Question

Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$-algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, are there other ones? Does the smooth structure of $M$ endow $A$ with additional structure such that $M$ can be completely recovered from $A$? In other words, is there some kind of smooth Gelfand-duality?

Answer 1

Perhaps I should post this as an answer (even if I don't really know that theory): in

Juan A. Navarro González & Juan B. Sancho de Salas, C^∞-Differentiable Spaces, LNM 1824 Google Books Preview

a theory of "$C^{\infty}$ -differentiable spaces" is developped, and it would be something like the smooth analog to (possibly singular and nonreduced) complex analytic spaces.

Answer 2

I suspect you may be interested in the following nlab page: http://ncatlab.org/nlab/show/smooth+algebra. In particular, note the section on smooth function algebras on smooth manifolds and the remark immediately preceding it:

By the properties of $C^∞(X)$ for $X$ a smooth manifold discussed below, the $ℝ$-points of $C^∞(X)$ are precisely the ordinary points of the manifold $X$.

Answer 3

You don't need any additional (topological or whatsoever) structure on the $\mathbb{R}$-algebra $A=C^\infty(M)$. As an algebra alone it allows you to reconstruct the manifold completely. (And not just as a set or topological space, but with it's smooth structure).

This had been answered in other questions on MO, unfortunately I don't remember where. Anyway a good reference is the book: Jet Nestruev, Smooth manifolds and observables, as well as the above mentioned book C-infinty differentiable spaces.

Answer 4

The functor from the category of smooth manifolds to to the category of real algebras that sends a manifold M to C^∞(M) is fully faithful, hence it is an equivalence of categories of smooth manifolds and real algebras of certain type. The inverse functor sends a real algebra A to the real spectrum of A (homomorphisms of algebras from A to R) equipped with the standard Zariski topology (every ideal corresponds to a closed set) and the obvious structure sheaf of smooth functions (every element of A gives a function on the real spectrum of A).

The construction also works for manifolds with boundaries and/or corners.

Here is a link to a related question: http://mathoverflow.net/questions/14877/how-much-of-differential-geometry-can-be-developed-entirely-without-atlases/14890#14890

Answer 5

If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$.

A derivation in this case is an $\mathbb R$-linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$.

I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to prove that $C(M)$ is the closure of $C^\infty(M)$ under the compact-open topology. The differentiable manifold structure is given by the derivations.