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I have a question about multiplier algebras and commutants of $C^*$-algebras in general.

First of all, the question is related to this structure theorem about completely positive order zero maps (you can find the theorem in ``Completely positive maps of order zero", by Winter and zacharias, Theorem 2.3 ):

Theorem: Let $A$ and $B$ $C^*-$algebras and $\phi:A\to B$ be a completely postive map of order zero. Set $C:=C^*(\Phi(A))\subset B$.
Then there is a positive element $h\in M(C)\cap C'$ with $\|h\|=\|\phi\|$ and a $*-$homomorphism $$\pi_{\phi}:A\to M(C)\cap \{h\}'$$ such that $$\phi(a)=\pi_{\phi}(a)h $$ for $a\in A$.

If $A$ is unital, then $\phi(1_A)=h\in C$.

Ma Problem: I have problems to understand why the intersection $M(C)\cap C'$, similar $M(C)\cap \{h\}'$, does make sense.

$M(C)$ is the multiplier algebra of $C$ and the elements are maps $m:C\to C$ such that there is a map $m^*:C\to C$ with $(m(a))^*b=a^*m^*(b)$ for all $a,b\in C$, so called multipliers of $C$. $C'$ is the commutant of $C$: $C'=\{y\in L(H); xy=xy\; \text{ for all $x\in$ C}\}$, this means that this is a subset of $L(H)$ for a suitable Hilbert space $H$.

Why it is possible to intersect this sets? They seem very different. And is $H$ here the universal Hilbert space with $C$ acts nondegenerately on $H$? I appreciate your help. Regards

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If you represent $C $ in $B (H) $ faithfully and nondegenerately, then it is well-known that $$ M (C)\simeq\{x\in B (H):\ xc\in C,\ cx\in C,\ \forall c\in C\}. $$

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  • $\begingroup$ I totally forgot it, thank you! $\endgroup$ – user197416 Jun 18 '15 at 8:47
  • $\begingroup$ $C$ should be represented faithfully and nondegenerately for this to hold. $\endgroup$ – Mike F Oct 5 '15 at 16:52
  • $\begingroup$ Indeed. $\ \ \ $ $\endgroup$ – Martin Argerami Oct 5 '15 at 17:47

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