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The following question naturally originates from this question and this one.

While the usual $C^{0}$ Gelfand duality involves a topology on the function algebras considered (it relates compact Hausdorff topological spaces to unital $C^{*}$-algebras, which in particular are Banach algebras), why the "smooth Gelfand duality" seems, according to what I understood from the above questions, to see only the "pure" algebraic structure of certain algebras over $\mathbb{R}$ ?

Edit: I've just read the introduction of this. The topology actually enters the picture, but not in the form of a structure of topological algebra on the function spaces that locally model those $C^{\infty}$-differentiable spaces; it enters the picture when defining a "differentiable algebra" as the quotient of the algebra of smooth functions on $\mathbb{R}^n$ by a Fréchet-closed ideal.

But a question still stands: would it be possible to define compact Hausdorff topological spaces in the analogous way? Perhaps the answer is "no because of a lack of a universal local model $C^{0}(...)$", but "yes in the case of topological manifolds". Does it make sense?

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In the case of Gelfand duality, one also does not need topology in the following sense: all the maximal ideals are closed, and they all give points of the corresponding topological space. So is the situation really so different? –  Emerton Apr 13 '10 at 17:37
    
Well, technically a $C^{*}$-algebra is, in particular, a topological algebra, while the $\mathbb{R}$-algebras that are local models for C-infty differentiable spaces are just algebras with no further structure. Or maybe we can consider the latter as endowed with a topology induced by the Fréchet topology on $C^{\infty}(\mathbb{R}^n)$ ?... –  Qfwfq Apr 13 '10 at 19:13
    
Well a C*-algebra is, in a sense, just an algebra. That is, a -algebra on which A->sqrt(spectral radius of A A) is a complete Banach-algebra norm. –  George Lowther Apr 13 '10 at 21:21
    
@Emerton: but in order to recover the Gelfand topology on the maximal ideal space of a Banach algebra, doesn't the topology of the algebra enter here? or am I misunderstanding your point. –  Yemon Choi Apr 13 '10 at 23:02
    
@yenon: the topology is not involved (but useful for arguments). –  Martin Brandenburg Apr 14 '10 at 8:51
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1 Answer

up vote 3 down vote accepted

We don't need a topology because the proofs show that the space can be recovered without a topology. I think it's just that simple. Perhaps this is not satisfactory for you?

Continuing George's comment, there is an important result in the theory of $C^\*$-algebras: the norm is unique. The reason is that every $C^\*$-homomorphism is of norm $\leq 1$. In particular, the category of $C^\*$-algebras is a full subcategory of the category of $\mathbb{C}$-algebras with involution and basically I could also ask for a pure algebraic characterization of the essential image... In particular, the category of compact hausdorff spaces is a full subcategory of the opposite category of commutative $\mathbb{C}$-algebras with involution. Thus we have the same situation as in the smooth Gelfand duality.

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I see . –  Qfwfq Apr 14 '10 at 7:41
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