78 lines
3.9 KiB
TeX
78 lines
3.9 KiB
TeX
\section{The Gelfand-Naimark Theorem}
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\label{section:gelfand-naimark}
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\begin{theorem}[Gelfand-Naimark]
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\label{theorem:gelfand-naimark}
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Let $A$ be a commutative unital $C^*$-algebra, then the Gelfand transform
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\[
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\Gamma_A: A \to C(\Omega(A); \complex) \quad \Gamma_A(x)(\phi) = \phi(x)
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\]
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is a unital $C^*$-isomorphism.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 9.4]{Zhu}}}. ]
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By construction $\Gamma_A$ is a unital algebra homomorphism.
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To see that $\Gamma_A$ preserves involutions, let $y \in A$ be self-adjoint. By \autoref{proposition:gelfand-transform-gymnastics} and \autoref{proposition:self-adjoint-spectrum}, $\Gamma_A(y)(\Omega(A)) = \sigma_A(y) \subset \real$, 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
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\begin{align*}
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\Gamma_A(x^*) &= \Gamma_A(\text{Re}(x) - i\text{Im}(x)) \\
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&= \Gamma_A(\text{Re}(x)) - i\Gamma_A(\text{Im}(x)) \\
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&= \overline{\Gamma_A(\text{Re}(x)) + i\Gamma_A(\text{Im}(x))} = \overline{\Gamma_A(x)}
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\end{align*}
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so $\Gamma_A(x^*) = \Gamma_A(x)^*$.
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Now, for each $x \in A$, \autoref{corollary:c-star-unique-norm} and \autoref{proposition:gelfand-transform-gymnastics} imply that
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\begin{align*}
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\norm{x}_A^2 &= \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)} \\
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&= \sup\bracs{|\Gamma_A(x^*x)(\phi)|\ | \phi \in \Omega(A)} \\
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&= \sup\bracs{|\Gamma_A(x)(\phi)|^2\ | \phi \in \Omega(A)} \\
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\norm{x}_A &= \norm{\Gamma_A(x)}_u
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\end{align*}
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Thus $\Gamma_A$ is an isometry, and $\Gamma_A(A)$ is a closed subalgebra of $C(\Omega(A))$.
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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 \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
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\end{proof}
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\begin{corollary}
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\label{corollary:gelfand-naimark-converse}
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Let $A$ be a unital $C^*$-algebra, then the following are equivalent:
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\begin{enumerate}
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\item $A$ is commutative.
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\item $\Gamma_A$ is a *-isomorphism.
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\item $\Gamma_A$ is injective.
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\end{enumerate}
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\end{corollary}
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\begin{corollary}
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\label{corollary:spectrum-characterisation-iff}
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Let $A$ be a commutative unital $C^*$-algebra and $x \in A$ be normal, then:
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\begin{enumerate}
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\item $x$ is self-adjoint if and only if $\sigma_A(x) \subset \real$.
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\item $x$ is unitary if and only if $\sigma_A(x) \subset \partial B_\complex(0, 1)$.
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\item $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$.
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\item $x$ is a projection if and only if $\sigma_A(x) \subset \bracs{0,1}$.
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\end{enumerate}
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\end{corollary}
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% NEEDS WORK
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\begin{corollary}
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\label{corollary:stonean-commutative-algebra}
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Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
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\end{corollary}
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\begin{proof}
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By \autoref{theorem:gelfand-naimark}, $A$ 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 \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
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\end{proof}
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\begin{corollary}
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\label{corollary:linfinity-extremely-disconnected}
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Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected.
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\end{corollary}
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\begin{proof}
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By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected.
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\end{proof}
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