The ring $C^{\infty}(M)$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:53:27Z http://mathoverflow.net/feeds/question/36965 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36965/the-ring-c-inftym The ring $C^{\infty}(M)$? Topologieee 2010-08-28T12:50:59Z 2010-08-28T18:23:01Z <p>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 $M$. </p> <p>(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.)) </p> <p>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. </p> <p>Are there any well-known concrete description about the ring $C^\infty(M)$ for some manifold $M$ with simple topology? </p> http://mathoverflow.net/questions/36965/the-ring-c-inftym/36966#36966 Answer by Robin Chapman for The ring $C^{\infty}(M)$? Robin Chapman 2010-08-28T13:06:07Z 2010-08-28T13:06:07Z <p>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 $I$.</p> <p>See the book <a href="http://books.google.co.uk/books?id=N5mHmFiovgkC&amp;printsec=frontcover&amp;dq=Jet+nestruev&amp;source=gbs_similarbooks_s&amp;cad=1#v=onepage&amp;q&amp;f=false" rel="nofollow">Smooth Manifolds and Observables</a> by "Jet Nestruev" to see these ideas fully worked out.</p> http://mathoverflow.net/questions/36965/the-ring-c-inftym/36986#36986 Answer by MTS for The ring $C^{\infty}(M)$? MTS 2010-08-28T18:23:01Z 2010-08-28T18:23:01Z <p>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.</p> <p>Also, in the paper <a href="http://arxiv.org/abs/0810.2088" rel="nofollow">On the Spectral Characterization of Manifolds</a>, 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.</p>