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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?

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First of all, what do you mean by "the spectrum of the $\mathbb{R}$-algebra $A$"? I don't think it's useful to take the prime spectrum in the sense of (algebraic) scheme theory. On the other hand, if you considered $C^{\infty}(M)$ as a $C^{*}$-algebra with trivial involution, perhaps you would get some compact Housdorff topological space different from $M$ ?... –  Qfwfq Apr 12 '10 at 12:03

5 Answers 5

up vote 3 down vote accepted

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.

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this is exactly what I'm looking for. so let me answer my question: yes there is a smooth Gelfand duality and the theory is very nice :-) –  Martin Brandenburg Apr 12 '10 at 16:32

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 $X$.

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nice. I'll look for a proof.. –  Martin Brandenburg Apr 12 '10 at 12:18
    
The key reference is Moedijk and Reyes Models for Smooth Infinitesimal Analysis. (Same Reyes as the reference that "unknown" mentions.) –  Loop Space Apr 12 '10 at 12:41

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.

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I had a look to Nestruev's book a lot of time ago. I don't remember: didn't you need to consider closed (w.r.t. some -perhaps Fréchet- topology on $A$) maximal ideals to have the points of $M$? –  Qfwfq Apr 12 '10 at 12:47
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To recover the points of M you should not look at all maximal ideals but only those whose residue field are real numbers. Those are called the $\mathbb{R}$-points as Andrew Stacey wrote. But no topology on A is needed. –  Michael Bächtold Apr 12 '10 at 14:02
    
Ok, thank you. –  Qfwfq Apr 12 '10 at 16:24

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: How much of differential geometry can be developed entirely without atlases?

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yes this is covered in the book about C^infty-differentiable spaces cited above. –  Martin Brandenburg Apr 13 '10 at 0:00
    
Does that mean we can totally change geometic problems into algebraic ones, like we did in algebraic geometry? Is there any difference between these?Thank you! –  HYYY Aug 28 '10 at 13:49
    
@HYYY: Yes. But rings of smooth functions are much bigger than typical rings occuring in algebraic geometry, and we use different tools to study them. Perhaps the first important difference occurs in the definition of the cotangent bundle: In algebraic geometry you use Kähler differentials, but for smooth manifolds you use the dual of the module of derivations. –  Dmitri Pavlov Aug 28 '10 at 14:30

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.

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