Timeline for The holomorphic version of Galois theory
Current License: CC BY-SA 3.0
10 events
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| Dec 17, 2016 at 16:28 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Dec 11, 2013 at 15:32 | history | edited | Loïc Teyssier | CC BY-SA 3.0 |
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| Dec 11, 2013 at 15:27 | comment | added | Loïc Teyssier | I'm no specialist, but Riemann manifolds of algebraic functions are compact for the Zariski topology, and thus compact as projective manifolds by Chow's theorem. I hope I'm not too much mistaken saying things in that way (again, I'm no specialist of these questions). The reference I found is a paper by Zariski himself : "The compactness of the Riemann manifold of an abstract field of algebraic functions." (1944) with MR number MR0011573. I hope this covers your (essential) objection. | |
| Dec 11, 2013 at 15:01 | comment | added | Ali Taghavi | There is no any obstruction for compactification to obtain a complex manifold again? There is no any obstruction for holomorphic extension? I would appreciate if you explain more clear. Where you used the polynomial ness of $\sigma$. Could you please give me a reference for some background? | |
| Dec 11, 2013 at 13:25 | comment | added | Loïc Teyssier | @AliTaghavi: See edited answer. I removed all my comments since they were incorporated in the edit. | |
| Dec 11, 2013 at 13:25 | history | edited | Loïc Teyssier | CC BY-SA 3.0 |
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| Dec 11, 2013 at 13:12 | comment | added | Ali Taghavi | I think that the radical formula has a Riemann surface so this Riemann surface is the desired M | |
| Dec 11, 2013 at 13:02 | comment | added | Ali Taghavi | According to the original formulation of my question we search for complex manifold M (Not orbifold)So do you think that the answer of my question is "No"? I mean that is it possible to prove that there is no a complex manifold with those properies? | |
| Dec 11, 2013 at 12:41 | comment | added | Ali Taghavi | Thank you.puting $a_{n}=1$ we simplify the question as follows: we identify all monic polynomial of degree n with $\mathbb{C}^{n}$. We search for covering space $M \rightarrow \mathbb{C}^{n}$. It seems that I donot underestand something. Does your covering map cover wholle C^n? could you please more explain? | |
| Dec 11, 2013 at 10:54 | history | answered | Loïc Teyssier | CC BY-SA 3.0 |