## How to classify the algebras C^∞(M)?

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This continues my question about smooth Gelfand-duality. In the book

Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824

it is shown that $M \mapsto C^\infty(M)$ is a fully faithfull contravariant functor from the category of manifolds (smooth, separable and without boundary) to the category of $\mathbb{R}$-algebras. Isn't this nice? It would be even more nice if there is an algebraic description of the essential image of this functor, so that we have an antiequivalence of categories between manifolds and certain $\mathbb{R}$-algebras. Thus my question is:

• Which $\mathbb{R}$-algebras $A$ are isomorphic to $C^{\infty}(M)$ for some manifold $M$?

Of course, you could just formulate that $Spec_r(A)=Hom(A,\mathbb{R})$ with the obvious structure sheaf is a manifold and that the canonical map $A \to C^{\infty}(M)$ is an isomorphism in terms of the ring structure of $A$. But this does not seem to be handy at all. I want some nontrivial purely algebraic formulation. If possible avoiding structure sheaves at all.

Here are some necessary conditions:

• If $f \neq g$ in $A$, then there is some $\mathbb{R}$-homomorphism $\phi : A \to \mathbb{R}$ such that $\phi(f) \neq \phi(g)$. In particular, $A$ is reduced.
• For every $p \in Spec_r(A)$ with corresponding maximal ideal $m_p$, then the maximal ideal $\overline{m_p}$ of $A_{\mathfrak{m}_p}$ is finitely generated, say by elements $f_1,...,f_n$, and the canonical map $\mathbb{R}[t_1,...,t_n] / (t_1,...,t_n)^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}, t_i \mapsto f_i$ is an isomorphism for all $r \geq 0$.
• With the notation above, the canonical map $A/m_p^{r+1} \to A_{\mathfrak{m}_p} / \overline{m_p}^{r+1}$ is an isomorphism.
• The function $Spec_r(A) \to \mathbb{N}, p \to \dim_\mathbb{R} \mathfrak{m}_p/{\mathfrak{m}_p}^2$ is locally constant.

Are they sufficient [no, see Michael's answer]? Finally [solved by Dmitri's answer]:

• How can we characterize the algebras (at least within all the $C^{\infty}(M)$'s), that come from compact manifolds?

You might admit that "$Spec_r(A)$ is compact with the Gelfand topolgy" is not a satisfactory answer ;-).

Addendum: At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods. I don't claim that this applies to my problem. But at least I invite you to think about it. The properties of the algebras above are just an approximation. Even if we add some of the conditions in the answers (such as $\cap_{r} \overline{m}_p^{r+1} \neq 0$), it would be a great surprise that the conditions are sufficient. But I'm not convinced of the contrary as soon someone provides a counterexample. It is fun trying to deduce some of the differential geometric theorems such as IFT from the properties above (if $A \to B$ is an isomorphism in one tangent space, then it is a local isomorphism). Perhaps a first step is to characterize the local rings $C^{\infty}_p(\mathbb{R}^n)$.

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Shouldn't the right condition be something like "is locally isomorphic to C^{\infty}(U) for some open subset of R^n"? I don't know if one can avoid mentioning R^n somewhere in the answer... – Qiaochu Yuan Apr 13 2010 at 0:38

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Does this help? arxiv.org/abs/math.GT/9404228 – Jonas Meyer Apr 13 2010 at 0:39

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I asked something similar earlier - you may find the references there helpful. mathoverflow.net/questions/5344/… – Jason DeVito Apr 13 2010 at 0:44

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@Jason: ok you already posted the same question (but already assuming $M$ to be compact) but the answer basically consists of abstraction typical for MO ... I want something concrete, you know. – Martin Brandenburg Apr 13 2010 at 0:47

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"At first glance, it appears too optimistic to find an algebraic characterization. But many famous problems started like that and involved unexpected methods." I don't know how to convince you that this is the problem Connes has solved! He has characterised function algebras of compact smooth manifolds by saying that these algebras have a canonical projective module whose properties one can recognise. The cohomology of a compact manifold satisfies Poincare duality, and $C^\infty(M)$ knows this via the structure of its Hochschild/cyclic homology - hence Connes' invocation of Hochschild chains. – Tim Perutz Apr 15 2010 at 18:15
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## 4 Answers

There is a strongly geometric characterization of those algebras which arise as $C^\infty(M)$ for $M$ compact and orientable, recently proved by Connes, see here. This has come up on MO before, e.g. in Joel Fine's answer to this question:

http://mathoverflow.net/questions/5344/algebraic-description-of-compact-smooth-manifolds

Like the proofs of most major theorems in differential topology, Connes's approach invokes (a) Riemannian metrics, and (b) hard analysis. Spectral geometry is not my area, so this will be an amateurish explanation...

If $M$ is a compact smooth manifold, $C^\infty(M)$ is represented faithfully on the Hilbert space $H$ of $L^2$ sections of any hermitian vector bundle $S$. If $D$ is a first-order differential operator acting (unboundedly) on $H$ then we can recover the projective $C^\infty(M)$-module of sections $C^\infty(M;S)$ within $H$ as $\bigcap_{k>0}{dom(D^k)}$ (these domains will actually by Sobolev spaces, I believe). The algebraic counterpart of being first-order is that $[[D,f],g]=0$ for any $f,g \in C^\infty(M)$.

$D$ has particularly nice properties when it's elliptic. There's no canonical elliptic operator over a smooth manifold until one chooses a Riemannian metric; there's then the signature operator $D=d+d^\ast$ acting on the complexified differential forms. This is an example of a Dirac operator (its square is a Laplacian - this is a condition on the symbol of the operator). As such, it's formally self-adjoint, Fredholm, and its (real) spectrum has known growth rate depending on $\dim(M)$.

