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.
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$\begingroup$ It comes down to whether the centre has a faithful state or not, and an abelian von Neumann algebra has a faithful state iff it is $\sigma$-finite, i.e. iff every disjoint family of non-zero projections is countable. A finite factor is of course tracial because its centre is isomorphic to $\mathbb{C}$. $\endgroup$– Robert FurberFeb 10, 2023 at 0:48
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$\begingroup$ raum-brothers.eu/sven/data/teaching/2015-16/… it says something in section 1.2. I think i am misunderstanding something $\endgroup$– Anupam AhFeb 10, 2023 at 4:39
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$\begingroup$ 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. $\endgroup$– Anupam AhFeb 10, 2023 at 5:16
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$L^\infty[0,1]$ is a finite von Neumann algebra. Every state on it is tracial.
$l^\infty(\mathbb{R})$ is also a finite von Neumann algebra, and it has no faithful states.