1

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?

math112358
  • 3,183

1 Answers1

3

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 $M(A)/A$.

Martin Argerami
  • 205,756
  • I wonder whether there is another method without using free ultra filter. – math112358 Sep 14 '18 at 23:15
  • 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). – Martin Argerami Sep 15 '18 at 07:49
  • 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? – math112358 Sep 16 '18 at 06:04
  • Yes. As for ultrafilters, there is little I know beyond what I wrote here. – Martin Argerami Sep 16 '18 at 12:19
  • 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. – math112358 Sep 18 '18 at 15:49
  • Yes, you are right. My argument does not apply to such elements. – Martin Argerami Sep 18 '18 at 22:35
  • Pro Argerami, It means that if $a$ is nonzero in$ M(A)/A$,we cannot conclude that there must exist tracial states $\tau$ on $M(A)/A$ s.t $\tau(a)$ is nonzero? – math112358 Oct 05 '18 at 06:38
  • I'm not sure. The straightforward way to construct traces is with $\tau_\omega$ as above. But those traces will be zero in the elements with $\tau_n(a_n)\to0$, so I cannot tell if there is another way to construct a nonzero trace for those guys. – Martin Argerami Oct 05 '18 at 06:47
  • But in your construction ,for nonzero element $a$ ,if $\tau(a)=0$is true for any tracial state,then$\tau_n(a_n)\to0$.We can not get a contradiction .since$\tau_n(a_n)\to0$ does't imply $a \in A$ – math112358 Oct 05 '18 at 07:56
  • Pro Argerami,I also have two quesions about the above proof:
    1.I searched the definitions of principal ultrafilter and free ultrafilter online.The ultrafilter generated by the sets $E_j$ is called principal.Principal filter is not free.But in the begining,you assume that $\omega$ be any free filter on $\beta\mathbb N\setminus\mathbb N$     – math112358 Oct 05 '18 at 14:40
  • No, it's not principal. Read the definition more carefully. – Martin Argerami Oct 05 '18 at 14:46
  • 2.For each $r$ there exists ultrafilter $\omega_r$,but how to show that different $r$s have different $\omega_r$s – math112358 Oct 05 '18 at 14:58
  • If you take $r_1\ne r_2$, eventually the corresponding $E_j$ will be disjoint. – Martin Argerami Oct 05 '18 at 23:08
  • Let us continue this discussion in chat. – math112358 Oct 06 '18 at 09:32
  • Pro Argerami.I have another question.You define $\tau_\omega(a)=\lim_\omega\operatorname{tr}(a_j)$,the limit dependend on the accumulation point of tr(a_j),if it has only one accumulation point,the limit is the same for different free ultrafilters.In this case,how to assure that the tracial states are uncountable. – math112358 Oct 06 '18 at 09:38
  • Some sequences have one accumulation point, but many have uncountably many. It is enough to take a single ${a_n} $ such that ${\operatorname {tr}(a_n)} $ is dense in $[0,1] $, and then for each $r $ there exists $\omega $ with $\lim_\omega\operatorname {tr}(a_n)=r $. – Martin Argerami Oct 06 '18 at 12:00
  • Pro Argerami,can we define a trace $\tau$ on $M(A)$ as follows:$\tau(x)=\sum_{j=1}^{\infty} k_j tr(x_j)$,where $\sum_{j=1}^{\infty} k_j=1$? – math112358 Oct 06 '18 at 13:04
  • Yes, of course. But such a trace is not zero on $A$, so it doesn't induce a trace on the corona. – Martin Argerami Oct 06 '18 at 15:12
  • So the tracial states you construct are not faithful.I wonder whether there is some methods of constructing faithful tracial states through tracial states. – math112358 Oct 07 '18 at 14:41