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Mar
6 |
comment |
Is the Mendeleev table explained in quantum mechanics?
According to V.I.Arnold, springer.com/us/book/9780387968902, Classical Mechanics can be at least considered as an axiomatic system of second order, i.e. an axiomatic system inside another axiomatic system, like General Topology, or Theory of Real numbers, or Probability Theory, etc. Hilbert's axiomatization of Euclid's geometry, en.wikipedia.org/wiki/Hilbert%27s_axioms, is another example. |

Mar
6 |
revised |
Is the Mendeleev table explained in quantum mechanics?
excuse me, I corrected some grammar mistakes |

Mar
5 |
revised |
Is the Mendeleev table explained in quantum mechanics?
I edited the link to physics.stackexchange.com |

Feb
7 |
comment |
Stone-Weierstrass theorem for holomorphic functions?
Alex, you can contact me directly by e-mail: mathnet.ru/php/… |

Feb
7 |
comment |
Stone-Weierstrass 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/Functional-Analysis-Walter-Rudin/dp/0070542368 |

Feb
6 |
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Unifying (& “justifying”) the various definitions for differential operators
Ah, yes! $\phantom{*}$ |

Feb
6 |
comment |
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 |
comment |
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/Geometric-Invariant-Differential-Operators-Spherical/… 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
3 |
awarded | Good Question |

Feb
3 |
revised |
Stone-Weierstrass theorem for holomorphic functions?
deleted 23 characters in body |

Feb
3 |
revised |
Stone-Weierstrass theorem for holomorphic functions?
added 528 characters in body |

Feb
2 |
revised |
Stone-Weierstrass theorem for holomorphic functions?
added 331 characters in body |

Feb
2 |
comment |
Stone-Weierstrass 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
2 |
revised |
Stone-Weierstrass theorem for holomorphic functions?
I added the description of topology on ${\mathcal C}^\infty(M)$ |

Dec
11 |
comment |
Stone-Weierstrass 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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Stone-Weierstrass 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 |
comment |
Stone-Weierstrass 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
10 |
comment |
Stone-Weierstrass 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 Stone-Weierstrass 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
10 |
comment |
Stone-Weierstrass theorem for holomorphic functions?
Yes, of course! |

Dec
10 |
comment |
Stone-Weierstrass theorem for holomorphic functions?
@SimonHenry, as far as I understand, even in the case of $M={\mathbb C}$ there is no answer. |