Timeline for Hom between Brody hyperbolic varieties
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2021 at 17:30 | comment | added | Ariyan Javanpeykar | ...$Hom(X,Y) \to Y$ given by $f\mapsto f(c) $ is finite. Let me leave it at that for now. | |
| Mar 5, 2021 at 17:29 | comment | added | Ariyan Javanpeykar | It is another theorem that Hom and Hol are the same. Indeed, in Kobayashi's book it is also shown that every holomorphic map from an algebraic variety $Y$ to a projective Brody hyperbolic variety is algebraic; see arxiv.org/abs/1806.09338 for more on this. Let me also add that the Hom-scheme you are interested in (namely, the one that parametrizes morphisms from a projective variety $X$ to a Brody hyperbolic projective variety $Y$) is in fact projective and Brody hyperbolic. See arxiv.org/abs/1807.03665 for similar statements. The idea is that the evaluation map.... | |
| Mar 2, 2021 at 14:02 | comment | added | diverietti | I didn't want to be pushy of course! It was just in case as a new user you didn't know exactly how it works here! Take your time, and feel free not to "accept" my answer of course! :) | |
| Mar 2, 2021 at 13:59 | comment | added | user175135 | Absolutely. I just need some time to absorb it. | |
| Mar 2, 2021 at 13:58 | comment | added | diverietti | P.S. Since you are a new user and maybe you don't know exactly how it works here, may I say that once you get an answer which is satisfactory to you, then you should click to "accept" it. | |
| Mar 2, 2021 at 13:50 | comment | added | diverietti | Yes, I think the same. | |
| Mar 2, 2021 at 13:50 | history | edited | diverietti | CC BY-SA 4.0 |
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| Mar 2, 2021 at 13:44 | comment | added | user175135 | I meant the $\mathrm{Hom}$ functor which is representable by a scheme under the assumptions. But I think it coincides with what you wrote. | |
| Mar 2, 2021 at 13:38 | history | answered | diverietti | CC BY-SA 4.0 |