# Can we recover a compact smooth manifold from its ring of smooth functions?

It is well-known that if $X$ is a reasonably nice topological space (compact Hausdorff, say) then we can recover $X$ from the ring $C(X)$ of continuous functions $X\to\mathbb R$; see this MSE question for a discussion and problem 26 in the first chapter of Atiyah MacDonald for the construction. Is the same true for a compact smooth manifold $M$ and its ring $C^\infty(M)$ of smooth functions? More specifically,

1. Let $M$ and $N$ be compact smooth manifolds. If $C^\infty(M)$ and $C^\infty(N)$ are isomorphic, then are $M$ and $N$ necessarily diffeomorphic?
2. Can we recover the topological space $M$ from $C^\infty(M)$? If so, can we also recover the smooth structure on $M$?
• See Milnor-Stasheff, exercise 1C. 2(b) is slightly subtle and it's not yet completely obvious to me how to recover charts from the structure of the ring $C^\infty(M)$ alone.
– user98602
Apr 30, 2016 at 4:11
• This question on MSE refers to that problem (for reference). Apr 30, 2016 at 13:59
• If I'm not wrong you can find the proofs in the paper Ordinary differential equations on vector bundles and chronological calculus (R. V. Gamrelidze; A. A. Agrachev; S. A. Vakhrameev). I'll try to write an answer later..
– PtF
Jan 1, 2018 at 0:16

[I assume all "smooth manifolds" are Hausdorff and paracompact.]

Yes, you can recover $M$ as a smooth manifold from the ring $C^\infty(M)$. Here's a quick sketch.

First, note that we can recover the set of connected components of $M$, since each connected component $N\subseteq M$ corresponds to a minimal nonzero idempotent in $C^\infty(M)$, and the ideal generated by such an idempotent is isomorphic as a rng to $C^\infty(N)$. Thus we can recover each of the rings $C^\infty(N)$ from $C^\infty(M)$, so we may assume without loss of generality that $M$ is connected.

Now note that every ring-homomorphism $\varphi:C^\infty(M)\to\mathbb{R}$ is evaluation at a point of $M$, which lets us recover the set of points of $M$ from $C^\infty(M)$. For details, see the answers to this question.

So we've recovered the set $M$, and we also know how to think of elements of $C^\infty(M)$ as functions $M\to\mathbb{R}$ (since we identify points of $M$ with their evaluation homomorphisms). We can now also recover the smooth structure: we know exactly which functions $M\to\mathbb{R}$ are smooth, so we also know exactly which functions $M\to\mathbb{R}^n$ are smooth. Since every connected manifold $N$ embeds in $\mathbb{R}^n$ for some $n$, we also know exactly which functions $M\to N$ are smooth for any such $N$. This means we have recovered the entire functor $\operatorname{Hom}(M,-)$ on the category of connected smooth manifolds. By Yoneda, this is enough to recover $M$.

• sorry for reviving, may you elaborate on how one use Yoneda to recover $M$? So I guess you have shown that $C^\infty(M)$ uniquely determines $Hom(M,-)$, so if $C^\infty(M) \cong C^\infty(N)$, then so are their Hom functors, hence by Yoneda's, $M \cong N$ in category of connected smooth manifolds? Oct 11, 2018 at 9:58
• Yes, that's correct. Oct 11, 2018 at 14:48

I believe this is proven in Chapter 7 of Nestruev's Smooth Manifolds and Observables, but I haven't checked carefully. More precisely, the functor $M \to C^{\infty}(M)$ from smooth manifolds to the opposite of real commutative algebras is fully faithful, meaning that smooth maps $M \to N$ are precisely algebra maps $C^{\infty}(N) \to C^{\infty}(M)$.

• You shouldn't even need the $\mathbb{R}$-algebra structure. If $M$ is connected, the constant functions can be described as the unique maximal subfield of $C^\infty(M)$ (since every non-constant function takes a rational value, and hence some rational polynomial in it is non-invertible but not identically $0$). If $M$ is disconnected, you can apply this to each component separately, since you can recognize the components are determining the idempotents in $C^\infty(M)$. Apr 30, 2016 at 6:01
• Alternatively, you can say there is only one ring-homomorphism $f:\mathbb{R}\to C^\infty(M)$, since such a homomorphism must be order preserving (you can prove this by thinking about square roots, though you have to be a bit careful how you state things since $\sqrt{x}$ is only smooth on $(0,\infty)$, not $[0,\infty)$). Apr 30, 2016 at 6:05
• Ah, that's very nice. So in fact the embedding into affine schemes is fully faithful. Apr 30, 2016 at 6:24

I want to give full steps to this problem. The answer is that one can recover $$M$$ from $$\mathscr{C}^{\infty}(M)$$ for arbitrary manifold $$M$$ (assuming second countability and Hausdorff).

