I hope my question is ok for mathoverflow. I first asked on math.stackexchange but received no answer and then delated it.

I want to understand the proof of the theorem (which you can find in the paper ``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 $\phi(1_A)=h\in C$.

The beginning of the proof: First we assume that $A$ is unital to apply the result for the unital case, as you can find in Wolff's paper about "disjointness preserving operators on $C^*$-algebras". Set $\phi(1_A)=h\in C$. Let $C$ acts nondegenerate on its universal Hilbert space $H$. Wollf's theorem gives us that $h$ lies in the center of $C$. Now it is claimed: (i) h is strictly positive ( this means $\overline{hCh}=C$) and (2) the support projection of $h$ is $1_H$. These claims I want to understand

ad. (i): Let $c_1\in C,\epsilon >0$, why there is $c_2\in C$ such that $\|c_1-hc_2h\|<\epsilon$? I know if you consider $g\in hCh$, it is $g=\phi(1_A)t\phi(1_A)$ with $t\in C$ and t commutes with $\phi(1_A)$. But I'm stuck here, $\phi$ isn't multiplicative.

ad. (ii): What is the exact definition of a support projection here and why is the support projection $1_H$? Maybe the support projection of h can be understood as $\Phi(\chi_{\sigma(h)\setminus\{0\}})=\chi_{\sigma(h)\setminus\{0\}}(h)$, with $\Phi$ is the borel measurable functional calculus of $h$ and $\chi_{\sigma(h)\setminus\{0\}}$ the characteristic function on $\sigma(h)\setminus \{0\}$ and $\chi_{\sigma(h)\setminus\{0\}}(h)$ is an orthogonal projection, which is the identity on $\sigma{(h)}\setminus\{0\}$. But what is the definition/ characteristic property of a support projection in detail and why is it $1_H$? Here it should be important that $C$ acts nondegenerate on $H$.

My third problem is how Stinespring's Theorem is apllied. First they define $$\pi_{\phi}:A\to C^{**}\subseteq B(H),\; a\mapsto s.o. \lim\limits_{n\to\infty}(h+\frac{1}{n}1_H)^{-1}\varphi(a)$$ (s.o.-lim is the strong operator topology limit) which is complitely positve, therefore you can apply Stinespring ( https://en.wikipedia.org/wiki/Stinespring_factorization_theorem ). They say that there is a unital $C^*-$algebra $D$ containing $C^{**}$ and a $*$-homomorphism $$\rho :A\to D$$ such that $$\pi_\phi(a)=1_{C**}\rho(a)1_{C^{**}}$$ for $a\in A$.

My questions: Why does $D$ contain $C^{**}$ (With "$D$ containing $C^{**}$" they mean there is an embedding $i:C^{**}\to D$ which is a nonunital $*-$homomorphism I think)? I think $D$ corresponds to $B(K)$ as stated in Stinespring. Because $A$ and $\pi_\phi$ are unital, $H$ can be considered as a sub Hilbert space of $K$ and it is, up to an unitary, $\pi_{\phi}(a)=P_H\rho(a)_{|H}$ for all $a\in A$. Here is $P_H$ the projection of K onto H. But why is it $\pi_\phi(a)=1_{C**}\rho(a)1_{C^{**}}$ for $a\in A$?

I hope that the topic is specific enough for MO. I appreciate your help. Regards

Edit: I found a criterion for (i). For positive elements p in a $C^*$-algebra $A$ it is equivalent:
-$p$ is strictly positive
-Let $u_n=p(\frac{1}{n}+p)^{-1}$, $n\in\mathbb{N}$. It is $(u_n)$ a approximative unit for $A$

h is clearly positive and maybe with this criterion it is easier to prove (i)? I only obtain $\|hu_n-h\|\to 0$ and $\|u_nh-h\|\to 0$, but I need $\|xu_n-x\|\to 0$ and $\|u_nx-x\|\to 0$ for all $x\in A$. Here is $u_n=h(\frac{1}{n}+h)^{-1}$


1 Answer 1



Step 1: The kernel of $h$ is trivial:

Step 1.1: If $hv=0$, then $0=\langle v,hv\rangle = \langle v\phi(1)v\rangle$ and therefore $0=\langle v,\phi(c1)v\rangle$ for all $c\in\mathbb{C}$. Since $a\mapsto \langle v,\phi(a)v\rangle$ is positive, it follows $0\leq a\leq \|a\|1 \implies 0\leq\langle v,\phi(a)v\rangle\leq \langle v,\phi(\|a\|1)v\rangle=0$ so that $\langle v,\phi(a)v\rangle=0$ for all $a\in A_+$.

