Jerry's Digital Garden

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

41.5 Multiplicative Domains

Definition 41.5.1 (Module Map).label Let $A, B, C$ be $C^{*}$-algebras with $C \subset A, B$ and $1_{A} = 1_{B} = 1_{C}$, then $A$ and $B$ are $C$-bimodules with respect to multiplication.

Let $\phi: A \to B$ be a linear map, then $\phi$ is a $C$-bimodule map if $\phi(c_{1}ac_{2}) = c_{1}\phi(a)c_{2}$ for all $a \in A$ and $c_{1}, c_{2} \in C$.

Lemma 41.5.2.label Let $A$ be a $C^{*}$-algebra and $x, y \in A$, then

\[\begin{bmatrix}0&x \\ x^{*}&y\end{bmatrix} \ge 0\]

if and only if $x = 0$ and $y \ge 0$.

Proof. ($\Rightarrow$): Using the Gelfand-Naimark Theorem, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$ is a $C^{*}$-subalgebra. In which case, for any $\xi, \eta \in H$,

\begin{align*}\angles{\begin{bmatrix}0&x \\ x^{*}&y\end{bmatrix}\begin{bmatrix}\xi \\ \eta\end{bmatrix}, \begin{bmatrix}\xi \\ \eta\end{bmatrix}}_{H^2}&= \angles{\begin{bmatrix}x\eta \\ x^{*}\xi + y\eta\end{bmatrix}, \begin{bmatrix}\xi \\ \eta\end{bmatrix}}_{H^2}\\&= \dpn{x^*\xi, \eta}{H}+ \dpn{x\eta, \xi}{H}+ \dpn{y\eta, \eta}{H}\\&= 2\text{Re}(\dpn{x^*\eta, \xi}{H}) + \dpn{y\eta, \eta}{H}\end{align*}

If $x \ne 0$, then there exists $\xi_{0}, \eta_{0} \in H$ such that $\dpn{x^*\eta_0, \xi_0}{H}\ne 0$. Therefore

\[\inf_{\xi \in H}\text{Re}(\dpn{x^*\eta_0, \xi}{H}) = -\infty\]

and the given matrix is not positive.$\square$

Definition 41.5.3 (Multiplicative Domains).label Let $A, B$ be unital $C^{*}$-algebras and $\phi: A \to B$ be a unital[1] $4$-positive map, then

  1. (1)

    For any $a \in A$, $\phi(a)^{*}\phi(a) = \phi(a^{*}a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space

    \[R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)}\]

    is a subalgebra of $A$, and $\phi|_{R}$ is a homomorphism.

  2. (2)

    For any $a \in A$, $\phi(a)\phi(a)^{*} = \phi(aa^{*})$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space

    \[L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)}\]

    is a subalgebra of $A$, and $\phi|_{L}$ is a homomorphism.

  3. (3)

    $C = L \cap R$ is a $C^{*}$-subalgebra of $A$ with $1_{C} = 1_{A}$, and $\phi|_{C}$ is a unital *-homomorphism.

The spaces $R$ and $L$ are the right and left multiplicative domains of $\phi$, respectively. The $C^{*}$-subalgebra $C$ is the multiplicative domain of $\phi$.

Proof, [Theorem 3.18, Po02]. (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^{*}) = \phi(a)^{*}\phi(a)$ after plugging in $b = a^{*}$.

(1, $\Rightarrow$): Suppose that $\phi(a^{*}a) = \phi(a)^{*}\phi(a)$. By the Kadison-Schwarz inequality,

\begin{align*}\phi_{2}\begin{bmatrix}a&b^{*} \\ 0&0\end{bmatrix}^{*}\phi_{2}\begin{bmatrix}a&b^{*} \\ 0&0\end{bmatrix}&\le \phi_{2}\begin{bmatrix}a^{*}a&a^{*}b^{*} \\ ba&bb^{*}\end{bmatrix} \\ \begin{bmatrix}\phi(a^{*})\phi(a)&\phi(a^{*})\phi(b) \\ \phi(b)\phi(a)&\phi(b)\phi(b^{*})\end{bmatrix}&\le \begin{bmatrix}\phi(a^{*}a)&\phi(a^{*}b) \\ \phi(ba)&\phi(bb^{*})\end{bmatrix}\end{align*}

so

\[\begin{bmatrix}0&\phi(a^{*}b^{*}) - \phi(a^{*})\phi(b) \\ \phi(b)\phi(a) - \phi(ba)&\phi(b)\phi(b^{*}) - \phi(bb^{*})\end{bmatrix} \ge 0\]

By Lemma 41.5.2, $\phi(b)\phi(a) - \phi(ba) = 0$.$\square$

  1. For without this assumption the domains will not be populated.keyboard_return

Post a Comment

Name:Email:
Please enter the tag of the current page (1KD) to post the comment.
Tag:

Direct References

Jerry's Digital Garden

BibliographyComments

Direct References