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Definition 2.3.1. A C*-algebra $A$ is nuclear if the identity map $id_{A}: A\rightarrow A$ is nuclear.

Definition 2.3.2. A C*-algebra $A$ is exact if there exists a faithful representation $\pi:A \rightarrow B(H)$ such that $\pi$ is nuclear.

There is quotation below:

Let $\pi: A\rightarrow B(H)$ be a faithful representation, then ,

(1). $A$ is nuclear if and only if $\pi$ is nuclear when regarded as taking values in $\pi(A)$.

(2). While $A$ is exact if and only if $\pi$ is nuclear when regarded as taking values in $B(H)$.

My question is how to explain regarding as taking values in $\pi(A)$ and $B(H)$ above, do they make any difference in estimating nuclearness and exactness?

Proof. (1). "only if" Since $A$ is nuclear, then we have c.c.p. $\bar{\phi}_{n}: A \rightarrow M_{k(n)(C)}$ and $\bar{\psi}_{n}: M_{k(n)}(C)\rightarrow A$ such that $\bar{\psi}_{n} \circ \bar{\phi}_{n} \rightarrow I_{A}$ in point-norm topology. Then, we consider $\phi_{n}=\bar{\phi}_{n}$ and $\psi_{n}=\pi \circ \bar{\psi}_{n}$, we can check $\psi \circ \phi_{n} \rightarrow \pi$ in point-norm topology.

"if " Since $\pi$ is nuclear (and we take values in $\pi(A)$), we have c.c.p. $\bar{\phi}_{n}: A \rightarrow M_{k(n)(C)}$ and $\bar{\psi}_{n}: M_{k(n)}(C)\rightarrow \pi(A)$ such that $\bar{\psi}_{n} \circ \bar{\phi}_{n} \rightarrow \pi$ in point-norm topology. Then we consider $\phi_{n}=\bar{\phi}_{n}$ and $\psi_{n}=\pi^{-1} \circ \bar{\psi}_{n}$, we can check $\psi \circ \phi_{n} \rightarrow I_{A}$ in point-norm topology.

(Because of utilizing the invertibility of $\pi$, we need to consider the value in $\pi(A)$, right?)

(2). "if" is clear from the definition. But how to verify the "only if"? Could you give me some hints?

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The difference between nuclearity and exactness is that in the range of the maps $\psi_n$: when $\psi_n:M_{k(n)}(\mathbb C)\to \pi(A)$, the algebra is nuclear. When the range of $\psi_n$ is allowed to be bigger than $\pi(A)$, then $A$ is exact.

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  • $\begingroup$ Yeah, and I edited my question. And I have two questions: 1. I give a brief proof of (1), is there anything wrong in my proof? 2. how to verify the "only if" in (2) above? $\endgroup$ – Yan kai Apr 1 '14 at 10:57
  • $\begingroup$ Your proof of 1 is correct. There is nothing to prove in 2, it is exactly the definition. $\endgroup$ – Martin Argerami Apr 1 '14 at 12:06
  • $\begingroup$ But if $A$ is exact, there exists a faithful $\pi_{1}: A \rightarrow B(H_{1})$ such that $\pi_{1}$ is nuclear. I suppose this representation $\pi_{1}$ is different from the $\pi$ in the question above. $\endgroup$ – Yan kai Apr 1 '14 at 13:10
  • $\begingroup$ You are right. But you just compose the second map in the nuclear decomposition of $\pi_1$ with $\pi\circ\pi_1^{-1}$ to get a nuclear decomposition for $\pi$. $\endgroup$ – Martin Argerami Apr 1 '14 at 13:25
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    $\begingroup$ You are right. What you do is you extend $\pi\circ\pi_1^{-1}:\pi_1(A)\to\pi( A)$ to a cp map $\rho:B(H_1)\to B(H)$. Then you replace $\psi_n$ with $\rho\circ\psi_n$. $\endgroup$ – Martin Argerami Apr 1 '14 at 19:41

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