The following question naturally originates from [this question](https://mathoverflow.net/questions/21168/how-to-classify-the-algebras-cm)
and [this one](https://mathoverflow.net/questions/21090/smooth-gelfand-duality).


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](http://books.google.it/books?id=79bCwXbVnHAC&printsec=frontcover&dq=C-infinity+differentiable+spaces&source=bl&ots=xjtnnn8bSq&sig=BnSizsqq1t7wT8deQfeE--Ts4Ps&hl=it&ei=8w_DS92DFo7U7APZiu2tCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q&f=false). 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](http://books.google.it/books?id=79bCwXbVnHAC&printsec=frontcover&dq=C-infinity+differentiable+spaces&source=bl&ots=xjtnnn8bSq&sig=BnSizsqq1t7wT8deQfeE--Ts4Ps&hl=it&ei=8w_DS92DFo7U7APZiu2tCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q&f=false) 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? 



Why "smooth Gelfand duality" does not involve a topology on the algebras?
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.