1. First step, recover the space. Let $$\Sigma=\{\textrm{non-zero \mathbb{R}-algebra homomoprhoism F:\mathscr{C}^{\infty}(M)\to \mathbb{R}}\}$$ Actually, any $$F\in \Sigma$$ is a calculation, say $$f\mapsto f(x)$$ for some fixed $$x$$.

Proof. Pick a "step" smooth function $$f_0: M\to \mathbb{R}$$ such that for any $$\lambda\in \mathbb{R}$$, $$f_0^{-1}(\lambda)$$ is compact (with using the assumption of existence of countable partition of unity). Assume $$F$$ is not a calculation, then $$\forall x\in M, \exists f_x\in\mathscr{C}^{\infty}(M),\quad \textrm{such that}\quad f_x(x)\neq F(f_x)$$ replace $$f_x$$ by $$f_x-F(f_x)$$, one can assume that $$f_x(x)\neq 0$$ and $$Ff_x=0$$. Then $$U_x=\{a\in M: f_x(a)\neq 0\}\ni x$$ gives a open cover of $$f_0^{-1}(F(f_0))$$, thus there exists finite subset $$X_0$$ such that $$\{U_x: x\in X_0\}$$ cover $$f_0^{-1}(F(f_0))$$, construct $$g=(f_0-Ff_0)^2+\sum_{x\in X_0} f_x^2\qquad Fg=0$$ but for any $$y\in M$$, $$g(y)>0$$. $$\square$$

1. Second steo, recover the topology. Topologize $$\Sigma$$ with the weak topology such that for any $$f\in \mathscr{C}^\infty(M)$$, $$F\mapsto F(f)$$ is continuous. Then $$\varphi: M\to \Sigma\qquad x\mapsto [f\mapsto f(x)]$$ is homoemorphism.

Proof. Firstly, the map is surjective by above. By Urysohn's Lemma of smooth version, the map is injective. To show the map is continuous, it suffices to show $$[F\mapsto F(f)]\circ \varphi=f$$ is continuous, which is trivial. To show the map is closed map, for any closed $$A$$, by Urysohn Lemma of smooth version, there exists an $$f:M\to \mathbb{R}$$ such that $$f^{-1}(0)=A$$. This shows $$\varphi(A)=[F\mapsto F(f)]^{-1}(A)$$ is closed. $$\square$$

1. Third step, recover its smooth structure 1. Now pick $$p\in M=\Sigma$$, One can defines a equivalent relation on $$\mathscr{C}^{\infty}(M)$$ $$f\sim g\iff f|_U=g|_U\textrm{ for some open neighborhood U around p}$$ Then the corresponding quotient space is exactly the germ of $$\mathscr{C}^{\infty}_p(M)$$.

Proof. Let $$(f,U)$$ be a representation of some germ at $$x$$, then by times a bump function, one can show that there exists some $$g$$ defining all over $$M$$ such $$g|_V=f|_V$$ for some more small neighborhood of $$x$$. $$\square$$

1. Fourth step, recover its smooth structure 2. Now let $$\mathfrak{m}_p=\{f\in \mathscr{C}^{\infty}_p(M): f(p)=0\}$$ One knows that $$\mathfrak{m}_p/\mathfrak{m}_{p}^2$$ is exactly the cotangent space at $$p$$. Let $$x^1,\ldots,x^n \in \mathscr{C}^{\infty}(M)$$, such that their image in $$\mathfrak{m}_p/\mathfrak{m}_{p}^2$$ forms a basis. It must form a local coordination for $$x$$.

Proof. That means $$dx^1, \ldots, dx^n$$ forms a basis of cotangent space at $$x$$. It must form a local coordination for $$x$$. It is a consequence of inverse maps theorem. $$\square$$

Once the local coordination is recovered, the proof is complete.

As Eric Wofsey points out, because $M$ as a topological space is the space of homomorphisms $C^\infty(M) \to \Bbb R$, appropriately topologized, we know precisely what the elements of $C^\infty(M)$ are as functions on $M$. So we can recover the space $C^\infty(M)_p$ of germs at $p$, and hence we can recover the dimension of $M$ as the dimension of the space of derivations. Now we can pick a set of $n$ functions $M \to \Bbb R$ such that these functions induce an isomorphism $T_pM \to T_p \Bbb R^n$; restricting the functions to an appropriate subset of $M$, these are charts. So we can explicitly construct the charts from $C^\infty(M)$.

A deeply fancy way of explaining what's going on (even though you can do the above in all dimensions, and smoothing theory proper only in dimension $n \geq 5$) is smoothing theory. Smoothing theory equips a topological manifold $M$ with a "tangent microbundle", and if you can lift this microbundle structure to an honest vector bundle, this lift provides your manifold with a smooth structure. The point of the above is that, through $C^\infty(M)$ alone, we can construct the tangent bundle $TM$, and hence invoke smoothing theory.