Connes (see Theorem 11.4) shows that a commutative $\mathbb{R}$-algebra $A$ arises as $C^\infty(M)$ for a smooth manifold structure on the space $M$ Gelfand-dual to $A$ provided that it's part of a "spectral triple" $(A,H,D)$ of the right kind. This means that $A$ should act on a Hilbert space $H$ carrying an unbounded symmetric operator $D$ satisfying various properties. I've hinted at some of these; the most sophisticated property is an "orientation" condition invoking a Hochschild cycle $c\in Z_{\dim M}(A,A)$. This cycle is something like a volume form, and from its components Connes rebuilds local charts.

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Tim, I'm no geometer, but doesn't Connes' construction assume more and (hence) obtain more? The original question asks about $M$ with only a smooth structure and not a choice of Riemannian metric. (For what it's worth, I'm not claiming this approach is better or worse; just slightly different.) – Yemon Choi Apr 13 2010 at 21:02
Yemon - this issue was raised earlier on MO, but I don't quite see the objection. Slightly different to what? Is there a comparable theorem that invokes only the bare algebra? I don't see one mentioned in Connes' paper. The roles of the metric here - in defining a Hilbert-space completion of a projective module of smooth sections of a vector bundle, and in defining a differential operator from which we can find our original module inside that Hilbert space - seem very natural to me, both geometrically and analytically. – Tim Perutz Apr 13 2010 at 21:31
Tim - I actually agree that the spectral triple formulation is very natural, and I have no problem with it. It just seemed to me that the original post's wording, "I want some nontrivial purely algebraic formulation, if possible avoiding structure sheaves at all", was ambitiously looking for something only using the bare algebra. In other words, a bunch of conditions that can be formulated solely in terms of ${\mathbb C}$-algebra notions. But perhaps I misunderstood. – Yemon Choi Apr 14 2010 at 20:04
Yemon: I agree with your reading of Martin's question. All the same, and notwithstanding my previous comment, Connes' condition ARE conditions on the algebra! No sheaves are invoked. They aren't conditions on the structure of its ideals, but instead on its representation theory. – Tim Perutz Apr 14 2010 at 20:41
very impressive theorem, though I won't understand it. thanks Tim for the reference! as yemon pointed out, I'm hoping for a purely algebraic formulation. and yes, this seems to be only a dream. – Martin Brandenburg Apr 16 2010 at 14:44

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This is to expand further on something I wrote in the comments. Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:

Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\lim_{r \geq 1} A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.

In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).

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Note also that, if $M$ is positive dimensional, then $C^{\infty}(M)$ is not Noetherian, since if $\mathfrak m$ is a maximal ideal corresponding to a point $x \in M$, then $\bigcap_{n \geq 0} \mathfrak m^n$ is non-zero (since there are non-zero smooth functions whose Taylor series is identically zero).  This also rules out polynomial or power-series rings. – Emerton Apr 14 2010 at 17:12
thanks michael and emerton. I've added some remarks in my original question. – Martin Brandenburg Apr 14 2010 at 23:46

I think the appropriate category here is not that of algebras but algebras with derivations (linear maps satisfying Leibniz's product rule). If you don't look at the derivations you're forgetting the differentiable structure of the manifold and all the manifolds homeomorphic (possibly not diffeomorphic) to your manifold support your algebra of functions. I posted an answer along the same lines in your previous questions.

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the fact that $C^{\infty}$ is full faithful implies that the manifold can be reconstructed from the algebra and every additional structure needed on this algebra can be recovered from the algebra structure. your claim that homeomorphic manifolds yield the same algebra is wrong. $C^{\infty}(M)$ and $C^{\infty}(N)$ are isomorphic as algebras iff $M,N$ are diffeomorphic. – Martin Brandenburg Apr 13 2010 at 11:32

How can we characterize the algebras (at least within all the C^∞(M)'s), that come from compact manifolds?

An algebra of the form C^∞(M) corresponds to a compact manifold if and only if all of its maximal ideals have codimension 1.

If the manifold is non-compact, consider the ideal of all functions with compact support and extend it to some maximal ideal using Zorn's Lemma. Clearly, this ideal cannot correspond to any point of the manifold, hence its codimension must be greater than 1.

To prove the converse, choose a maximal ideal I in C^∞(M) for some compact M. Denote by A the set of all points x∈M such that all elements of I vanish at x. A must be nonempty, because otherwise we can cover M by preimages of R \ {0} of functions in I, choose a finite subcover of this open cover, and then observe that the corresponding finite set of functions generates C^∞(M) as an ideal. A cannot consist of more than one point, because otherwise I is not maximal. For the same reason I consists of all functions that vanish at A, hence it has codimension 1.

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what is the codimension of an ideal? perhaps the codimension of the corresponding zero set? – Martin Brandenburg Apr 13 2010 at 2:14

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Martin: I assume Dmitri means codimension as a linear subspace – Yemon Choi Apr 13 2010 at 2:49

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@Martin: Yemon is right. – Dmitri Pavlov Apr 13 2010 at 3:27
alright. thus the maximal ideals of codim. 1 are exactly the points of the real spectrum (aka the rational points). – Martin Brandenburg Apr 13 2010 at 3:32
@dmitri: ok this is just as in usual Gelfand-duality. fine :) – Martin Brandenburg Apr 13 2010 at 3:33
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