Trace in a finite dimensional $C^*$-Algebra

Question

The following questions is inspired by Exercise 6.2.2 and 6.2.3 from $C^*$-Algebra and Finite-dimensional Approximations written by Nathanial P. Brown and Narutaka Ozawa. Let $B = M_{n_1}(\mathbb{C}) \bigoplus M_{n_2}(\mathbb{C})$ and $\operatorname{tr}_1, \operatorname{tr}_2$ (resp.) be the normalized trace in the first and second matrix algebra. Now fix $r\in(0, 1)$ and define a new trace on $B$ as the following:

$$ \tau_r(S\bigoplus T) = r\operatorname{tr}_1(S) + (1-r)\operatorname{tr}_2(T) $$

The questions are:

1. Show that $B$ can be embedded into $M_{n_3}(\mathbb{C})$ such that the normalized trace $\operatorname{tr}_3$ in $M_{n_3}(\mathbb{C})$ extends $\tau_r$ (i.e. $\operatorname{tr}_3\Big|_B = \tau_r$)

2. Given a finite dimensional $C^*$-Algebra $\mathfrak{A}$ (which can be viewed as a finite direct sum of matrix algebras), is every trace in $\mathfrak{A}$ is equal to a convex combination of normalized trace on each summand matrix algebra? Further, given a trace $\tau$ in $\mathfrak{A}$ and $\epsilon>0$, show that there is a matrix algebra $M_N(\mathbb{C})$ and a $\ast$-homomorphism $\pi: B\rightarrow M_N(\mathbb{C})$ such that for each $a\in\mathfrak{A}$:

$$ \vert\,\tau(a) - \operatorname{tr}[\pi(a)]\,\vert \leq \epsilon\|a\| $$

I have no ideas how to work on the second question. For the first question, what blocks me is that the embedding needs to satisfy: each $S\in M_{n_1}(\mathbb{C})$, $\operatorname{tr}_3(S) = \tau_r(S\bigoplus 0) = r\operatorname{tr}_1(S)$, and $\operatorname{tr}_3(T) = \tau_r(0\bigoplus T) = (1-r)\operatorname{tr_2}(T)$ for each $T\in M_{n_2}(\mathbb{C})$. My attemps is to use the Proposition 3.8.3 from the same book:

(Radon-Nikodym). Assume $f$ is a state on a $C^*$-Algebra $A$ and let $\pi_f$ be the GNS-representation. If $\xi$ is another positive linear functional on $A$ such that $\xi(a)\leq f(a)$ for each positive element $a$, then we can find a unique $y_{\xi}\in\pi_f(A)'$ such that $0\leq y_{\xi} \leq 1$ and: $$ \xi(a) = \langle\,\pi_f(a)y_{\xi}\overset{\wedge}{1}, \overset{\wedge}{1}\,\rangle $$ for each $a\in A$ where $\overset{\wedge}{1}$ is the canonical cyclic vector in $L^2(A, f)$

We use $\pi_r$ to denote the GNS representation of $B$ associated to $\tau_r$. Since both $\xi_1: (S\bigoplus T)\mapsto r\operatorname{tr}_1(S), \xi_2: (S\bigoplus T)\mapsto (1-r)\operatorname{tr}_2(T)$ satisfy conditions in the above result, we can find $y_1, y_2 \in \pi_r(B)'$ such that

$$ \xi_1(S\bigoplus T) = \langle\,\pi_r(S\bigoplus T)y_1\,\overset{\wedge}{1}, \overset{\wedge}{1}\,\rangle = \langle\,\pi_r(S, 0)\sqrt{y_1}\,\overset{\wedge}{1}, \sqrt{y_1}\,\overset{\wedge}{1}\,\rangle \\ \xi_2(S\bigoplus T) = \langle\,\pi_r(S\bigoplus T)y_2\,\overset{\wedge}{1}, \overset{\wedge}{1}\,\rangle = \langle\,\pi_r(S, T)\sqrt{y_2}\,\overset{\wedge}{1}, \sqrt{y_2}\,\overset{\wedge}{1}\,\rangle $$