• It's not true that $M$ is the maximal spectrum of $C^\infty(M)$ if $M$ is not compact; it's just the $\mathbb{R}$-points. (If $M$ is not compact, there are maximal ideals which contain all functions with compact support.) Apr 30, 2016 at 18:22
• @EricW Thanks for the correction. The rest of the answer should still hold with the appropriate modifications.
– user98602
Apr 30, 2016 at 18:25

What I'll describe next comes from the paper

Ordinary differential equations on vector bundles and chronological calculus (R. V. Gamrelidze, A. A. Agrachev and S. A. Vakhrameev).

In the following $M$ must be second countable (it doesn't have to be compact). For a point $p\in M$ consider the map $\mathsf{ev}_p: C^\infty(M)\longrightarrow \mathbb R$ given by:

$$\mathsf{ev}_p(f):=f(p).$$

It is clearly an $\mathbb R$-algebra morphism.

Theorem 1. There exists a bijection:

$$M\longrightarrow \mathsf{Hom}_{\mathsf{Rng}}(C^\infty(M), \mathbb R)-\{0\},$$

which is given by:

$$p\longmapsto \mathsf{ev}_p.$$

The proof can be found in the forementioned paper. However, using this we can characterize easily the smooth functions on $M$.

Theorem 2. There exists a bijection:

$$C^\infty(M, N)\longrightarrow \mathsf{Hom}_{\mathsf{Alg}}(C^\infty(N), C^\infty(M))$$

which is given by:

$$f\longmapsto (f^*: C^\infty(N)\longrightarrow C^\infty(M), g\longmapsto g\circ f),$$ which preserves composition.

It is easy to see this map is well defined (that is, $f^*$ is a homomorphism of $\mathbb R$-algebras). On the other hand, given $g\in \mathsf{Hom}_{\mathsf{Alg}}(C^\infty(N), C^\infty(M))$ define:

$$f(p):=q\quad (p\in M),$$ where $q\in N$ is the unique point such that:

$$\mathsf{ev}_p\circ g=\mathsf{ev}_q.$$

Such point $q$ exists by Theorem 1. applied to the nonzero ring homomorphism $$\mathsf{ev}_p\circ g:C^\infty(M)\longrightarrow \mathbb R.$$

The smoothness of $f$ follows from the following criterium:

$$f\in C^\infty(M, N)\Leftrightarrow h\circ f\in C^\infty(M)\ \forall h\in C^\infty(N).$$ In fact: $$f^*(h)(p)=h(f(p))=h(q)=\mathsf{ev}_q(h)=\mathsf{ev}_p\circ g(h)=g(h)(p),$$ for every $h\in C^\infty(N)$ and for every $p\in M$, that is $$f\circ h=g(h)\in C^\infty(M)\ \forall h\in C^\infty(N),$$ and we are done. The above computation also shows that $f^*=g$.

Now using the above you can show easily $f$ is a diffeomorphism if and only if $f^*$ is an isomorphism of algebras.

I believe the above constructions gives an equivalence (maybe an isomorphism) of categories between the category of second countable manifolds $\mathsf{Man}$ and the category $\mathsf{Alg}_{\mathbb R}$ of $\mathbb R$-algebras. Just check how the categories $\mathsf{Man}$ and $\mathsf{Alg}_{\mathbb R}$ are defined in order to this equivalence to hold.

• This is definitely not an equivalence between $\mathsf{Man}$ and $\mathsf{Alg}_{\mathbb R}$: most $\mathbb{R}$-algebras are not isomorphic to $C^\infty(M)$ for any manifold $M$. Also, it's contravariant. What it does give is an equivalence between $\mathsf{Man}$ and a full subcategory of $\mathsf{Alg}_{\mathbb R}^{op}$. Jan 1, 2018 at 2:03
• @EricWofsey: This is interesting. Do you know a good reference for this equivalance? Oct 29, 2019 at 18:50
• @Bumblebee: Not sure what you mean. The equivalence is exactly what Theorem 2 of this answer says. Oct 29, 2019 at 18:53
• @EricWofsey: I was thinking about this full subcategory of $\mathsf{Alg}_{\mathbb R}.$ What are the properties of it? How can we identify it ? Explicitly, is there any way we can identify whether a given algebra is isomorphic to an algebra of smooth functions over a manifold? Oct 29, 2019 at 18:59
• @Bumblebee There is a nice discussion of this in Chapter 7 of Nestruev's Smooth Manifolds and Observables, mentioned in Qiaochu Yuan's answer. An $\mathbb{R}$-algebra $F$ is of the form $C^{\infty}(M)$ if and only if 1) it is geometric, meaning that if an element $f$ of $F$ is sent to zero by any $\mathbb{R}$-algebra map $F \to \mathbb{R}$ then it is already zero, 2) it is complete (the definition of which doesn't fit in this comment) and 3) it is locally isomorphic to $C^{\infty}(\mathbb{R}^n)$ for fixed $n$. Jul 8, 2022 at 14:08