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Let M be a 2nd countable Hausdorff space. Let F be a set of functions from open sets of M to R. Does there exist a smooth structure on M that makes these functions smooth? Is this an equivalent way to define a manifold? Similar to how you can uniquely define a topological space by having a set of functions to be consider continuous?

Arctic Char
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Spencer Kraisler
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    Yes, this is the alternative way to do it. The classical way is to define a smooth structure. A more modern way is to define what a "regular function" is. Therefore, you define a sheaf structure on the manifold instead. A morphism between manifolds then becomes a morphism between manifolds-with-a-sheaf. – Nicolas Bourbaki Dec 25 '21 at 00:19
  • You also need locally homeomorphic to R^n, but even then that might not be enough. For example, take $S^1 v S^2$. This is 2nd countable, Hausdorff space as it's a finite CW complex. Yet, no set of functions will make this a smooth manifold. – Daniel H. Hartman Dec 25 '21 at 00:40
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    No. You didn't even assume $M$ to be a topological manifold yet. Even then, you certainly can't find a smooth structure such that, for example, all continuous functions become smooth. There's an approach to manifolds that takes the sheaf of smooth functions as primitive data, but you will need additional conditions to recover an equivalent definition. One positive answer, though, is that a smooth manifold $M$ can be recovered from its ring of smooth functions $C^{\infty}(M)$. – Thorgott Dec 25 '21 at 00:43
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    The comment by @Thorgott probably constitutes an answer, but I might add that the only time we can choose a smooth structure on a topological space $X$ such that all continuous functions are smooth is when $X$ is discrete. The second question has been asked here (as well as the topological version here). – Kajelad Dec 25 '21 at 08:12
  • An extreme approach, where you don’t even start with a space, can be found in this paper and it’s references. It’s a smooth analogue of the modern scheme theoretic approach to algebraic geometry. https://arxiv.org/pdf/1001.0023.pdf – Deane Dec 25 '21 at 18:28

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$\newcommand{\R}{\mathbb{R}}$ No matter what you have to use the concept of $C^\infty(\R)$ to impose additional assumptions on the set $F$ of functions on a set $M$ for $M$ to be a smooth manifold and for $F$ to be a set of smooth functions for a manifold structure. The additional assumptions should simply imply the existence of an atlas. This can be expressed without any assumption on $M$ itself and only on a set $F$ of scalar functions. Here is one possible set of assumptions.

An obvious necessary condition is that $F$ is a ring over $C^\infty(\R)$. In other words, the following have to hold:

  1. Given any $f, g \in F$ and $h \in C^\infty$, $$ hf + g \in F.$$
  2. The constant functions are in $F$.

Another obvious necessary condition is:

  1. For any $f \in F$ and $h \in C^\infty(\R)$, $h\circ f \in F$.

At least one more assumption is needed to get the local Euclidean structure. Here, I think, is one sufficient assumption:

  1. There exists a collection of maps $\Phi = (\phi^1, \dots, \phi^n): M \rightarrow \R^n$, which we'll call coordinate maps, such that the following hold:

    a) $\phi_1, \dots, \phi_n \in F$

    b) For each coordinate map $\Phi$, there exists a set $O \subset M$ such that the map $$\Phi = (\phi^1, \dots, \phi^n): O \rightarrow \Phi(O) $$ is a bijection and $\Phi(O) \subset \R^n$ is open.

    c) There exists a countable collection of coordinate maps $\Phi_k: O_k \rightarrow \R^n$ such that $$ \bigcup\limits_k O_k = M. $$

    d) If $\Phi_1: O_1\rightarrow \R^n$ and $\Phi_2: O_2 \rightarrow \R^n$ are coordinate maps such that $O_1 \cap O_2 \ne \emptyset$, then the map $$\Phi_2\circ\Phi_1: O_1\cap O_2 \rightarrow \R^n$$ is smooth and bijective.

    e) Given two different points $p_1, p_2 \in M$, there exist coordinate maps $\Phi_1: O_1 \rightarrow \R^n$, $\Phi_2: O_2 \rightarrow \R^n$ such that $p_1 \in O_1$, $p_2 \in O_2$, and $O_1 \cap O_2 = \emptyset$.

Deane
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  • This is in fact how I like to define a manifold. The topology and topological properties (Hausdorff and second countable) should be consequences of the definition and not a priori assumtions – Deane Dec 25 '21 at 19:43