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