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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).

I have two questions:

(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?

(2) Can one caracterize these locally convex topologies on $O(V)$ that are induced by a differential structure?

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What do you mean by "categorise": do you mean "caracterize"? –  Qfwfq Oct 11 '10 at 17:03
    
Yes, of course. –  John McCarthy Oct 11 '10 at 17:34
    
As Pete Clark explains in his answer below, (1) is definitely false when $V$ is complete e.g. projective. On the positive side, it ought to true that when $\mathcal{O}(V)$ separates points (e.g. if $V$ is affine) that $\mathcal{O}(V)+\overline{\mathcal{O}(V)}$ is dense in $\C^\infty(V)$, in the topology of uniform convergence on compact sets, by the Stone-Weierstrass theorem. –  Donu Arapura Oct 11 '10 at 23:20
    
@Donu: Be aware, that $\mathcah{O}(V) + \overline{\mathcal{O}}(V)$ is not an algebra, and therefore Stone-Weierstrass does not apply. One needs to take (at least) the $\IR$-algebra generated by these. –  Heinrich Hartmann Oct 11 '10 at 23:56
    
Right, I stand corrected. –  Donu Arapura Oct 12 '10 at 1:05

2 Answers 2

up vote 2 down vote accepted

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.

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.

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.

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.

(This answer doesn't feel all that creative, so maybe the intended question is different?)

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Your answer to the first question is great, just what I wanted. For the second part, centre arounds my guess that different structures on a variety will induce different topologies (in the manner discussed above, ie sups of derivatives on compact subsets) and so different completions of the regular functions. I was just wondering if it might be possibile to classify possible (exotic) differential structures in terms of different completions of the regular functions. (This is a wild guess, but I'm curious to see if it makes sense.) –  John McCarthy Oct 14 '10 at 17:27
    
A clarification of one point: If you exhaust a locally compact space $X$ by compact sets, you want an increasing sequence whose elements eventually contain every compact subset. These sequences are then cofinal with each other, and every one gives you the same Fréchet topology on $C^\infty(X)$. So there are no choices there. Beyond that, I don't know what you mean by an exotic differential structure. –  Greg Kuperberg Oct 14 '10 at 20:05
    
Rephrasing: Take a real variety (eg the 7-sphere) considered as topological space only. This variety has a canonical differential structure $C$. However, it might also have a non-standard structure $E$ (eg the exotic spheres of Milnor). Now surely the Frechet completion of the REGULAR functions wrt one diff structure will not be the same as the completion wrt the other? That is, $C^{\infty}(V)_C \neq C^{\infty}(V)_E$. If so, does there exist some identifiable subset of all the Frechet completions of the regular functions that corresponds to the set of non-isomorphic diff structures? –  Jean Delinez Oct 14 '10 at 22:51
    
But $V$ induces a specific smooth structure. If it's not the same smooth structure, then it's not the same real algebraic variety structure either. Even if you do just have two smooth structures on one topological manifold (that could even be the same structure conjugated by a non-differentiable homeomorphism), in general the Fréchet topologies are miles apart and I don't see a connection. Also I only see one relevant Fréchet space for each smooth structure, not a big collection of them. –  Greg Kuperberg Oct 15 '10 at 4:37
    
Ok. I think I was making the tacit assumption that a function which was regular wrt one smooth structure would smooth wrt any other structure. From your noting that the identifying homeomorphism need not be non-differentiable I see that this is of course not true. Thanks alot for your help in clarifying this. –  John McCarthy Oct 20 '10 at 17:53

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.

Added: 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).

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