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?

topologicalalgebra, 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-algebra on which A->sqrt(spectral radius of AA) is a complete Banach-algebra norm. – George Lowther Apr 13 '10 at 21:21