Support projection vs closed support projection of a normal state in enveloping von Neumann algebra

Question

I preface this by saying that I am fairly new to the enveloping von Neumann algebra scene, so there may be some gaps in my understanding.

Given a $C^*$-algebra $A$ and a state $\phi$ on $A$, one may consider $\phi$ as a normal state on the universal enveloping von Neumann algebra $A''$ of $A$. In this case, there should be a support projection $\text{supp}(\phi) \in A''$ for $\phi$, that is, a minimal projection such that $\phi(\text{supp}(\phi))=1$.

For a (concerning) example, if $A = C([0,1])$ and $\phi$ is the Lebesgue integral, this seems to give us a projection in $C([0,1])''$ which is smaller than the operator of multiplication by the indicator function of any set of full Lebesgue measure. This is a confronting possibility, so have I made a mistake somewhere or is this just evidence of how complicated $C([0,1])''$ is?

On the other hand, we could restrict to closed projections (in the sense of Akemann, The General Stone-Weierstrauss problem, 1969). Then for commutative $C^*$-algebras there is a smallest closed projection of full "measure", since (normal) states correspond to regular Borel probability measures and closed projections correspond to closed subsets of the spectrum. Furthermore, these "closed support projections" are much less pathological than the support projection in the enveloping von Neumann algebra.

I have a proof sketch via the universal representation that "closed support projections" do exist in the non-commutative case too, but nowhere do I use closedness, so I will not feel confident in its validity until I know what is going on with the support projection of the Lebesgue integral on $C([0,1])$.

I would also be interested to know if people have already thought about closed support projections, for example if it is known whether they are the closure of the support projection.

Thanks in advance.

Answer 1

I am not quite sure what the question is, so let me try to understand what your unease is. In the example we take $A=C([0,1])$ and for the state $\phi$ take normalised Lebesgue measure. What is the support projection in $A''$? When you write the following:

... which is smaller than the operator of multiplication by the indicator function of any set of full Lebesgue measure.

I think your intuition is proceeding as follows: we think of $A''$ as somehow being a space of functions on $[0,1]$, and then guess that the support projection of $\phi$ is the indicator function of $[0,1]$. But then we could adjust this function to be zero at some points, and it would still be a projection, and still give all of $\phi$. In this way, we seem to conclude that there can be no minimal projection.

The fault in this reasoning is to believe that $A''$ can in any way be thought of as functions on $[0,1]$. As $A''$ is a commutative von Neumann algebra, it is isomorphic to $C(X)$ for some compact hyperstonian $X$. In the literature, $X$ is called the hyperstonian cover of $[0,1]$. A recent book on this subject is "Banach spaces of continuous functions as dual spaces" by Dales, Dashiell, Jun, Lau and Strauss Zbl 1368.46003

How to think about $A''$? Let $z$ be the support projection of $\phi$, a central projection. One can show that $A'' \cong zA'' \oplus_\infty (1-z)A''$ the $\ell^\infty$-direct sum of two von Neumann algebras. In this case, $zA'' \cong L^\infty([0,1])$ while $(1-z)A''$ is very complicated, given by all probability measures on $[0,1]$ (= states on $A$) which are singular with respect to Lebesgue measure. As $zA''\cong L^\infty([0,1])$ we see that the indicator function of $[0,1]$ is minimal in $L^\infty([0,1])$, with the property that it has full Lebesgue measure.

Thus one picture of $A''$ is to take some maximal family $(\mu_i)$ of mutually singular probability measures on $[0,1]$ and then $A''$ is isomorphic to the $\ell^\infty$-direct sum of $L^\infty([0,1], \mu_i)$.

I do not know about closed projections, and so will let someone else answer there.