34.6 The Gelfand-Naimark Theorem
Theorem 34.6.1 (Gelfand-Naimark).label Let $A$ be a commutative unital $C^{*}$-algebra, then the Gelfand transform
is a unital $C^{*}$-isomorphism.
Proof, [Theorem II.9.4, Zhu93]. By construction $\Gamma_{A}$ is a unital algebra homomorphism.
To see that $\Gamma_{A}$ preserves involutions, let $y \in A$ be self-adjoint. By Proposition 33.8.2 and Proposition 34.3.6, $\Gamma_{A}(y)(\Omega(A)) = \sigma_{A}(y) \subset \real$, so $\Gamma_{A}(y) \in C(\Omega(A); \real)$. For any $x \in A$, write $x = \text{Re}(x) + i\text{Im}(x)$, where $\text{Re}(x)$ and $\text{Im}(x)$ are both self-adjoint, then
so $\Gamma_{A}(x^{*}) = \Gamma_{A}(x)^{*}$.
Now, for each $x \in A$, Corollary 34.3.5 and Proposition 33.8.2 imply that
Thus $\Gamma_{A}$ is an isometry, and $\Gamma_{A}(A)$ is a closed subalgebra of $C(\Omega(A))$.
Since $\Gamma_{A}(1_{A}) = 1$, $\Gamma_{A}(A)$ contains constants. As $\Gamma_{A}(A)$ separates points and is closed under complex conjugation, $\Gamma_{A}(A) = C(\Omega(A))$ by the Stone-Weierstrass Theorem.$\square$
Corollary 34.6.2.label Let $A$ be a unital $C^{*}$-algebra, then the following are equivalent:
- (1)
$A$ is commutative.
- (2)
$\Gamma_{A}$ is a *-isomorphism.
- (3)
$\Gamma_{A}$ is injective.
Corollary 34.6.3.label Let $A$ be a commutative unital $C^{*}$-algebra and $x \in A$ be normal, then:
- (1)
$x$ is self-adjoint if and only if $\sigma_{A}(x) \subset \real$.
- (2)
$x$ is unitary if and only if $\sigma_{A}(x) \subset \partial B_{\complex}(0, 1)$.
- (3)
$x$ is positive if and only if $\sigma_{A}(x) \subset [0, \infty)$.
- (4)
$x$ is a projection if and only if $\sigma_{A}(x) \subset \bracs{0,1}$.
Corollary 34.6.4.label Let $A$ be a unital $C^{*}$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
Proof. By Theorem 34.6.1, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^{*}$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the Stone-Nakano Theorem implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.$\square$
Corollary 34.6.5.label Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^{\infty}(X))$ is extremely disconnected.
Proof. By Corollary 22.6.8, $L^{\infty}(X; \real)$ is order complete. By Corollary 34.6.4, $\Omega(L^{\infty}(X))$ is extremely disconnected.$\square$
Post a Comment