# A question about nuclear C*-algebra

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?

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.
• 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. – Yan kai Apr 1 '14 at 13:10
• 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 $\pi$. – Martin Argerami Apr 1 '14 at 13:25
• 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 $\rho\circ\psi_n$. – Martin Argerami Apr 1 '14 at 19:41