36.5 *-Homomorphisms
Definition 36.5.1 (*-Homomorphism).label Let $A, B$ be involutive algebras over $\complex$ and $\phi: A \to B$, then $\phi$ is a *-homomorphism if:
- (SH1)
For each $x, y \in A$ and $\lambda \in \complex$, $\phi(\lambda x + y) = \lambda \phi(x) + \phi(y)$.
- (SH2)
For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$.
- (SH3)
For each $x \in A$, $\phi(x^{*}) = \phi(x)^{*}$.
If $A$ and $B$ are unital, then $\phi$ is unital if:
- (U)
$\phi(1_{A}) = 1_{B}$.
Proposition 36.5.2.label Let $A, B$ be unital $C^{*}$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$,
- (1)
$\sigma_{B}(\phi(x)) \subset \sigma_{A}(x)$.
- (2)
$\norm{\phi(x)}_{B} \le \norm{x}_{A}$.
Proof. (1): Since $\phi$ is unital, $\phi(G(A)) \subset G(B)$, so $\sigma_{B}(\phi(x)) \subset \sigma_{A}(x)$.
(2): By (1) and Corollary 36.4.5,
$\square$
Theorem 36.5.3.label Let $A, B$ be unital $C^{*}$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed.
Proof, [Theorem 11.1, Zhu93]. Let $y \in \ol{\Phi(A)}\cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_{B} \le \norm{y}_{B}/2$. Let
then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_{B} \le \norm{y}_{B} + \norm{y - \Phi(x)}_{B} \le 2\norm{y}_{B}$, $\sigma_{B}(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the continuous functional calculus, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the Spectral Mapping Theorem, $\sigma_{A}(f(x)) = f(\sigma_{A}(x))$. By Theorem 36.4.3, $\norm{f(x)}_{A} = [f(x)]_{sp}\le \norm{f}_{u} = 2\norm{y}_{F}$.
The above setup implies that for every $y \in \ol{\Phi(A)}\cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B}\le \norm{y}_{B}/2$, and $\norm{z}_{A} \le 2\norm{y}_{B}$. By the method of successive approximations, $\phi(A_{sa}) = \ol{\Phi(A)}\cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.$\square$
Definition 36.5.4 (Representation of $C^{*}$-Algebra).label Let $A$ be a $C^{*}$-algebra, then a representation of $A$ is a pair $(H, \pi)$, where $H$ is a Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism.
Definition 36.5.5 (Unitary Equivalence).label Let $A$ be a $C^{*}$-algebra and $(H_{1}, \pi_{1}), (H_{2}, \pi_{2})$ be representations of $A$, then $(H_{1}, \pi_{1})$ and $(H_{2}, \pi_{2})$ are unitarily equivalent if there exists an isometry $U \in L(H_{1}; H_{2})$ such that the following diagram commutes
for all $x \in A$.
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