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To be honest, I don't really know, whether or not the following is a research level question:

Let $M$ be a smooth manifold, $C^\infty(M)$ the smooth function ring on $M$ and suppose $R\subset C^\infty(M)$ is a subring. What are conditions, such that $R$ is the smooth function ring of a smooth manifold ?

On a first impression I would say, that this is a huge question and an exhaustive answer is unlikely. Nevertheless I don't really know where to start looking for something like that...

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Have you looked at Gillman/Jerrison's book "Rings of Continuous Functions"? Maybe they shed a little light on your situation. – LMN Dec 15 '12 at 0:43
just a small remark: subrings would rather correspond to quotient manifolds than to submanifolds. – Michael Bächtold Dec 15 '12 at 10:20
yeah right. I should have just written manifolds defined by subrings... – Nevermind Dec 17 '12 at 0:58
up vote 7 down vote accepted

You have to be careful: let $Y\subset V$ be a submanifold of $V$, you have a restriction map $$C^{\infty}(V)\rightarrow C^{\infty}(Y)$$ whose kernel is an ideal $p_Y$, and if $Y$ is a closed submanifold: $$C^{\infty}(Y)\cong C^{\infty}(V)/p_Y$$

Thus the question is rather what ideals of $C^{\infty}(V)$ are of the form $p_Y$ for $Y$ a submanifold of $V$.

1) if $Z$ is any subset of $V$ it makes sense to define the ideal $p_Z$ of functions of $C^{\infty}(V)$ that vanish on $Z$. And one can prove that a closed subset $Z$ of $V$ is a submanifold if and only if $p_Z$ is regular.

Now the question is what are the ideals of $C^{\infty}(V)$ of type $p_Z$ with $Z$ closed.

2) you consider $C^{\infty}(V)$ as a Fréchet space and notice that $p_Z$ is a closed ideal.

Hence the question is what are the closed ideals of $C^{\infty}(V)$ of type $p_Z$ with $Z$ closed?
Let me give you an answer when $V=\mathbb{R}^n$.

Let $I$ be a closed ideal of $C^{\infty}(\mathbb{R}^n)$ the quotient $C^{\infty}(\mathbb R^n)/I$ is called a differentiable algebra. $I$ is of the form $p_Z$ if this quotient algebra is reduced ($O$ is the unique element vanishing at any point of the real spectrum $Spec_r(C^{\infty}(\mathbb{R}^n)/I)$).

Reference: $C^{\infty}$-Differentiable spaces (LNM) Juan A. Navarro González, Juan B. Sancho de Salas

Edit: @Nevermind, if you have a smooth surjective map $\pi:V\rightarrow Y$, then you will have a ring map $\pi^*:C^{\infty}(Y)\rightarrow C^{\infty}(V)$ and this map is injective.
Now I recommand you to look at Dominic Joyce's survey "Algebraic Geometry over $C^{\infty}$-rings" Corollary 3.4 He explains that the category of smooth manifolds embeds (fully, faithfully) as a subcategory of the category of finitely presented $C^{\infty}$-rings. Thus if you have a sub-$C^{\infty}$-ring $$R\rightarrow C^{\infty}(V)$$ this morphism of $C^{\infty}$-ring will be realizable by a smooth map $$V\rightarrow \mathfrak{R}$$ such that $C^{\infty}(\mathfrak{R})=R$ if and only if $R$ is an algebra of smooth functions and the way to recognize these algebras is exactly P. Michor's theorem in the note cited in your post. You notice that condition 1) of this theorem is satisfied for any subring of $C^{\infty}(V)$. Thus you are left with two criteria: "finitely generated" and "germ determined". Germ dertermined (related to condition 3) in Michor's theorem) is related to "fair $C^{\infty}$-rings" in D. Joyce's papers and related to reduced in the book on differentiable spaces cited above.

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Thanks David. I'll think about it – Nevermind Dec 15 '12 at 19:11
You're welcome. Moerdijk-Reyes's book is also a very good reference, Dominic Joyce has written a vey nice survey paper (available on arXiv) on $C^{\infty}$-schemes (how you can think of differentiable manifolds from the point of view algebraic geometry). – David C Dec 15 '12 at 19:30
Why are you considering quotient rings and not subrings? – Will Sawin Dec 15 '12 at 19:41
@ Will. Because in the case of a closed submanifold $Y$ of $M$, $C^{\infty}(Y)$ is a quotient ring of $C^{\infty}(M)$ and not a subring. Maybe I have misunderstood your comment. I don't know if there is a nice geometric interpretation of subrings of $C^{infty}(M)$ and I will be very happy to learn about it. – David C Dec 15 '12 at 21:22
For example, I think it is valid to say that the subring of constant functions (i.e. multiples of $1_{C∞(M)}$) is the function ring of the point? It arise as a pullback by the terminal morphism. – Nevermind Dec 17 '12 at 1:15

Seems like I found something useful by myself:

In case someone else is interested, here a short article of Peter Michor on that topic


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