Varieties, Frechet Completions, and Regular Functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:53:50Z http://mathoverflow.net/feeds/question/41804 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41804/varieties-frechet-completions-and-regular-functions Varieties, Frechet Completions, and Regular Functions John McCarthy 2010-10-11T16:18:26Z 2010-10-14T08:15:25Z <p>Take an algebraic variety $V$, and its set of smooth functions $C^{\infty}(V)$. One can endow $C^{\infty}(V)$ with a canonical locally convex topology (the seminorms are defined using the local coordinate patches of the variety). With respect to this topology the space is a Frechet space (this means that, amongst other properties, that it is metrisable and complete with respect to the metric). </p> <p>I have two questions:</p> <p>(1) If $O(V)$ is the set of regular functions of $V$, is $O(V)$ dense in $C^{\infty}(V)$ with respect to the Frechet topology?</p> <p>(2) Can one caracterize these locally convex topologies on $O(V)$ that are induced by a differential structure?</p> http://mathoverflow.net/questions/41804/varieties-frechet-completions-and-regular-functions/41819#41819 Answer by Pete L. Clark for Varieties, Frechet Completions, and Regular Functions Pete L. Clark 2010-10-11T18:21:01Z 2010-10-11T22:51:40Z <p>For (1), consider the case of $V = \mathbb{P}^1$, the projective line. The only regular functions are the constants, which are obviously not dense in the space of all smooth $\mathbb{C}$-valued functions on the $2$-sphere. </p> <p><b>Added</b>: For that matter, consider the affine line $\mathbb{A}^1$. I think to make your question more reasonable, you should also throw in complex conjugates of the regular functions (c.f. Hodge theory). </p> http://mathoverflow.net/questions/41804/varieties-frechet-completions-and-regular-functions/42129#42129 Answer by Greg Kuperberg for Varieties, Frechet Completions, and Regular Functions Greg Kuperberg 2010-10-14T08:15:25Z 2010-10-14T08:15:25Z <p>First, $V$ needs to be the type of variety for which the ring of regular functions $O(V)$ separates points. If it doesn't, then that establishes an equivalence relation on $V$ and you might as well pass to the quotient in which points are separated. Thus $V$ is an affine algebraic variety. Moreover, if $V$ is complex, then you have to take its realification to have any hope of approximating smooth functions; the complex regular functions are all holomorphic and would at best approximate the space of holomorphic functions. So, we can suppose that $V$ is an affine real algebraic variety, or in scheme-speak, the real points of an affine real algebraic variety.</p> <p>If $V$ is compact, then the Stone-Weierstrass theorem directly tells you that $O(V)$ is dense in $C^\infty(V)$. Okay, you have to extend the Stone-Weierstrass theorem to derivatives, but in finite dimensions you can do that. And you have to consider derivatives at singularities of $V$ if you want to allow singularities. But I don't think that there is any real problem with that either.</p> <p>If $V$ is not compact, then I have a little trouble interpreting the question, but I think that it comes to the same thing. Every algebraic variety is locally compact in the analytic topology, and when you say "local coordinate patches", the reasonable interpretation is coordinate patches whose closures are compact. If $X$ is any suitably tame locally compact space and you make a Fréchet topology on $C^\infty(X)$ by exhausting $X$ by a sequence of compact subsets, then you can extend the Stone-Weierstrass theorem by diagonalization.</p> <p>I don't know what is meant by characterizing topologies on $O(V)$ that come from differentiable structure. I can't think of another equally natural Fréchet topology on all of $C^\infty(V)$, other than to take the sup of each derivative on each compact set. But, if you are willing to work with only part of $C^\infty(V)$, and if you are willing to give $V$ a Riemannian metric, then you can use $L^p$ norms instead of sup norms. Or you can make a Banach space instead of a Fréchet space by making some algebraic combination of the seminorms instead of listing them separately. I think that the resulting space is typically smaller than all of $C^\infty(V)$, but it can still be big enough to include all of $O(V)$. It is then a different locally convex topology on $O(V)$ because it has a different completion.</p> <p>(This answer doesn't feel all that creative, so maybe the intended question is different?)</p>