Timeline for Why does a complex linear normalization of a real algebraic surface inherit a real structure?
Current License: CC BY-SA 4.0
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| Nov 22, 2019 at 9:30 | comment | added | Mikhail Skopenkov | The real one is another projection onto $X$, corresponding to the morphism $\bar A\otimes_\mathbb{R}\mathbb{C}\to A\otimes_\mathbb{R}\mathbb{C}$. Luckily, by the universality, the latter projection must be a composition of the former one with a complex isomorphism, which gives the required lift. Just an unimportant comment to your very nice proof. Edited the question accordingly. (This shows that a reference, if any, would be preferrable.) | |
| Nov 22, 2019 at 9:19 | comment | added | Mikhail Skopenkov | Just a minor comment for future users of your answer. The projection $\bar X\to X$ needs not to be real at all because the complex normalization is only defined up to a complex automorphism. It is another projection, | |
| Nov 21, 2019 at 7:18 | comment | added | Angelo | Yes, sure, the projection is also defined by real polynomial maps. | |
| Nov 21, 2019 at 4:24 | comment | added | Mikhail Skopenkov | Thank you very much again! Personally, would include this type of argument into a paper, if there is no reference available, especially if the rest of the paper is far from algebraic geometry and commutative algebra. Just to make sure: does your argument imply in particular that the projection $\bar X\to X$ is given by linear polynomials with real coefficients? | |
| Nov 20, 2019 at 17:28 | comment | added | Angelo | I added a different, more algebraic proof. It's very standard commutative algebra, not something I would put in a paper. | |
| Nov 20, 2019 at 17:26 | history | edited | Angelo | CC BY-SA 4.0 |
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| Nov 20, 2019 at 9:21 | comment | added | Mikhail Skopenkov | (cont'd) And $\tau:X\to X$ is not a morphism and cannot be anti-analytic in the usual sense, once $X$ is not smooth. Thank you also for the explanation what is meant by anti-analytic involution in the books; this comment is really helpful (because the authors seem to find this too obvious to mention:). Finally, a bit offtopic remark: imho generalization to locally ringed spaces does not prove (1) but just moves the issue to a deeper level - the required properties of the spaces seem to be harder to prove than (1) itself. But let me apologize for too much muttering, and thank you again! | |
| Nov 20, 2019 at 9:13 | comment | added | Mikhail Skopenkov | Thank you very much for a clear explanation, especially for the last down-to-Earth part avoiding schemes. Notice that the question is primarily about a reference, not a proof (a reference to mathoverflow is not reliable enough for a research paper). Could you recommend a reference to the mentioned universal property, if you have time? So far looked up Liu's Algebraic Geometry and Arithmetic Curves - but Definition 1.19 there asserts universality only for dominant morphisms, not for `involutions of locally ringed spaces, that are antilinear with respect to complex scalars'. | |
| Nov 20, 2019 at 7:43 | history | answered | Angelo | CC BY-SA 4.0 |