Can you uniquely define a smooth manifold M by taking the set of functions from M to R you consider smooth?

Question

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?

Answer 1

$\newcommand{\R}{\mathbb{R}}$ 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:

3. 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:

4. 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 $O_1 \cap O_2 = \emptyset$.