Let $A$ be a tracial von Neumann algebra with trace $\tau$. If I'm not mistaken, this trace is unique. For $X=[x_{ij}]\in M_n(A)$, define $$\tau_n(X):=\frac{1}{n}\sum_{i=1}^n\tau(x_{ii}).$$ I find this a very natural construction.

My questions are then: Does this make $M_n(A)$ a tracial von Neumann algebra? If so, is it the unique one?


1 Answer 1


You don't say what "trace" is, so it is a bit hard to discuss uniqueness.

With the mostly usualy meaning "state with the tracial property" (and this leaves out whether you want your trace to be normal), traces are not unique. For starters, in any commutative von Neumann algebra, any state is a trace. Even in non-commutative von Neumann algebras it is easy to find multiple traces. In fact, every time that the centre is not trivial, if there is one trace there are infinitely many others. For instance in $M_2(\mathbb C)\oplus \mathbb C$ given any $t\in[0,1]$ you have that $$ \tau_t(A\oplus\lambda)=\tfrac12\Big[t\,\operatorname{Tr}(A)+(1-t)\lambda\Big] $$ is a tracial state for any $t$.

Tracial states are unique on factors, because in such case any two projections are comparable and the factors is the norm-closure of the linear span of the projections.

Your $\tau_n$ is a tracial state if $\tau$ is. It's uniqueness depends on the uniqueness of $\tau$.

As a final comment, your construction is as natural as it can get, because what you are doing is define $\tau_n=\operatorname{tr}\otimes \tau$ on the algebra $M_n(\mathbb C)\otimes A$.

  • $\begingroup$ The notes I'm using defines it as follows: $\tau$ is a linear functional on $A$ with (i) $\tau(x^*x)\geq 0$ for all $x\in A$, (ii) $\tau(x^*x)=0\rightarrow x=0$, (iii) $\tau(1)=1$, $\tau(xy)=\tau(yx)$, (iv) restriction to the unit ball is continuous in WOT. $\endgroup$
    – chhro
    Mar 15, 2021 at 19:48
  • $\begingroup$ Ok, that's what one would usually call a faithful normal tracial state. $\endgroup$ Mar 15, 2021 at 20:21
  • $\begingroup$ Sorry, I thought accepting the answer is enough. So in that case, $\tau_n$ Is the unique faithful normal tracial state on $M_n(A)$? $\endgroup$
    – chhro
    Mar 15, 2021 at 20:26
  • $\begingroup$ If $A$ has nontrivial centre, then there are infinitely many. If $A$ has a unique tracial state, then so does $M_n(A)$. $\endgroup$ Mar 15, 2021 at 20:28
  • $\begingroup$ In what I'm looking at, the center is trivial. Thanks for your help! $\endgroup$
    – chhro
    Mar 15, 2021 at 20:35

You must log in to answer this question.

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