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StoneWeierstrass theorem for holomorphic functions?
Alex, you can contact me directly by email: mathnet.ru/php/… 
2d

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StoneWeierstrass theorem for holomorphic functions?
Yes, it's obvious, if $M$ is $\sigma$compact: then ${\mathcal C}^\infty(M)$ is a Fréchet space (I don't know, perhaps people consider now non$\sigma$compact manifolds as well). For the case when $M$ is an open subset in $\mathbb{R}^m$ this topology on ${\mathcal C}^\infty(M)$ is described in Rudin's "Functional analysis": amazon.com/FunctionalAnalysisWalterRudin/dp/0070542368 
Feb
6 
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Unifying (& “justifying”) the various definitions for differential operators
Ah, yes! $\phantom{*}$ 
Feb
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Unifying (& “justifying”) the various definitions for differential operators
You can add another definition: a linear operator $D:C^\infty(X)\to C^\infty(X)$ is a differential operator of order $n$ if for any sequence of functions $f_0,f_1,...,f_n\in C^\infty(X)$ the commutator with the operators of multiplication by $f_i$ vanishes: $[...[[D,f_0],f_1]...,f_n]=0$. 
Feb
6 
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Unifying (& “justifying”) the various definitions for differential operators
No, you don't need the topology on ${\mathcal O}_X$. You can look at the proof of J.Peetre's theorem in S.Helgason's book: amazon.com/GeometricInvariantDifferentialOperatorsSpherical/… He also gives an equivalent definition: the operator $D:C^\infty(X)\to C^\infty(X)$ must not extend the support of function: $\text{supp}Df\subseteq \text{supp}f$. And these operators indeed have locally finite order. 
Feb
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awarded  Good Question 
Feb
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StoneWeierstrass theorem for holomorphic functions?
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StoneWeierstrass theorem for holomorphic functions?
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StoneWeierstrass theorem for holomorphic functions?
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StoneWeierstrass theorem for holomorphic functions?
@AlexM. I edited. There is no need to take into account the stereotype theory, when you define topology on ${\mathcal C}^\infty(M)$. This is equivalent to your condition with vector fields $X_i$. 
Feb
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StoneWeierstrass theorem for holomorphic functions?
I added the description of topology on ${\mathcal C}^\infty(M)$ 
Dec
11 
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StoneWeierstrass theorem for holomorphic functions?
Simon, I don't understand you. Why meromorphic functions? You can consider a narrower class, the class of entire functions on ${\mathbb C}$, and even in this situation the problem is unsolved: if $A$ is a subalgebra in ${\mathcal O}({\mathbb C})$, nobody knows which conditions $A$ should satisfy for being dense in ${\mathcal O}({\mathbb C})$. 
Dec
10 
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StoneWeierstrass theorem for holomorphic functions?
Necessary, but I would be surprised if it were essential in the answer. I believe, there must be a much more strong condition instead of separating tangent vectors. 
Dec
10 
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StoneWeierstrass theorem for holomorphic functions?
David, actually, I would not expect that the criteria for ${\mathcal O}(M)$ and for ${\mathcal C}^\infty(M)$ will be similar. I would suppose that separating points will be their common part, but separating tangent vectors sounds strange for me. 
Dec
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StoneWeierstrass theorem for holomorphic functions?
David, your example is interesting, but I think, you did not understand my question. First, I have no hypotheses on what the answer should be. Second, separating points does not mean the equality of spectra. For example, from the StoneWeierstrass theorem it follows that the algebra $A$ of almost periodic functions is dense in ${\mathcal C}({\mathbb R})$, but $\text{Spec}(A)\cong\text{Bohr compactification of} \ {\mathbb R}\ne{\mathbb R}\cong\text{Spec}({\mathcal C}({\mathbb R}))$ (if $\text{Spec}$ is defined somewhat reasonably, say, if it means continuous involutive characters). 
Dec
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StoneWeierstrass theorem for holomorphic functions?
Yes, of course! 
Dec
10 
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StoneWeierstrass theorem for holomorphic functions?
@SimonHenry, as far as I understand, even in the case of $M={\mathbb C}$ there is no answer. 
Dec
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StoneWeierstrass theorem for holomorphic functions?
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Dec
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StoneWeierstrass theorem for holomorphic functions?
Ah, I see. Maybe... 