Theorem 41.1.5: Stinespring's Dilation Theorem

Theorem 41.1.5 (Stinespring’s Dilation Theorem).label Let $A$ be a unital $C^{*}$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a completely positive map, then there exists a complex Hilbert space $K$, a unital *-homomorphism $\pi: A \to B(K)$, and $V \in L(H; K)$ such that

(1)
$\normn{\phi(1_A)}_{B(H)}= \norm{V}_{L(H; K)}^{2}$.
(2)
For all $a \in A$, $\phi(a) = V^{*}\pi(a)V$.

Moreover, if $\phi$ is unital, then $V$ is an isometry, and

(3)
Under the identification of $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$.

If $\phi$ is a state, then $(K, \pi)$ is the GNS representation of $A$ associated with $\phi$.

The triple $(K, V, \pi)$ is a Stinespring representation for $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is minimal.

Proof, [Theorem 4.1, Po02]. ($K$): For each $(a \otimes \xi), (b \otimes \eta) \in K_{0}$, let

\[\dpn{a \otimes \xi, b \otimes \eta}{\phi}= \dpn{\phi(b^*a)\xi, \eta}{H}\]

and extend by linearity. For each $\seqf{(a_j \otimes \xi_j)}\subset A \otimes H$, let

\[T = \sum_{i, j = 1}^{n} E_{ij}\otimes a_{i}^{*}a_{j} = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^{*}\braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}\in M_{n}(A)\]

Then $T \ge 0$. As $\phi$ is completely positive,

\[\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_{\phi}= \angles{\phi_n(T)\begin{bmatrix}\xi_{1} \\ \vdots \\ \xi_{n}\end{bmatrix}, \begin{bmatrix}\xi_{1} \\ \vdots \\ \xi_{n}\end{bmatrix}}_{H^n}\ge 0\]

so $\dpn{u, u}{\phi}$ is a pseudo inner product. Let

\[N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}\]

then by the Cauchy-Schwarz inequality,

\[N = \bracsn{u \in A \otimes H| \dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}\]

Let $K_{0} = (A \otimes H)/N$ and $K$ be the completion of $K_{0}$, then $K$ is a Hilbert space.

($\pi$): For each $a \in A$, let

\[\pi_{0}(a): K_{0} \to K_{0} \quad \pi_{0}(a)(b \otimes \xi + N) = (ab) \otimes \xi + N\]

For any $b \in A$, $b^{*}a^{*}ab \le b^{*}\norm{a}_{A}^{2}b = \norm{a}_{A}^{2}b^{*}b$. As such, for any $u = \sum_{j = 1}^{n} b_{j} \otimes \xi_{j} \in A \otimes H$,

\begin{align*}\angles{\pi(a)u, \pi(a)u}_{\phi}&= \sum_{i, j = 1}^{n} \dpn{\pi(a)(b_i \otimes \xi_i), \pi(a)(b_j \otimes \xi_j)}{\phi}\\&= \sum_{i, j = 1}^{n} \dpn{(ab_i) \otimes \xi_i, (ab_j) \otimes \xi_j}{\phi}= \sum_{i, j = 1}^{n} \dpn{\phi(b_j^*a^*ab_i) \otimes \xi_i,\xi_j}{H}\\&\le \sum_{i, j = 1}^{n} \norm{a^*a}_{A}\dpn{\phi(b_j^*b_i)\xi_i, \xi_j}{H}= \norm{a}_{A}^{2} \dpn{u, u}{\phi}\end{align*}

Thus $\pi_{0}(a)$ is well-defined with $\norm{\pi_0(a)}_{B(K_0)}\le \norm{a}_{A}$, and extends into a continuous linear operator $\pi(a) \in B(K)$. The map $\pi: A \to B(K)$ is then a *-homomorphism.

($V$): Let $V: H \to K$ be defined by $\xi \mapsto 1_{A} \otimes \xi + N$.

(1): For each $\xi \in H$,

\[\dpn{V\xi, V\xi}{K}= \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi}= \dpn{\phi(1_A)\xi, \xi}{H}\]

so $\normn{\phi(1_A)}_{B(H)}= \norm{V}_{L(H; K)}^{2}$. In particular, if $\phi$ is unital, then $V$ is an isometry by polarisation.

(2): Finally, for any $\xi, \eta \in H$,

\begin{align*}\dpn{V^*\pi(a)V\xi, \eta}{H}&= \dpn{\pi(a)V\xi, V\eta}{K}= \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi}\\&= \dpn{\phi(a)\xi, \eta}{H}\end{align*}

$\square$

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