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

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

2010-08-28T12:50:59Z

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?

Answer by Robin Chapman for The ring $C^{\infty}(M)$?

2010-08-28T13:06:07Z

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.

Answer by MTS for The ring $C^{\infty}(M)$?

2010-08-28T18:23:01Z

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.