A tracial von Neumann algebra $(M,\tau)$ is a von Neumann algebra with a faithful normal tracial state $\tau$ on $M$. That is, $\tau$ is a function from $M \to \mathbb{C}$ such that it is a faithful normal state and $\tau(xy)=\tau(yx)$. I m confused with tracial and finite von Neumann algebras. I could see references saying that a finite von Neumann algebra $M$ has a unique centre valued $Z(M)$ trace. But this need not be scalar valued no? My definition of trace is a positive linear functional $\tau$ satisfying $\tau(xy)=\tau(yx)$. Does a finite von Neumann algebra has a faithful tracial states? That is a scalar valued one?are they unique? I know a finite factor has a unique one.


1 Answer 1


If you have a faithful tracial state, then the algebra is finite (if $v^*v=1$ then $0\leq \tau(1-vv^*)=\tau(1-v^*v)=0$, so $1$ is finite).

But the converse is not true. An algebra with a faithful state has to be "countably decomposable", it can only admit countably many pairwise orthogonal projections. But there are finite von Neumann algebras that have uncountably many pairwise orthogonal projections. For instance $\ell^\infty[0,1]$.

  • $\begingroup$ My doubt is different. I could see that every finite von Neumann algebra has a unique faithful center valued trace( $\tau: M \to Z(M)$ ). Infact I saw a result that a von Neumann algebra is finite iff it has a unique faithful center valued trace. Out of a center valued trace on a finite von Neumann algebra can we get construct a scalar valued faithful trace? $\endgroup$
    – budi
    Feb 9 at 15:50
  • $\begingroup$ No. There is an example in my answer. $\endgroup$ Feb 9 at 16:12
  • $\begingroup$ But I could see some result relating this in Kadison,s book, Fundamentals of the theory of operator algebras, volume 2, chapter 8, Proposition 8.3.10. I couldn't fully understand the proposition. Also I could see the following which makes me confusing. raum-brothers.eu/sven/data/teaching/2015-16/… section 1.2 $\endgroup$
    – budi
    Feb 10 at 4:38
  • $\begingroup$ Also in Theory of operator algebras 1, chapter 5 theorem 2.4 says about existance of sufficiently many normal traces (but doesn't say about faithfulness)but a unique center valued faithful trace in theorem 2.6. I am confused with these facts totally. $\endgroup$
    – budi
    Feb 10 at 5:16
  • $\begingroup$ Not sure what your aim is. I already gave you an example of a finite von Neumann algebra without faithful states. The proposition in Kadison's book will produce a faithful state if you can first have a faithful state in the centre of the algebra; but that's not always the case, as the example shows. $\endgroup$ Feb 10 at 18:26

You must log in to answer this question.

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