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I'm trying to show that manifolds are affine, i.e. $\text{Man}(M,N)\cong \mathbb{R}\text{-Alg}(C^\infty(N),C^\infty(M)) $. If I could show this for $N=\mathbb{R}^n$, then I know how to do the rest using the strong Whitney's embedding theorem. But it seems hard to show the surjectivity of $ \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M)):f\mapsto f^*$.

I found this post Are all manifolds are affine?, but it didn't mention any explicit material about this, and I failed finding anything other than that post. I learned that $\text{Man}(M,\mathbb{R}^n)\cong C^\infty\text{-Ring}(C^\infty(\mathbb{R}^n),C^\infty(M))$ from Models for Smooth Infinitesimal Analysis, but I don't know how to show that $C^\infty\text{-Ring}(C^\infty(\mathbb{R}^n),C^\infty(M))\cong \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$.

Thanks in advance for any solution or material about this.

EDIT: I had a further investigation (but no success) and I'm putting it here with supplementary details of my question.

The images of the coordinates functions on $\mathbb{R}^n$ consist to give a map $ \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))\to \text{Man}(M,\mathbb{R}^n)$, which tells the injectivity of $ \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$. To show the desired surjectivity, we still need to verify that the composition $$\mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))\to \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$$ is the identity.

I find that it suffices to show that, for any $\mathbb{R}$-algebra homomorphism $\varphi:C^\infty(\mathbb{R}^n)\to C^\infty(M)$, if $\rho\in C^\infty(\mathbb{R}^n)$ satisfies $\rho|_V=0$ for some open subset $V\subset\mathbb{R}^n$, then $\varphi(\rho)|_{f^{-1}(V)}=0$, where $f:M\to \mathbb{R}^n$ is given by the images of the coordinate functions on $\mathbb{R}^n$ under $\varphi$. If this is true, then we can define the restriction of $\varphi$ to any open subset $U\subset \mathbb{R}^n$,

$$ \varphi|_U :C^\infty(U)\to C^\infty(f^{-1}(U)),$$

by putting, for any $g\in C^\infty(U)$, $\varphi|_U(g)$ to be the unique smooth function such that

$$\varphi|_U(g)|_{f^{-1}(V)}=\varphi(G)|_{f^{-1}(V)} $$

for any open set $V\subset U$ and $G\in C^\infty(\mathbb{R}^n)$ such that $G|_V=g|_V$; to show that $\varphi(G)|_{f^{-1}(V)} $ does not depend on the choice of $G$ requires exactly the property I claimed above.

If the restrictions of $\varphi$ can be defined, then we obtain a sheaf morphism. Investigating the germs and we can find that $\varphi$ is indeed the pullback by $f$.

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  • $\begingroup$ For an $\mathbb{R}$-algebra map, what are the images in $C^\infty(M)$ of the coordinate functions on $\mathbb{R}^n$? When you put those coordinate functions together, does that give a manifold morphism to $\mathbb{R}^n$? $\endgroup$
    – Jason Starr
    Commented Apr 17, 2022 at 16:26
  • $\begingroup$ @JasonStarr Surely it does. But to show that $\text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$ is surjective, we need to verify that the image of this manifold morphism under $C^\infty$ is indeed the given $\mathbb{R}$-algebra map. This is exactly where I get stuck. $\endgroup$
    – Shana
    Commented Apr 17, 2022 at 16:29
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    $\begingroup$ You might want to compare this paper by Mrčun, though his technique of proof goes in a rather different direction to yours: arxiv.org/abs/math/0309179 $\endgroup$
    – Branimir Ćaćić
    Commented Apr 17, 2022 at 17:36

1 Answer 1

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The proof of this fact is available in modern textbooks.

For example, see Theorem 7.16 in Jet Nestruev's Smooth Manifolds and Observables (Second Edition, 2020).

In fact, the cited book contains a lot of material that explains how to pass between differential geometric objects and the corresponding algebraic objects, e.g., vector bundles and modules, differential forms and Kähler differentials, etc.

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    $\begingroup$ That's an amazing book! The key is in fact Theorem 7.2, which identifies a manifold with the dual of its algebra of smooth functions; under this identification, the composition $\mathbb{R}\text{-Alg}(C^\infty(N),C^\infty(M))\to \text{Man}(M,N)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(N),C^\infty(M)) $ is exactly "taking twice duals", and Theorem 3.8 tells that it is indeed the identity. Thanks! $\endgroup$
    – Shana
    Commented Apr 18, 2022 at 7:21
  • $\begingroup$ Would you mind giving me some materials about the fact that "supermanifolds are affine"? I once thought that I can prove that by using "manifolds are affine", but it turns out that the same gap in my question stops me from doing so. $\endgroup$
    – Shana
    Commented Apr 18, 2022 at 12:47
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    $\begingroup$ @Shana: This follows from the affineness of ordinary manifolds combined with Batchelor's theorem: every supermanifold is isomorphic to the supermanifold given by the sheaf of sections of the exterior algebra bundle of a finite-dimensional vector bundle. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 18, 2022 at 16:35
  • $\begingroup$ I see! So the Batcherlor's theorem gives a projection from a supermanifold to its underlying ordinary manifold, and an argument using the trick of bump and the affineness of ordinary manifolds gives my desired local property. Thanks so much! $\endgroup$
    – Shana
    Commented Apr 20, 2022 at 7:09

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