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It is the quotation below:

Exploiting the duality between completely positive map $A \rightarrow M_{n}(C)$ and states on $M_{n}(A)$, it is not too hard to deduce the next result from Glimm's lemma.

(Glimm's lemma) Let $A\subset B(H)$ be a separable C*-algebra containing no nonzero compact operators on $H$. If $\phi$ is a state on $A$, then there exist orthonormal vectors $\{\xi_{n}\}$ such that $\langle a\xi_{n}, \xi_{n} \rangle \rightarrow \phi(a)$ for all $a\in A$.

Proposition 1.7.1. Let $H$ be separable Hilbert space, $1\in A\subset B(H)$ be a separable C*-algebra and $\phi: A\rightarrow M_{n}(C)$ be a unital completely positive map such that $\phi|_{A\cap K(H)}=0$. Then there exist isometries $V_{k}:l_{n}^{2}\rightarrow H$ with the following properties:

(1) the ranges of the $V_{k}$ 's are pairwise orthogonal;

(2) $||\phi(a)-V_{k}^{*}aV_{k}||\rightarrow 0$ for every $a\in A$.

(Here, $K(H)$ denotes the compacts and $l_{n}^{2}$ denotes the n-dimensional Hilbert space.)

I do not know how to utilise the duality to deduce the proposition, could someone show me more details?

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You can read about such duality in the first pages of Chapter 6 in Paulsen. Given $\phi\in\text{UCP}(A,M_n(\mathbb C))$, it can be identified with the state $s_\phi$ on $M_n(A)$ by $$ s_\phi([a_{kj}])=\frac1n\,\sum_{k,j}\phi(a_{kj})_{kj}. $$ The hypothesis that $\phi$ is zero on the compacts implies that so is $s_\phi$. So we can apply Glimm's Lemma to $M_n(A)\subset B(H^n)$ to obtain orthonormal vectors $\xi_k=(\xi_k^1,\ldots,\xi_k^n)\in H^n$ with $s_\phi(A)=\lim_k\langle B\xi_k,\xi_k\rangle$ for all $B\in M_n(A)$.

(this construction does not yield that the ranges of the $V_k$ are pairwise orthogonal; I have some doubts that such a thing is possible).

Now define $W_k:l_n^2\to H$ by $W_ke_j\mapsto \sqrt n\xi_k^j$. We have, for any $a\in A$, $$ \phi(a)_{ij}=n\,s_\phi(a\otimes E_{ij})=n\,\lim_k\,\langle(a\otimes E_{ij})\xi_k,\xi_k\rangle=n\,\lim_k\langle a\,\xi_k^j,\xi_k^i\rangle\\ =\lim_k\,\langle a\,W_ke_j,W_ke_i\rangle=\lim_k\, (W_k^*aW_k)_{ij}. $$ As these are limits of matrices, an entry-wise limit implies norm limit. So $\|\phi(a)-W_k^*aW_k\|\to0$.

Note also that $$ \|W_k(\sum_{s=1}^nc_se_s)\|^2=n\,\|\sum_{s=1}^nc_s\xi_k^s\|^2 $$ doesn't look like an isometry. But $$ \lim_k\,\langle W_ke_s,W_ke_t\rangle= \lim_k\,\langle \xi_k^s,\xi_k^t\rangle=\lim_k\,\langle E_{ts}\xi_k,\xi_k\rangle=\frac1n\,s_\phi(I\otimes E_{st})=\phi(I)_{st}=\delta_{st}. $$ So the components of $\xi_k$ are almost orthonormal and, more importantly, $W_k^*W_k\to I$. Again, the weak convergence implies norm convergence, as $W_k^*W_k\in M_n(\mathbb C)$; so $\|W_k^*W_k-I\|\to0$. So, for big enough $k$, $W_k^*W_k$ is invertible.

Write $W_k=V_k\,|W_k|$ the polar decomposition. From the previous paragraph, for big $k$ we have that $V_k$ is an isometry, and $\|V_k-W_k\|\to0$ (because $W_k^*W_k\to I$). So we get $$ \|\phi(a)-V_k^*aV_k\|\to0 $$ for isometries $V_k$.


