41.3 Completely Positive and Completely Bounded Maps

Proof. After applying a unitisation, assume without loss of generality that $A$ is unital.

By the Gelfand-Naimark Theorem, there exists a faithful representation $(H, \pi)$ of $A$. Under the identification that $H^{n} = \complex^{n} \otimes H$, $\pi$ induces a faithful unital *-representation

whose image is a uniformly closed subalgebra of $B(H^{n})$. As such, $B(H^{n})$ induces a norm on $M_{n}(A)$, making it a $C^{*}$-algebra.

By Corollary 38.4.5, this norm is unique.$\square$

Proof. For any $n \in \natp$ and $1 \le k < n$, there is a natural *-homomorphism

such that the following diagram commutes:

(1): Since $e \otimes \text{Id}_{S}$ is the restriction of a *-homomorphism, it is positive. On the other hand, for any $x \in M_{k}(B)$, $x \ge 0$ if and only if $(e \otimes \text{Id}_{B})(x) \ge 0$. Given that $\phi_{n}$ is positive, $\phi_{k}$ must also be positive.

(2): The maps $e \otimes \text{Id}_{S}$ and $e \otimes \text{Id}_{B}$ are both isometric. As such, the norm of $\phi_{k}$ is bounded above by the norm of $\phi_{n}$.$\square$

Proof, [Lemma 3.1, Po02]. Using the Gelfand-Naimark Theorem, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$,

($\Rightarrow$): If $y \ge x^{*}x$, then by the Cauchy-Schwarz inequality,

($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$,

so $y \ge x^{*}x$.$\square$

Proof, [Proposition 3.2, Po02]. Let $x \in S$ with $\norm{x}_{A} \le 1$, then

By Lemma 41.3.6, $\norm{\phi(x)}_{B} \le 1$.$\square$

Proof. Since

and $\phi$ is unital and $2$-positive,

By Lemma 41.3.6, $\phi(x)^{*}\phi(x) \le \phi(x^{*}x)$.$\square$

Proof. ($\Leftarrow$): Suppose that $z \ge 0$ and $x - yz^{-1}y^{*} \ge 0$, then since $z$ is invertible,

Since the middle factor is positive, the given matrix is positive as well.

($\Rightarrow$): Now suppose that the given matrix is positive, then for any $\xi \in H$,

so $z \ge 0$. As such,

Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^{*} \ge 0$ as well.$\square$

Proof. After taking a unitisation, assume without loss of generality that $B$ is unital.

It is sufficient to show that $\norm{\phi}_{\text{cb}}\le \norm{\phi(1_A)}_{B}$. To this end, let $T \in M_{n}(S)$ with $\norm{T}_{M_n(S)}\le 1$, then

in $M_{2}(M_{n}(S)) = M_{2n}(S)$ by Lemma 41.3.6. Since $\phi$ is completely positive,

First suppose that $\phi_{n}(1_{M_n(A)})$ is invertible, then

By the Schur complement condition,

and $\normn{\phi_n(T)}_{M_n(B)}\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}}\le \normn{\phi(1_A)}_{B}$.

Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_{B}\psi$ is also a completely positive map. In which case,

$\square$

Proof. Let

then

is invertible, with negative determinant, so $T_{2}(C)$ is not positive.$\square$

Proof. For each $n \in \natp$, the induced map $\phi_{n}: M_{n}(A) \to M_{n}(B)$ is a *-homomorphism.$\square$

Proof. (1): For each $n \in \natp$ and $T \in M_{n}(A)$,

$\square$

Proof. (1): For any $f \in C(X; M_{n}(\complex))$,

(3): If $f \ge 0$, then there exists $g \in C(X; M_{n}(\complex))$ such that $f(x) = g^{*}(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$.

On the other hand, if $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the continuous functional calculus, $g$ is continuous from $X$ to $M_{n}(\complex)$. Therefore there exists $g \in C(X; M_{n}(\complex))$ such that $f = g^{*}g$.$\square$

Proof, [Proposition 3.8, Po02]. By Proposition 41.3.14, the norm and order relation on $C(X; M_{n}(\complex)) \iso M_{n}(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$.

Let $a = (a_{ij}) \in M_{n}(S)$. For each $\xi, \eta \in \complex^{n}$,

where

(1):

(2): If $\xi = \eta$, then

so

$\square$

Proof. Let $P \in C(X; M_{n}(\complex))$ with $P \ge 0$ and $\eps > 0$. By Proposition 6.4.5 and Theorem 5.20.10, there exists a partition of unity $\seqf{\psi_j}\subset C(X; [0, 1])$ and positive matrices $\seqf{p_j}\subset M_{n}(\complex)$ such that

for all $x \in X$. For each $1 \le j \le n$, $\phi_{n}(p_{j}\psi_{j}) = p_{j}\phi(\psi_{j}) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_{n}(P) \ge 0$.$\square$

Proof. (2) $\Rightarrow$ (3): The matrix $E = (E_{ij})$ is positive in $M_{n}(\complex)$.

(3) $\Rightarrow$ (1):

$\phi, \psi \in S(G)$, $K \subset G$ be compact, and $V \in \cn_{G}(1_{G})$ be relatively compact, then

$\square$