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)
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)
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)
$C = L \cap R$ 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,
so
By Lemma 41.5.2, $\phi(b)\phi(a) - \phi(ba) = 0$.$\square$
- For without this assumption the domains will not be populated.keyboard_return
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