Now set $V_1 = \pi_r(B)\sqrt{y_1}\,\overset{\wedge}{1}$ and $V_2 = \pi_r(B) \sqrt{y_2}\,\overset{\wedge}{1}$, then for the following embedding of $B$:

$$ (S\bigoplus T) \mapsto \begin{bmatrix} \pi_r(S\bigoplus 0) & 0 \\ 0 & \pi_r(0\bigoplus T)\end{bmatrix} \in \mathbb{B}(V_1\bigoplus V_2) $$

I am not sure if I can find ONE matrix algebra to contain $\mathbb{B}(V_1\bigoplus V_2)$. If I could, then the trace $\operatorname{tr}_3$ defined on that matrix algebra will be equal to:

$$ \operatorname{tr}_3(S\bigoplus T) = \langle\,\pi_r(S, 0)\sqrt{y_1}\,\overset{\wedge}{1}, \sqrt{y_1}\,\overset{\wedge}{1}\,\rangle + \langle\,\pi_r(0, T)\sqrt{y_2}\,\overset{\wedge}{1}, \sqrt{y_2}\,\overset{\wedge}{1}\,\rangle = \tau_r(S\bigoplus T) $$

Answer 1

1. You are missing a crucial part of the statement from the book, which is that $r$ is rational. If $r$ is irrational, then the embedding you are looking for does not exist; the reason is that a matrix algebra cannot have projections with irrational trace. And if you take the identity of $M_{n_1}(\mathbb C)$ as an element of $M_{n_1}(\mathbb C)\oplus M_{n_2}(\mathbb C)$, it is a projection of trace $r$. To find $n_3$ given that $r=p/q$ as in the book, all you care is about the trace of $P=I_{n_1}\oplus 0_{n_2}$, as all the rest adjusts on its own due to the way the traces are defined. On a first take, we want $P$ to be diagonal, $q\times q$, with $p$ ones and $q-p$ zeroes. The problem is that this may not allow us to embed $M_{n_1}(\mathbb C)$. That is, we want an embedding $$ S\oplus T\longmapsto \operatorname{diag}(\overbrace{S,\ldots,S}^{x\ \text{times}},\overbrace{T,\ldots,T}^{y\ \text{ times}})\subset M_{n_3}(\mathbb C). $$ We want the trace of $\operatorname{diag}(S,\ldots,S,0,\ldots,0)$ to be $p/q$. That is, we want, looking at $S=I_{n_1}$, $$\tag1 \frac{xn_1}{xn_1+yn_2}=\frac pq. $$ With a tiny bit of algebra we can rewrite $(1)$ as $$\tag2 \frac yx=\frac{(q-p)n_1}{pn_2}. $$ Which shows that the easiest choice is to take $$ x=pn_2,\qquad\qquad y=(q-p)n_1. $$ So $n_3=xn_1+yn_2=pn_1n_2+(q-p)n_1n_2=qn_1n_2$, with the embedding $$ S\oplus T\longmapsto \operatorname{diag}(\overbrace{S,\ldots,S}^{pn_2\ \text{times}},\overbrace{T,\ldots,T}^{(q-p)n_1\ \text{ times}})\subset M_{qn_1n_2}(\mathbb C). $$
2. What you do is notice that given any finite dimensional C$^*$-algebra $B$, you can consider the central minimal projections $P_1,\ldots,P_m$. These give you precisely the decomposition $$ B=\bigoplus_{k=1}^m M_{n(k)}(\mathbb C), $$ and each block is precisely $P_k A$. Then any trace on $A$ is of the form $$ \tau(b)=\sum_{k=1}^m \tau(P_k)\,\operatorname{tr}(b). $$ If the numbers $\tau(P_k)$ are rational, you can apply part 1. When they are not rational, they can be approximate as closed as you want by rationals and you can apply part 1 in an approximate way.