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/Part 7: Operator Algebras/Chapter 41: Operator Systems

41.1 Dilations

Definition 41.1.1 (Compression).label Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a compression of $T$ if $S = PT|_{K}$.

Definition 41.1.2 (Dilation).label Let $H$ be a complex Hilbert space and $T \in B(H)$, then a dilation of $T$ is a pair $(K, S)$ where:

  1. (1)

    $K$ is a Hilbert space containing $H$.

  2. (2)

    $S \in B(K)$.

  3. (3)

    If $P \in B(K)$ is the orthogonal projection onto $H$, then $T = PS|_{H}$.

If $T^{n} = PS^{n}|_{H}$ for all $n \in \natz$, then $S$ is a power dilation of $T$.

Lemma 41.1.3.label Let $H$ be a complex Hilbert space, $R \in B(H)$, $(K, S)$ be a dilation of $R$, and $(L, T)$ be a dilation of $S$, then $(L, T)$ is a dilation of $R$.

Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$.

Theorem 41.1.4 (Sz.-Nagy’s Dilation Theorem).label Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)}\le 1$, then there exists a unitary power dilation of $T$.

Proof, [Theorem 1.1, Po02]. Using Lemma 41.1.3 it is sufficient to show the following statements:

  1. (i)

    For any $T \in B(H)$ with $\norm{T}_{B(H)}\le 1$, there exists an isometric power dilation of $T$.

  2. (ii)

    For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$.

(i): Since $\norm{T}_{B(H)}\le 1$, $1 - T^{*}T \ge 0$, and $(1 - T^{*}T)^{1/2}$ is well-defined. Let $D_{T} = (1 - T^{*}T)^{1/2}$, and

\[V: l^{2}(\natp; H) \to l^{2}(\natp; H) \quad (Vx)_{n} = \begin{cases}Tx_{1}&n = 1 \\ D_{T}x_{1}&n = 2 \\ x_{n-1}&n \ge 3\end{cases}\]

then for each $x \in H$,

\begin{align*}\norm{Tx}_{H}^{2} + \norm{D_Tx}_{H}^{2}&= \dpn{T^*Tx, x}{H}+ \dpn{(1 - T^*T)x, x}{H}\\&= \dpn{x, x}{H}= \norm{x}_{H}^{2}\end{align*}

Thus for each $x = \seq{x_n}\in l^{2}(H)$,

\begin{align*}\norm{Vx}_{l^2(\natp; H)}^{2}&= \norm{Tx_1}_{H}^{2} + \norm{D_Tx_1}_{H}^{2} + \sum_{n = 3}^{\infty} \norm{x_{n-1}}_{H}^{2} \\&= \norm{x_1}_{H}^{2} + \sum_{n = 2}^{\infty} \norm{x_{n}}_{H}^{2} = \norm{x}_{l^2(\natp; H)}^{2}\end{align*}

So $V$ is an isometry. Moreover, $[V^{n}(x, 0, \cdots)]_{1} = T^{n}x$ for all $n \in \natz$. Therefore under the identification that $H = \bracsn{(x, 0, \cdots)|x \in H}\subset l^{2}(\natp; H)$, $V$ is an isometric power dilation of $T$.

(ii): Let $P = I_{H} - VV^{*}$ be the projection onto $V(H)^{\perp}$, and $U \in B(H^{2})$ be defined by

\[U = \begin{bmatrix}V&P \\ 0&V^{*}\end{bmatrix}\]

then

\begin{align*}UU^{*}&= \begin{bmatrix}V&P \\ 0&V^{*}\end{bmatrix} \cdot \begin{bmatrix}V^{*}&0 \\ P&V\end{bmatrix} = \begin{bmatrix}VV^{*} + P&PV \\ V^{*}P&V^{*}V\end{bmatrix} = I_{H^2}\\ U^{*}U&= \begin{bmatrix}V^{*}&0 \\ P&V\end{bmatrix} \cdot \begin{bmatrix}V&P \\ 0&V^{*}\end{bmatrix} = \begin{bmatrix}V^{*}V&V^{*}P \\ PV&VV^{*} + P\end{bmatrix} = I_{H^2}\\\end{align*}

so $U$ is unitary. Moreover, for each $n \in \natp$,

\[U^{n} = \begin{bmatrix}V&P \\ 0&V^{*}\end{bmatrix}^{n} = \begin{bmatrix}V^{n}&\cdots \\ 0&(V^{*})^{n}\end{bmatrix}\]

Under the identification that $H = \bracsn{(x, 0)|x \in H}\subset H^{2}$, $QU^{n}|_{H}= V^{n}$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$.$\square$

Theorem 41.1.5 (Von Neumann’s Inequality).label Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)}\le 1$, then for any $p \in \complex[x]$,

\[\norm{p(T)}_{B(H)}\le \sup_{z \in B_\complex(0, 1)}|p(z)|\]

Proof, [Corollary 1.2, Po02]. By Sz.-Nagy’s Dilation Theorem, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^{n} = P_{H}U^{n}|_{H}$ for all $n \in \natp$, where $P_{H}$ is the orthogonal projection onto $H$.

Thus for any $p \in \complex[x]$, $p(T) = P_{H}p(U)|_{H}$. As $U$ is unitary, the continuous functional calculus shows that $\norm{p(U)}_{B(K)}\le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)}\le \sup_{z \in B_\complex(0, 1)}|p(z)|$.$\square$

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