Let $A$ be the $c_0$ direct sum of $M_{n}(\mathbb{C})$,I know the fact that the multiplier algebra of $A$ ,M($A$) is $\prod M_n(\mathbb{C})$.

Does the corona algebra $M(A)/A$ have uncountable tracial states?How to construct the tracial state?

I try to define $\tau:M(A)/A\rightarrow \mathbb{C}$ as following: $\tau((a_1,\cdots,a_n,\cdots)+\oplus M_n(\mathbb{C}))=tr(a_1)$,where $a_i \in M_{i}(\mathbb{C})$ for $i=1,2,\cdots$,$tr$ is the standard trace on $M_{1}(\mathbb{C})$.But this definition is not well defined.Would you mind giving me some help?


For any $a\in M(A)$, consider the sequence $t(a)=(\operatorname{tr}(a_1),\operatorname{tr}(a_2),\ldots)\in\ell^\infty(\mathbb N)$. Let $\omega\in\beta\mathbb N\setminus\mathbb N$ be any free ultrafilter. Then, using the normalized trace $\operatorname{tr}$ on each $M_n(\mathbb C)$, we have that $\tau_\omega(a)=\lim_\omega\operatorname{tr}(a_j)$ defines a trace on $M(A)$, and since $\tau_\omega(a)=0$ for all $a\in A$, it extends to $M(A)/A$.

Let $\{q_n\}$ be some enumeration of $\mathbb Q\cap [0,1]$. For each $r\in[0,1]$, there exists an ultrafilter $\omega_r$ such that $r=\lim_{\omega_r}q_n=r$ (let $\{n_j\}$ be a subsequence such that $|q_{n_j}- r|<1/j$; now let $\omega_r$ be the ultrafilter generated by the sets $E_j=\{n\geq n_j:\ |q_n-r|<1/j\}$). So each $\omega_r$ produces a different trace on $M(A)/A$.

  • $\begingroup$ I wonder whether there is another method without using free ultra filter. $\endgroup$ – math112358 Sep 14 '18 at 23:15
  • $\begingroup$ If you mean nonzero and positive, yes. Because you cannot have $\operatorname{tr}(a_n)\to0$ (that would put it in $A$); then infinitely many of $\operatorname{tr}(a_n)$ are nonzero, and one can construct an ultrafilter that converges to some nonzero accumulation point of the sequence. And no, I wouldn't know how to avoid ultrafilters (or an analog constructs, like Banach limits or Stone-Cech compactifications). $\endgroup$ – Martin Argerami Sep 15 '18 at 7:49
  • $\begingroup$ Pro Argerami,$\tau_{\omega}$ extends to $M(A)/A$ since we can define $\tau$ on $M(A)/A$ as following:$\tau(x+A)=\tau_{\omega}(x)$ for all $x\in M(A)$.It is well defined since $\tau_{\omega}(x)=0,x\in A$. Is my thought true? $\endgroup$ – math112358 Sep 16 '18 at 6:04
  • $\begingroup$ Yes. As for ultrafilters, there is little I know beyond what I wrote here. $\endgroup$ – Martin Argerami Sep 16 '18 at 12:19
  • $\begingroup$ Pro Argerami,I felt a little confused about this statement"Because you cannot have tr(an)→0 (that would put it in A); ".If $a \in A$,we have $tr(a_n)\rightarrow 0$,I think the converse may not be true. $\endgroup$ – math112358 Sep 18 '18 at 15:49

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