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Let $M$ be a smooth paracompact manifold. I think that the ring $C^{\infty}(M)$ contains many (possibly almost all?) geometric or topological information about $M$.

(e.g. Let $E$ be a vector bundle over $M$,$\Gamma(E)$ be a set of smooth section of $E$. Then, $\Gamma(E)$ is a $C^\infty(M)$-module. (Actually, I think $\Gamma(E)$ is projective $C^\infty(M)$-module because every a short exact sequence of vector bundle splits.))

But I have a feeling that $C^\infty(M)$ is too large to change the problem of Manifold theory into an algebraic problem or Ring theoretic problem.

Are there any well-known concrete description about the ring $C^\infty(M)$ for some manifold $M$ with simple topology?

flag	asked Aug 28 2010 at 12:50 Topologieee 493●2●7

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To what extent to the responses to these questions answer the one you've asked: mathoverflow.net/questions/21090/… mathoverflow.net/questions/21168/… mathoverflow.net/questions/21217/… – Emerton Aug 28 2010 at 13:02

Oh, Thank you for helpful links. – Topologieee Aug 28 2010 at 13:23

Robin Chapman · Accepted Answer · 2010-08-28 13:06:07Z UTC

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You are correct: $C^\infty(M)$ does contain all the geometry and topology of $M$ (at least when it is considered as an $\mathbb{R}$-algebra). For example when $M$ is compact the points of $M$ correspond to the maximal ideals of $C^\infty(M)$ (this is quite easy to prove). If $M$ is not compact there are maximal ideals $I$ not corresponding to points, but these can be distinguished since $C^\infty(M)/I$ is a proper extension of $\mathbb{R}$ for such $I$.

See the book Smooth Manifolds and Observables by "Jet Nestruev" to see these ideas fully worked out.

answered Aug 28 2010 at 13:06

Robin Chapman
14.5k●2●30●50

I'm looking Chapter 10,11 of Smooth manifolds and observables, the book you mentioned. It contains the information what I'm looking for and interested. Thank you – Topologieee Aug 28 2010 at 13:27

and I think, that tere is also a topology on the space of all maximal ideals, which makes (in the compact case) the manifold homeomorphic to the space of all maximal ideals. – HenrikRüping Aug 28 2010 at 13:39

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Incidentally, if you allow manifolds that are "too big" (not second countable...), then you can no longer distinguish points by looking at the residue fields. In particular, consider a zero-dimensional manifold (i.e. a set) with cardinality strictly greater than the cardinality of the reals; then there are points in max-spec with residue field precisely $\mathbb R$ that do not correspond to evaluating at a point in your set. – Theo Johnson-Freyd Aug 28 2010 at 16:55

MTS · Answer 2 · 2010-08-28 18:23:01Z UTC

Connes proved an analogue of the Hochschild-Kostant-Rosenberg theorem which asserts that for a compact manifold $M$, there is a canonical isomorphism between the continuous Hochschild cohomology groups of $C^\infty(M)$ and the spaces of de Rham currents on $M$, which are the dual to differential forms. You can find this in Chapter 8 of the book Elements of Noncommutative Geometry, by Varilly, Gracia-Bondia, and Figueroa.

Also, in the paper On the Spectral Characterization of Manifolds, Connes shows how to reconstruct a manifold from a commutative spectral triple, i.e. take a commutative pre-$C^*$-algebra $A$ plus some extra data and build a manifold $M$ from it such that $A \simeq C^\infty(M)$. This is really more than you were asking for but I thought it might be interesting nonetheless.

The ring $C^{\infty}(M)$?

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