Edit: here is an example that shows that the condition of being zero on the compacts is essential. Let $n=1$, and $\{e_j\}$ an orthonormal basis for $H$. Let $A=B(H)$ (or $K(H)$, which would be enough). Define a ucp map $\phi:B(H)\to\mathbb C$ by $$\tag1 \phi(a)=\tfrac12\,\langle ae_1,e_1\rangle+\tfrac12\,\langle ae_2,e_2\rangle. $$ Suppose that $\phi(a)=\lim_k V_k^*aV_k$ for isometries $V_k:\mathbb C\to H$. It is easy to check that $V_k\lambda=\lambda v_k$ for some $v_k\in H$ with $\|v_k\|=1$, and that $V_k^*x=\langle x,v_k\rangle$. Thus $$ \phi(a)=\lim_k\langle av_k,v_k\rangle. $$ With $P=E_{11}+E_{22}$, we have that from $(1)$ that $\phi(a)=\phi(aP)=\phi(Pa)$. So $$ \phi(a)=\lim_k\langle aPv_k,Pv_k\rangle. $$ As $Pv_k=\lambda_{k,1}e_1+\lambda_{k,2}e_2$ with $1\geq\|Pv_k\|^2=|\lambda_{k,1}|^2+|\lambda_{k,2}|^2$ the bounded sequences $\{\lambda_{k,j}\}_k$ admit a convergent subsequence for each $j=1,2$. So we obtain a subsequence of $\{Pv_k\}$ that converges to a certain $v$, and thus $$\tag2 \phi(a)=\langle av,v\rangle. $$ Say $v=\lambda_1e_1+\lambda_2e_2$. Comparing $(1)$ and $(2)$ we get: $$ \tfrac12=\phi(E_{11})=\langle E_{11}v,v\rangle=\alpha, $$ and similarly $\beta=\tfrac12$. But then $$ 0=\phi(E_{21})=\langle E_{21}v,v\rangle=\tfrac12, $$ a contradiction.

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  • $\begingroup$ I have two questions on the details: 1.In the 11 lines of your answer, why does the equation $\langle aW_{k}e_{j}, W_{k}e_{i}\rangle=lim_{k}(W_{k}^{*}aW_{k})_{ij}$ holds? I suppose the rightside of the equation is a limit of an operator from $l_{n}^{2} \rightarrow l_{n}^{2}$; 2. Why an entry-wise limit implies norm limit? $\endgroup$ – Yan kai Mar 15 '14 at 16:08
  • $\begingroup$ And if we define $s_{\phi}([a_{k,j}])=\Sigma_{kj}\phi(a_{kj})_{kj}$ and $W_{k}$ by $W_{k} e_{j}\longmapsto \xi_{k}^{j}$, then the $W_{k}$ is an isometry? $\endgroup$ – Yan kai Mar 15 '14 at 16:14
  • $\begingroup$ If you define that way, $s_\phi(I)=n$, so not a state. As for that $W_k$, I see not reason at all why that would be an isometry. $\endgroup$ – Martin Argerami Mar 15 '14 at 16:55
  • $\begingroup$ And I still have two questions on the details of your answer: 1.In the 11 lines of your answer, why does the equation $\langle a W_{k}e_{j}, W_{k}e_{i}\rangle=lim_{k}(W_{k}^{*}aW_{k})_{ij}$ holds? I suppose the rightside of the equation is a limit of an operator from $l_{n}^{2} \rightarrow l_{n}^{2}$; 2. Why an entry-wise limit implies norm limit? $\endgroup$ – Yan kai Mar 16 '14 at 11:00
  • $\begingroup$ 1. I guess you didn't see the lim that was left hanging in the previous line; I have edited it now. 2. Because in a finite-dimensional space all norms are equivalent. $\endgroup$ – Martin Argerami Mar 16 '14 at 12:57

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