Gelfand-Naimark Theorem

The Gelfand–Naimark Theorem states that an arbitrary C*-algebra $A$ is isometrically *-isomorphic to a C*-algebra of bounded operators on a Hilbert space. There is another version, which states that if $X$ and $Y$ are compact Hausdorff spaces, then they are homeomorphic iff $C(X)$ and $C(Y)$ are isomorphic as rings. Are these two related anyway?

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2 Answers

The first result that you stated is commonly known as the Gelfand-Naimark-Segal Theorem. It is true for arbitrary C*-algebras, and its proof employs a technique known as the GNS-construction. This technique basically allows one to construct a Hilbert space $\mathcal{H}$ from a given C*-algebra $\mathcal{A}$ such that $\mathcal{A}$ can be isometrically embedded into $B(\mathcal{H})$ as a C*-subalgebra.

The Gelfand-Naimark Theorem, on the other hand, states that every commutative C*-algebra $\mathcal{A}$, whether unital or not, is isometrically *-isomorphic to ${C_{0}}(X)$ for some locally compact Hausdorff space $X$. When $X$ is compact, ${C_{0}}(X)$ and $C(X)$ become identical.

Note: The assumption of commutativity is essential for stating the Gelfand-Naimark Theorem. This is because we cannot realize a non-commutative C*-algebra as the commutative C*-algebra ${C_{0}}(X)$, for some locally compact Hausdorff space $X$.

What follows is a statement of the Gelfand-Naimark Theorem, with the utmost level of precision.

Gelfand-Naimark Theorem Let $\mathcal{A}$ be a commutative C*-algebra. If $\mathcal{A}$ is unital, then $\mathcal{A}$ is isometrically *-isomorphic to $C(X)$ for some compact Hausdorff space $X$. If $\mathcal{A}$ is non-unital, then $\mathcal{A}$ is isometrically *-isomorphic to ${C_{0}}(X)$ for some non-compact, locally compact Hausdorff space $X$.

This result is often first established for the case when $\mathcal{A}$ is unital. One basically tries to show that the compact Hausdorff space $X$ can be taken to be the set $\Sigma$ of all non-zero characters on $\mathcal{A}$, where $\Sigma$ is equipped with a special topology. Here, a character on $\mathcal{A}$ means a linear functional $\phi: \mathcal{A} \to \mathbb{C}$ satisfying $\phi(xy) = \phi(x) \phi(y)$ for all $x,y \in \mathcal{A}$. A rough outline of the proof is given below.

• Show that every character has sup-norm $\leq 1$. Hence, $\Sigma \subseteq {\overline{\mathbb{B}}}(\mathcal{A}^{*})$, where ${\overline{\mathbb{B}}}(\mathcal{A}^{*})$ denotes the closed unit ball of $\mathcal{A}^{*}$.

• Equip ${\overline{\mathbb{B}}}(\mathcal{A}^{*})$ with the subspace topology inherited from $(\mathcal{A}^{*},\text{wk}^{*})$, where $\text{wk}^{*}$ denotes the weak*-topology. By the Banach-Alaoglu Theorem, ${\overline{\mathbb{B}}}(\mathcal{A}^{*})$ then becomes a compact Hausdorff space.

• Prove that $\Sigma$ is a weak*-closed subset of $\left( {\overline{\mathbb{B}}}(\mathcal{A}^{*}),\text{wk}^{*} \right)$. Hence, $\Sigma$ becomes a compact Hausdorff space with the subspace topology inherited from $\left( {\overline{\mathbb{B}}}(\mathcal{A}^{*}),\text{wk}^{*} \right)$.

• For each $a \in \mathcal{A}$, define $\hat{a}: \Sigma \to \mathbb{C}$ by $\hat{a}(\phi) \stackrel{\text{def}}{=} \phi(a)$ for all $\phi \in \Sigma$. We call $\hat{a}$ the Gelfand-transform of $a$.

• Show that $\hat{a}$ is a continuous function from $(\Sigma,\text{wk}^{*})$ to $\mathbb{C}$ for each $a \in \mathcal{A}$. In other words, $\hat{a} \in C((\Sigma,\text{wk}^{*}))$ for each $a \in \mathcal{A}$.

• Finally, prove that $a \longmapsto \hat{a}$ is an isometric *-isomorphism from $\mathcal{A}$ to $C((\Sigma,\text{wk}^{*}))$.

Let us now take a look at the following theorem, which the OP has asked about.

If $X$ and $Y$ are compact Hausdorff spaces, then $X$ and $Y$ are homeomorphic if and only if $C(X)$ and $C(Y)$ are isomorphic as C*-algebras (not only as rings).

One actually does not require the Gelfand-Naimark Theorem to prove this result. Let us see a demonstration.

Proof

• The forward direction is trivial. Take a homeomorphism $h: X \to Y$, and define $h^{*}: C(Y) \to C(X)$ by ${h^{*}}(f) \stackrel{\text{def}}{=} f \circ h$ for all $f \in C(Y)$. Then $h^{*}$ is an isometric *-isomorphism.

