Arveson's Extension Theorem in C*-algebra

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The proof is making (liberal) use of the following identifications.

If $P$ is a finite-rank projection in $B(H) $, say of rank $n$, then $PH\simeq\mathbb C^n$ and $P\, B(H)\,P\simeq M_n(\mathbb C) $.

To see this last identification, construct a basis $\{e_i\}$ of $H$ such that $e_1,\ldots,e_n$ form a basis of $PH$. Construct the corresponding matrix units $E_{kj}\in B(H)$, i.e. $E_{kj}e_i=\delta_{ji}e_k$.

Note that $P=\sum_jE_{jj}$. Now consider a map $f:M_n(\mathbb C)\to P\,B(H)\,P$, by $$ f([c_{kj}])=\sum_{j,k=1}^nc_{kj}E_{kj}. $$ Exercise: show that $f$ is a $*$-homomorphism.

Exercise 2: show that $f$ is injective.

To see that $f$ is onto, show that for any $T\in B(H)$, $E_{kk}TE_{jj}=\langle Te_j,e_k\rangle\,E_{kj}$. So, if $T\in P\,B(H)\,P$, $$ T=PTP=\left(\sum_{k=1}^nE_{kk}\right)T\left(\sum_{j=1}^nE_{jj}\right)=\sum_{k,j=1}^nE_{kk}TE_{jj}=\sum_{k,j=1}^n\langle Te_j,e_k\rangle\,E_{kj}=f([\langle Te_j,e_k\rangle]). $$

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Yan kai
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Updated on August 01, 2022

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  • Yan kai
    Yan kai over 1 year

    I am reading a book C*-algebra and finite-Dimensional Approximations. There are two conclusions (in the book) below.

    Corollary 1.5.16. Let $E\subset A$ be an operator subsystem and $\phi: E \rightarrow M_{n}(C)$ be a completely positive map. THen $\phi$ extends to a completely positive map $A \rightarrow B(H)$. $~$($M_{n}(C)$ denotes the sets of all complex matrix.)

    Theorem 1.6.1 Let $A$ be a unital C*-algebra and $E\subset A$ be an operator subsystem. Then, every contractive completely positive map $\phi:E\rightarrow B(H)$ extends to a contractive completely positive map $\bar{\phi}: A\rightarrow B(H)$.

    Proof. Let $P_{i}\in B(H)$ be an increasing net of finite-rank projections which converge to the identity in the strong operator topology. For each $i$, we regard the contractive completely positive map $\phi_{i}: E \rightarrow P_{i}B(H)P_{i}$, $\phi_{i}(e)=P_{i}\phi(e)P_{ i}$ as taking values in a matrix algebra. Thus , by Corollary 1.5.16, we may assume that each $\phi_{i}$ is actually defined on all of $A$. Now we regard $\phi_{i}$ as taking values in $B(H)$ and apply compactness of the unit ball of $B(A,B(H))$ in the point-ultraweak topology to find a cluster point $\Phi:A:\rightarrow B(H)$. It is readily verified that $\Phi$ is completely positive and extends to $\phi$.

    My question is how to explain take values of $\phi_{i}$ in a matrix algebra and, meanwhile, take values in $B(H)$ in the proof? ($B(H)$ denotes all the linear bounded operators on $H$)

  • Yan kai
    Yan kai over 9 years
    I know $PH\simeq\mathbb C^n$, but why $P\, B(H)\,P\simeq M_n(\mathbb C)$ holds?
  • Martin Argerami
    Martin Argerami over 9 years
    Please see the edit.
  • Yan kai
    Yan kai over 9 years
    What is the matrix units $E_{kj} \in B(H)$? If $H$ is not a finite-dimensional space, I suppose, $B(H)$ is not isomorphic to $M_{n}(C)$.
  • Martin Argerami
    Martin Argerami over 9 years
    Of course not. But given an orthonormal basis $\{e_j\}$ on a Hilbert space you define $E_{kj}$ as the operator $E_{kj}x=\langle xe_j,e_k\rangle$. The operators $E_{kj}$ are called matrix units. They satisfy $$E_{st}E_{kj}=\delta_{t,k}\,E_{sj},\ \ E_{kj}^*=E_{jk},\ \ \sum_kE_{kk}=I, $$ where the sum is taken in the strong operator topology.
  • Yan kai
    Yan kai over 9 years
    I am sorry, I still do not know what does $E_{k,j}x=\langle xe_{j}, e_{k}\rangle$ mean. Is $x$ an element in $H$?
  • Martin Argerami
    Martin Argerami over 9 years
    I'm sorry, I completely mistyped that. It is $E_{kj}x=\langle x,e_j\rangle\, e_k$.
  • Yan kai
    Yan kai over 9 years
    And how can you ensure $\Sigma_{j,k}^{n}c_{kj}E_{kj}$ is an element in $P B(H) P$ ?
  • Martin Argerami
    Martin Argerami over 9 years
    Because $P\left(\Sigma_{j,k}^{n}c_{kj}E_{kj}\right)P=\Sigma_{j,k}^{n}c_{kj}E_{kj}.$
  • Yan kai
    Yan kai over 9 years
    Please, forgive my stupid, I still have a question about the proof of the theorem above: how to verify the $\Phi$ is completely positive and extends to $\phi$?
  • Martin Argerami
    Martin Argerami over 9 years
    Completely positive: you check that a point ultraweak limit of completely positive is completely positive. That it extends $\phi$: as you move towards the limit in the subnet given by the compactness, you will be dealing with bigger and bigger finite-dimensional projections that converge to the identity. Given $a\in A$ and $\xi\in H$, eventually the finite projections in the net will contain both vectors $\xi$ and $\phi(a)\xi$, so $\Phi(a)\xi=\phi(a)\xi$.
  • Yan kai
    Yan kai over 9 years
    I am sorry, I still do not know how to verify. Could you show me more details on these two verifications?
  • math112358
    math112358 almost 5 years
    If $S\subset B(H)$ and $Id_H\in S$,$PSP$ may be zero?
  • Martin Argerami
    Martin Argerami almost 5 years
    Only if $P=0$, since $P\in S $.