2 Last $C(M)$ changed to $C^\infty(M)$

If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$.

A derivation in this case is an $\mathbb R$-linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$.

I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to define "continuous" functions on prove that $M$ as C(M)$is the closure of$C(M)$C^\infty(M)$ under the compact-open topology. The differentiable manifold structure is given by the derivations.

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If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$.

A derivation in this case is an $\mathbb R$-linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$.

I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to define "continuous" functions on $M$ as the closure of $C(M)$ under the compact-open topology. The differentiable manifold structure is given by the derivations.