• The other direction is non-trivial. Let $\Sigma_{X}$ and $\Sigma_{Y}$ denote the set of non-zero characters of $C(X)$ and $C(Y)$ respectively. As $C(X)$ and $C(Y)$ are isomorphic C*-algebras, it follows that $\Sigma_{X} \cong_{\text{homeo}} \Sigma_{Y}$. We must now prove that $X \cong_{\text{homeo}} \Sigma_{X}$. For each $x \in X$, let $\delta_{x}$ denote the Dirac functional that sends $f \in C(X)$ to $f(x)$. Next, define a mapping $\Delta: X \to \Sigma_{X}$ by $\Delta(x) \stackrel{\text{def}}{=} \delta_{x}$ for all $x \in X$. Then $\Delta$ is a homeomorphism from $X$ to $(\Delta[X],\text{wk}^{*})$ (this follows from the fact that $X$ is a completely regular space). We will be done if we can show that $\Delta[X] = \Sigma_{X}$. Let $\phi \in \Sigma_{X}$. As $\phi: C(X) \to \mathbb{C}$ is surjective (as it maps the constant function $1_{X}$ to $1$), we see that $C(X)/\ker(\phi) \cong \mathbb{C}$. According to a basic result in commutative ring theory, $\ker(\phi)$ must then be a maximal ideal of $C(X)$. As such, $$\ker(\phi) = \{ f \in C(X) ~|~ f(x_{0}) = 0 \}$$ for some $x_{0} \in X$ (in fact, all maximal ideals of $C(X)$ have this form; the compactness of $X$ is essential). By the Riesz Representation Theorem, we can find a regular complex Borel measure $\mu$ on $X$ such that $\phi(f) = \displaystyle \int_{X} f ~ d{\mu}$ for all $f \in C(X)$. As $\phi$ annihilates all functions that are vanishing at $x_{0}$, Urysohn's Lemma implies that $\text{supp}(\mu) = \{ x_{0} \}$. Hence, $\phi = \delta_{x_{0}}$, which yields $\Sigma_{X} \subseteq \Delta[X]$. We thus obtain $\Sigma_{X} = \Delta[X]$, so $X \cong_{\text{homeo}} \Sigma_{X}$. Similarly, $Y \cong_{\text{homeo}} \Sigma_{Y}$. Therefore, $X \cong_{\text{homeo}} Y$ because $$X \cong_{\text{homeo}} \Sigma_{X} \cong_{\text{homeo}} \Sigma_{Y} \cong_{\text{homeo}} Y.$$

We actually have the following general categorical result.

Let $\textbf{CompHaus}$ denote the category of compact Hausdorff spaces, where the morphisms are proper continuous mappings. Let $\textbf{C*-Alg}$ denote the category of commutative unital C*-algebras, where the morphisms are unit-preserving *-homomorphisms. Then there is a contravariant functor $\mathcal{F}$ from $\textbf{CompHaus}$ to $\textbf{C*-Alg}$ such that

(1) $\mathcal{F}(X) = C(X)$ for all $X \in \textbf{CompHaus}$, and

(2) $\mathcal{F}(h) = h^{*}$ for all proper continuous mappings $h$. If $h: X \to Y$, then $h^{*}: C(Y) \to C(X)$, which highlights the contravariant nature of $\mathcal{F}$.

Furthermore, $\mathcal{F}$ is a duality (i.e., contravariant equivalence) of categories.

The role of the Gelfand-Naimark Theorem in this result is to prove that $\mathcal{F}$ is an essentially surjective functor, i.e., every commutative C*-algebra can be realized as $\mathcal{F}(X) = C(X)$ for some $X \in \textbf{CompHaus}$.

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where can I find the proof of last statement. – K.Ghosh Dec 31 '12 at 6:41

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– Qiaochu Yuan Dec 31 '12 at 6:49
@K.Ghosh: I apologize that I took so long to reply. I was actually trying to make my answer more complete. – Haskell Curry Dec 31 '12 at 9:51
@Qiaochu: If I had seen that you had already posted the link, I would have spared myself the pain of having to provide so much detail in my posted solution. Anyway, thanks for providing the link! – Haskell Curry Dec 31 '12 at 9:53

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– Martin Dec 31 '12 at 10:49
show 3 more comments

The second theorem you describe is the Banach-Stone theorem. The commutative Gelfand-Naimark theorem says something stronger, namely that every commutative (unital) C*-algebra is of the form $C(X)$ for some compact Hausdorff space $X$. The strongest version of the theorem says that the functor $X \mapsto C(X)$ is a contravariant equivalence of categories.

I don't know the history here, but both Gelfand-Naimark theorems are "Cayley theorems" for C*-algebras, one saying that commutative C*-algebras can be represented faithfully as function spaces and the other saying that noncommutative C*-algebras can be represented faithfully as spaces of operators.

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– Martin Dec 31 '12 at 10:35
@Martin: thanks for the correction. Admittedly I did not read the Wikipedia article too closely... – Qiaochu Yuan Dec 31 '12 at 11:37
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