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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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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.2012-12-31
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    The last statement is substantially easier to prove than Gelfand-Naimark. It is actually an exercise in Atiyah-MacDonald (which is solved here: http://qchu.wordpress.com/2009/11/24/spectra-of-rings-of-continuous-functions/).2012-12-31
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    @K.Ghosh: I apologize that I took so long to reply. I was actually trying to make my answer more complete.2012-12-31
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    @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!2012-12-31
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    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](http://math.stackexchange.com/q/226736/) for references.2012-12-31
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    @Martin: Automatic-continuity results are indeed one of my areas of interest. Anyway, the Gelfand-Kolmogorov result is quite non-trivial, and I don't think I could have included a proof here.2012-12-31
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    @K.Ghosh: I see that you have deleted your last comment. One can view the Fourier transform as a special case of the Gelfand transform. You can refer to the following Wikipedia article: http://en.wikipedia.org/wiki/Gelfand_representation.2012-12-31
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    Sure, I didn't mean to say that you should have included a proof. I wanted to clarify that the statement asked in the OP is true. Actually, if you are willing to restrict to real scalars and assuming that the isomorphism is an algebra isomorphism then the proof is of about the same level of difficulty as the one you present. The main point is that the order structure is determined by the algebra structure $f \geq 0$ iff $f = g^2$ and that $-\lVert f\rVert 1_X \leq f \leq \lVert f \rVert 1_X$. The proof on p. 365f in the notes of Nate's answer in the linked thread fits easily on one page.2012-12-31
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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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    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](http://math.stackexchange.com/q/226736/) 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.2012-12-31
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    @Martin: thanks for the correction. Admittedly I did not read the Wikipedia article too closely...2012-12-31