote by $\nu,\nu^{-1}$ the measures on all of $G$ induced by $\mu$ and Haar system $\{\lambda^x\}$.

I have a question regarding the dependance of the groupoid von Neumann algebra $\text{vN}_\mu(G)$ on $\mu$.

In the above setting, we can consider the space $L^2(G,\nu^{-1})$ and the left regular representation $\lambda_\mu: C_c(G) \rightarrow B(L^2(G,\nu^{-1})$ given by $\lambda(f)g = f*g$. By completing $C_c(G)$ w.r.t the norm $\|f\|_{r,\mu} := \|\lambda(f)\|$ we obtain a $C^*$-algebra denoted by $C_{r,\mu}^*(G)$ and by taking the double commutant we get the von Neumann algebra $\text{vN}_\mu(G) := C_{r,\mu}^*(G)'' \subseteq B(L^2(G,\nu^{-1}))$.

It can be seen that the $C^*$-algebra $C_{r,\mu}^*(G)$ does only depend on the support of $\mu$ [1, Proposition 3.1.2], however I do not know if the same holds for the von Neumann algebras since the space $B(L^2(G,\nu^{-1}))$ can vary a lot depending on $\mu$.

If $\mu_1, \mu_2 $ are two quasi-invariant measures on $G^0$ with the same support, is it true that $ \text{vN}_{\mu_1}(G) \cong \text{vN}_{\mu_2}(G) $?

Of course, if $\mu_1, \mu_2 $ are equivalent the above is true.

[1] Paterson, Alan L. T., Groupoids, inverse semigroups, and their operator algebras, Progress in Mathematics (Boston, Mass.). 170. Boston, MA: Birkhäuser. xvi, 274 p. (1999). ZBL0913.22001.

Improve this question
$\endgroup$
3
  • $\begingroup$ What do you mean by the support of $\mu$ if $G$ is a measured groupoid? $\endgroup$
    – R W
    Commented yesterday
  • $\begingroup$ A measured groupoid is a pair $(G,\mu)$ where $G$ is a locally compact groupoid (+ a Haar system) and $\mu$ a measure on the unit space $G^0$ that satisfies an invariance condition called quasi-invariance. The support of $\mu$ is the usual measure theoretic notion, i.e. the largest (closed) subset of $G^0$ for which every open neighbourhood of every point of the set has positive measure. $\endgroup$
    – Tomás Pacheco
    Commented yesterday
  • $\begingroup$ This is not the usual definition. A measured groupoid is not a topological one additionally endowed with a quasi-invariant measure. Its standard definition doesn't involve any topology. $\endgroup$
    – R W
    Commented 19 hours ago

1 Answer 1

Reset to default
6
$\begingroup$

No, the von Neumann algebra can vary a lot, even when the groupoid is a transformation groupoid for an action of a discrete group $\Gamma$ on a compact space $X$. For instance, take $X$ to be the Cantor set $(\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}}$ and let $\Gamma$ be the direct sum $\Gamma = (\mathbb{Z}/2\mathbb{Z})^{(\mathbb{N})}$ acting by translation. Whenever $\mu_n$ is a sequence of probability measures on the two-point set $\mathbb{Z}/2\mathbb{Z}$ with $p_n = \mu_n(0)$ satisfying $0 < p_n < 1$, the product measure $\mu = \prod_n \mu_n$ is quasi-invariant. The resulting crossed product von Neumann algebra is an infinite tensor product of $2 \times 2$ matrices w.r.t. a sequence of states given by the eigenvalues $p_n$, $1-p_n$. If for instance you take $0 < \lambda < 1$ and $p_n = (1+\lambda)^{-1}$, you get the Powers factors of type III$_\lambda$, which are all nonisomorphic.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What a neat example, thank you! Just to make sure I followed, if we pick different $\lambda$'s we obtain different probability measures, all of which give non-zero measure to points, meaning the support is always the whole Cantor set, correct? $\endgroup$
    – Tomás Pacheco
    Commented 13 hours ago
  • $\begingroup$ Most of the time, in particular when all $p_n$ are equal, the measure $\mu$ is nonatomic. But the support is indeed the entire Cantor set. If $U \subset (\mathbb{Z}/Z\mathbb{Z})^{\mathbb{N}}$ is an open subset, there exists an $n \in \mathbb{N}$ and $x_1,\ldots,x_n \in \mathbb{Z}/Z\mathbb{Z}$ such that $\{x_1\} \times \cdots \times \{x_n\} \times (\mathbb{Z}/Z\mathbb{Z})^{[n+1,\infty)}$ is a subset of $U$, with strictly positive measure $\mu_1(x_1) \cdots \mu_n(x_n)$. $\endgroup$
    – Stefaan Vaes
    Commented 3 hours ago

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .