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There is a classical result which says that the assignment $$M \mapsto C^{\infty}\left(M\right)$$ is an embedding of the category of (paracompact Hausdorff) smooth manifolds into the opposite category of $\mathbb{R}$-algebras. However, all of the proofs I have seen first establish this for manifolds of the form $\mathbb{R}^n$ and then use Whitney's embedding theorem. This of course uses the paracompact Hausdorff condtion in an essential way. My question is, is this functor still an embedding if we don't impose paracompactness conditions on our smooth manifolds? If so, is there a nice proof of this fact? If not, can someone provide a simple counterexample? Thanks!

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for manifolds like (open) subset of some $\mathbb{R}^n$ the proof (for the "faithful" part) is easy (considerind tha canonical proiection on the codomain). FOr the general case, a morphism between manifold is detected from its restrictions to open coordinate charts of a atlant. Then just because you can extend a germ of a (smooth real) function to the intere manifold, follow that this functor is faithful.Is this functor full too? –  Buschi Sergio Feb 20 '12 at 10:59
    
NAyway from "Models for Smooth Infinitesimal Analysis" (Moerdijk, Reyes) T.2.8 p.30, you have a proof (this functor is faithfull and full ) without using Whitney's theorems (I seems). –  Buschi Sergio Feb 20 '12 at 11:06
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You can do this without explicitly using Whitney's embedding theorem, if you are willing to use $C^\infty$-rings instead of $\mathbb{R}$-algebras, but the argument I know requires partitions of unity and the Hausdorff condition in an essential way. –  Justin Noel Feb 20 '12 at 12:20
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As Theo Johnson-Freyd has recently pointed out to me, the functor is full on paracompact manifolds that are not second countable (i.e., have more than a countable number of connected components) only if you require morphisms of rings to be continuous, or impose any other similar condition (e.g., sheaf locality). –  Dmitri Pavlov Feb 20 '12 at 12:42
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@Dmitri: Can you expand upon this comment? I am curious. Also, if instead of $\mathbb{R}$-algebras, we work with $C^{\infty}$-rings, would what Theo pointed out carry through? (Since this is would be an algebraic way of forcing continuity) –  David Carchedi Feb 20 '12 at 12:53
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2 Answers

up vote 14 down vote accepted

The functor is not an embedding if we remove the paracompactness assumption.

I will need some preliminary definitions. Let $R$ be the long ray, i.e., the topological space given by $\omega_1 \times [0, 1)$ equipped with the order topology induced by lexicographic order, and let $L$ be the long line obtained by gluing together two long rays. Topologically, we can think of $L$ as the colimit of spaces $L_\alpha$ where $\alpha$ is a countable ordinal and $L_\alpha$ is obtained by gluing together two rays $R_\alpha = \alpha \times [0, 1)$.

According to the Wikipedia article, not only does there exist a smooth ($C^\infty$) manifold structure on $L$, there exist infinitely many smooth ($C^{\infty}$) manifold structures on $L$ which extend a given $C^1$ manifold structure. On the other hand, there is just one smooth structure on the ordinary line which extends a given $C^1$ structure. We exploit these facts to show that $M \mapsto C^\infty(M)$ is not an embedding.

Let $L$ and $L'$ be distinct smooth structures which restrict to the same $C^1$ structure. If $C^\infty(-) = \hom(-, \mathbb{R})$ were an embedding on general Hausdorff not necessarily paracompact manifolds, then $L$ and $L'$ would be isomorphic if their smooth algebras are isomorphic. So it is enough to show that $C^\infty(L)$ and $C^\infty(L')$ are isomorphic. Now it is well-known that every continuous function on the long line is eventually constant (constant outside some bounded neighborhood). In that case, consider the algebra $C_\alpha$ of smooth functions on $L_\alpha$ which are eventually constant. For $\alpha \leq \beta$ there is an obvious extension $C_\alpha \to C_\beta$, and we have

$$C^\infty(L) = colim_\alpha C_\alpha$$

Similarly, we may write $C^\infty(L') = colim_\alpha C^\prime_\alpha$. However, notice that the identity function $L \to L'$ restricts to a diffeomorphism $L_\alpha \to L^\prime_\alpha$, because each $L_\alpha$ is topologically an ordinary line, where we had observed there is just one smooth structure extending the given $C^1$ structure. The isomorphisms $C_\alpha \to C^\prime_\alpha$ induce an isomorphism $C^\infty(L) \to C^\infty(L')$, as desired.

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Very nice. Thank you Todd. –  David Carchedi Feb 21 '12 at 17:50
    
By the way, does anyone know of a counterexample with $M$ paracompact, Hausdorff, but not second-countable? –  David Carchedi Feb 21 '12 at 22:44
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Hello,

I am not sure to fully understand the question. I understand it as: if two (possibly non paracompact or non Hausdorff) smooth manifolds have isomorphic algebra of smooth functions, are they isomorphic?

If I understand it well, then take the line with two origins (i.e. a quotient of the disjoint union of 2 real lines $\mathbb{R}\times\{0,1\}$ by the equivalence $(x,e)\sim(x',e') \Leftrightarrow (x=x' \text{ and } x\neq 0)$ ). Its algebra of smooth functions is isomorphic to $\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$.

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The hypothesis David wants to remove is paracompactness, not Hausdorffness. –  Mariano Suárez-Alvarez Feb 20 '12 at 5:26
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@Mariano: Actually, I would like to see what I can get away with. At any rate, both paracompactness and Hausdroffness are needed for partiions of unity, and also for Whitney's embedding theorem. I wanted to see how far away from (either of) these assumptions the theorem still holds. –  David Carchedi Feb 20 '12 at 11:21
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@Taladris: FYI, for a functor to me an embedding, it means that it is full and faithful, i.e.,I want to know if the natural map $$C^{\infty}\left(M,N\right) \to \Hom\left(C^{\infty}\left(N\right),C^{\infty}\left(N\right)\right)$$ is a bijection. Of course, what you wrote implies this cannot be the case, I just wanted to clarify for you. –  David Carchedi Feb 20 '12 at 11:24
    
Of course, now that I see that the Hausdorff condition cannot be removed, I am still curious if one can keep the Hausdorff condition, and remove paracompactness. –  David Carchedi Feb 20 '12 at 11:43
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