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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 2

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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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  • $\begingroup$ where can I find the proof of last statement. $\endgroup$
    – Koushik
    Dec 31, 2012 at 6:41
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    $\begingroup$ The last statement is substantially easier to prove than Gelfand-Naimark. It is actually an exercise in Atiyah-MacDonald (which is solved here: qchu.wordpress.com/2009/11/24/…). $\endgroup$ Dec 31, 2012 at 6:49
  • $\begingroup$ @K.Ghosh: I apologize that I took so long to reply. I was actually trying to make my answer more complete. $\endgroup$ Dec 31, 2012 at 9:51
  • $\begingroup$ @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! $\endgroup$ Dec 31, 2012 at 9:53
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    $\begingroup$ It suffices to assume that $C(X)$ and $C(Y)$ are isomorphic as rings to conclude that $X$ and $Y$ are homeomorphic. This is a bit harder to prove than the Gelfand-Naimark theorem since there is no simple automatic continuity result coming from assuming the homomorphism to be a *-homomorphism. The result is due to Gelfand and Kolmogorov. See here for references. $\endgroup$
    – Martin
    Dec 31, 2012 at 10:49
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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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    $\begingroup$ The Banach-Stone theorem says that $C(X)$ and $C(Y)$ are isometrically isomorphic as Banach spaces iff $X$ and $Y$ are homeomorphic. The subtlety is that there is no a priori assumption that there is an isometry which is a ring homomorphism. There's another theorem by Gelfand and Kolmogorov which says $C(X)$ and $C(Y)$ are isomorphic as rings iff $X$ and $Y$ are homeomorphic. There's yet another variant by Kaplansky showing that $C(X)$ and $C(Y)$ are isomorphic as Banach lattices iff $X$ and $Y$ are homeomorphic. $\endgroup$
    – Martin
    Dec 31, 2012 at 10:35
  • $\begingroup$ @Martin: thanks for the correction. Admittedly I did not read the Wikipedia article too closely... $\endgroup$ Dec 31, 2012 at 11:37

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