4
$\begingroup$

Let $M$ be a von Neumann with separable predual. It well known that one can write $M$ as a direct sum $M=M_I\oplus M_{II} \oplus M_{III}$ of von Neumann algebras of types $I$, $II$ and $III$. It is also known that one can write $M$ as a direct integral of factors $$ M=\int_X^\oplus M(x) d\mu(x) $$ and obtains that $Z(M)= L^\infty(X,\mu)$. Clearly one can partition $X$ into three parts $X_I$, $X_{II}$ and $X_{III}$ so that $M_j = \int_{X_j}^\oplus M(x) d\mu(x)$ where $j=I,II,III$. My question is whether a strengthening in the following sense is possible:

Is it possible to find factors $M_i$ and measure spaces $(X_i,\mu_i)$ indexed by some (necessarily countable) index set $I$, such that $$ M = \bigoplus_{i\in I} L^\infty(X_i,\mu_i) \bar\otimes M_i, \qquad \text{such that } i\ne j\implies M_i \not\simeq M_j ? $$ It would follow that $(X,\mu) \cong \bigoplus_{i\in I} (X_i,\mu_i)$ with $(X,\mu)$ as above.

My hope is that one can construct such a decomposition using the Choi-Effros Borel structure on the set of von Neumann algebras on a fixed separable Hilbert space $H$. Also, if the answer is yes I would be grateful for a reference.

$\endgroup$

1 Answer 1

7
$\begingroup$

No this is not always possible. For such a decomposition to exist, the family of factors $M(x)$ should in particular only contain countably many isomorphism classes, which need not be the case. It is of course never "cheap" to give an explicit uncountable family of nonisomorphic factors, but they do exist. For instance, you could fix a II$_1$ factor $N$ with trivial fundamental group, meaning that $p N p \not\cong N$ whenever $p$ is a nontrivial projection in $N$. Let $\tau$ be the trace on $N$ and choose a copy of $A=L^\infty([0,1])$ inside $N$ such that $\tau|_A$ is given by integration w.r.t. the Lebesgue measure. For every $x \in [0,1]$, denote $p_x = 1_{[0,x]}$ and $M(x) = p_x N p_x$. This is a concrete measurable family of two by two nonisomorphic II$_1$ factors. You cannot "simplify" the direct integral

$$\int_{[0,1]}^{\oplus} M(x) \, dx \; .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.