Here is a proposition in a book "C*-algebra and Finite-Dimensional Approximations" P239

Proposition 7.1.8 Every type I C*-algebra with a faithful tracial state is RFD.

Proof Let $\tau$ be a faithful trace on type I algebra $A$ and let $\pi_{\tau}: A\rightarrow B(L^{2}(A, \tau))$ be the associated GNS representation. Since $\tau$ is faithful, so is $\pi_{\tau}$.

By the structure theory of type I von Neumann algebras we have $$\pi_{\tau}(A)''\cong\prod\limits_{n}L^{\infty}(X_{n}, \mu_{n})\otimes B(H_{n}).$$

But $\pi_{\tau}(A)''$ has a faithful trace (the vector state in the GNS representation is tracial and faithful) and hence the dimensions of all the $H_{n}$'s must be finite. Since it is simple to show that $C(X)\otimes M_{k}(\mathbb{C})$ is RFD, the remainder of the proof is straightforward.

My question

  1. What is the faithful trace of $\pi_{\tau}(A)''$ in the proof? Is it $T \longmapsto <Tv_{\tau}, v_{\tau}>$? ($v_{\tau}$ is the cyclic vector of GNS.) Is it tracial?

  2. How to obtain that $H_{n}$ is finite dimensional?

  3. How to show that $C(X)\otimes M_{k}(\mathbb{C})$ is RFD.

Here is the definition of RFD:

Definition 7.1.6. A C*-algebra $A$ is called residually finite-dimensional (RFD) if there exist finite-dimensional *-homomorphisms $\pi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ s.t. $\oplus \pi_{n}: A\rightarrow \prod M_{k(n)}(\mathbb{C})$ is faithful.


1 Answer 1

  1. Yes. It is tracial because it is tracial in the dense subalgebra $\pi_\tau(A)$, and it extends to $\pi_\tau(A)''$ since it is weak operator continuous.

  2. $B(H)$, for infinite-dimensional $H$, admits no tracial state: because when it is infinite-dimensional, we can find pairwise orthogonal projections $p,q$ with $p\sim q\sim I=p+q$. So there are partial isometries with $v^*v=p$, $vv^*=I$, $w^*w=q$, $ww^*=I$. Then $$ 1=\tau(I)=\tau(p+q)=\tau(p)+\tau(q)=\tau(v^*v)+\tau(w^*w)=\tau(vv^*)+\tau(ww^*)=\tau(I)+\tau(I)=2, $$ a contradiction.

  3. The elements of $C(X)\otimes M_k(\mathbb C)$ are matrices with entries in $C(X)$. If $X\in C(X)\otimes M_k(\mathbb C)$ and $t\in X$, define $\pi_t(X)=[X_{kj}(t)]\in M_k(\mathbb C)$. The family of $*$-homomorphisms $\{\pi_t\}_{t\in X}$ satisfies that if $\pi_t(X)=0$ for all $t$, then $X=0$.

  • $\begingroup$ Er, but why does B(H) admits no tracial state when $H$ is infinite-dimensional. And in the end of the proof, is "the remainder of the proof" by that if $A_{n}$ is RFD, then $\prod A_{n}$ is also RFD? $\endgroup$
    – Yan kai
    Sep 29, 2014 at 1:51
  • 1
    $\begingroup$ I added the (trivial) proof that $B(H)$ admits no trace. "The remainder of the proof" is that if you have separating families of finite-dimensional representations for each of the $A_n$, putting them all together you get a separating family of finite-dimensional representations of $\prod A_n$. $\endgroup$ Sep 29, 2014 at 3:50

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