Step 1.2: But then $\phi(a)v = 0$ for all $a\in A_+$ since $\phi(a)\geq 0$. By linearity $\phi(a)v = 0$ for all $a\in A$.

Step 1.3: $\phi(A)v=0$ implies $Cv=0$. Since $C$ operates non-degenerated on $H$, there is no non-trivial subspace that is annihilated by $C$ (this is equivalent to operating non-degenerate). Thus $v=0$. This proves that $\ker(h)=0$.

Step 2: The support projection is $1$. Since $1=\chi_{\sigma(h)\setminus\{0\}} + \chi_{\{0\}}$, it suffices to show that $\chi_{\{0\}}(h) = 0$. But $\chi_{\{\lambda\}}(h)$ is the projection onto the $\lambda$-eigenspace of $h$ and we have just shown that the kernel of $h$ is trivial so that this projection must indeed be zero.


Now we can also prove that $\overline{hCh}=C$, because we have just shown that $\frac{h}{h+\frac{1}{n}} \xrightarrow{s.o.t} 1$ (because $\frac{x}{x+\frac{1}{x}} \xrightarrow{pointwise} \chi_{\sigma\setminus\{0\}}(x)$.)

This implies that $\frac{h}{h+\frac{1}{n}} x \frac{h}{h+\frac{1}{m}} \xrightarrow{n,m\to\infty} x$ for all $x\in C$. Since $C$ is an ideal in $C^+$, $\frac{1}{h+1/n}x\frac{1}{h+1/m} \in C$ for all $x\in C$ and therefore the left hand side expressions are elements of $hCh$ for all $n,m$.


In Springstine's theorem you have an operator $V: H\to K$ with $\pi_\phi(a)=V^\ast\rho(a)V$. Note $\rho$ is unital, since $\pi_\phi$ is defined on an unital $C^\ast$-algebra. Thus $\pi_\phi(1_A)=V^\ast \rho(1_A) V=V^\ast 1_{B(K)} V$ so that $V^\ast V = 1_{B(H)}$ since $\pi_\phi$ maps the unit to the unit. Thus $V$ is an isometry $H\hookrightarrow K$.

Via $T\mapsto VTV^\ast$ this induces a (non-unital!) injective $\ast$-homomorphism $B(H)\to B(K)$ that is used to establish an identification of operators $H\to H$ with operators $K\to K$.

(If you think of $K$ as $H\oplus H^\perp$ and $V$ as the inclusion $h\mapsto (h,0)$, then this identification maps an operator $T:H\to H$ to the operator $K\to K, (h,h')\mapsto(Th,0)$. In matrix language this is the embedding $T \mapsto \begin{pmatrix} T & 0 \\ 0 & 0 \end{pmatrix}$)

With this identification $\pi_\phi(a)=V^\ast \rho(a) V$ gets identified with $VV^\ast \rho(a) VV^\ast$ and $VV^\ast$ gets identified with $1_{B(H)} = 1_{C^{\ast\ast}}$ so that the equation reads $\pi_\phi(a) = 1_{C^{\ast\ast}} \rho(a) 1_{C^{\ast\ast}}$.

I agree, that it isn't the most clever way to simply ignore this identification.

  • 1
    $\begingroup$ @dr.mop: I've added my answer to the third question. $\endgroup$ Jun 20, 2015 at 22:20
  • $\begingroup$ Sorry, I'm repeating the proof of the theorem at the moment and I'm stuck again. Maybe I'm wrong, but why is it possible to indentify $VV^\ast$ with $1_{C^{\ast\ast}}$? This means that $V$ is unitary, but I don't think this is true in general. $\endgroup$
    – user62639
    Jul 24, 2015 at 12:41
  • 1
    $\begingroup$ Because the identification map is $T\mapsto VTV^\ast$ and $VV^\ast = V 1_{C^{\ast\ast}} V^\ast$. And no, $V$ does not need to be unitary. Note that $VV^\ast$ is not equal (with or without the identification) to $1_{L(K)}$ but only to an idempotent operator (which is the orthogonal projection onto $H$ when $H$ is considered as a subspace of $K$ via the embedding $V$) $\endgroup$ Jul 25, 2015 at 6:44
  • $\begingroup$ you are right, thanks. my question was overhasty. $\endgroup$
    – user62639
    Jul 26, 2015 at 11:26

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