Timeline for Quotient of a smooth projective surface by an involution
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2019 at 6:11 | comment | added | Daniel Loughran | @Kevin Casto: I'm not sure why you are telling me to read something which I have already read and has already been clarified in the comments. Anyway, Hironaka's example is much worse than you state; it gives a proper (non-projective) variety $X$ such that finite quotient $X/G$ is an algebraic space, but not a scheme. | |
| Feb 25, 2019 at 22:24 | comment | added | Kevin Casto | @DanielLoughran In Hironaka's example the variety quotiented by a finite group is not projective -- read the Wikipedia page you linked. | |
| Feb 25, 2019 at 16:31 | answer | added | Ariyan Javanpeykar | timeline score: 2 | |
| Feb 14, 2019 at 11:07 | comment | added | Jason Starr | @DanielLoughran. The argument by HYL is correct. Here is the elementary argument without GIT notation. For any ample invertible sheaf $L$ and any finite self-map $f$ of $X$, also $f^*L$ is ample. Any tensor product of ample invertible sheaves is ample, thus the tensor product $M$ of all pullbacks $f^*L$ over all elements $f$ of $G$ is ample and $G$-equivariant. For every $G$-invariant invertible sheaf $M$, for all sufficiently positive and divisible $n$, the invertible sheaf $M^{\otimes n}$ descends. Finally, by Chevalley, the descent invertible sheaf is ample. | |
| Feb 14, 2019 at 10:59 | comment | added | Daniel Loughran | Anyway, I agree the result should hold for surfaces. | |
| Feb 14, 2019 at 10:58 | comment | added | Daniel Loughran | @HYL: Are you sure this is true in general? Doesn't Hironaka's example give a counter-example for 3-folds? en.wikipedia.org/wiki/Hironaka%27s_example | |
| Feb 14, 2019 at 4:45 | review | Close votes | |||
| Feb 20, 2019 at 3:05 | |||||
| Feb 14, 2019 at 2:02 | comment | added | HYL | The quotient of a projective variety by a finite group is always projective. This follows from the GIT: Let $X$ be a projective variety and $G$ a finite group acting on it. Let $L$ be a very ample $G$-linearized invertible sheaf on $X$ (which always exists as $G$ is a linear reductive group). As $G$ is finite, every point of $X$ is stable. So we have a geometric quotient $X = X^s \to \mathrm{Proj}\bigoplus_{k \ge 0} H^0(X,L^{\otimes k})^G$ of $X$ by $G$. As $X$ is projective, the target $X/G$ is also projective. | |
| Feb 13, 2019 at 23:29 | comment | added | Donu Arapura | It should be true at least when the quotient $Y$ is smooth. To see this, first argue that the transcendence degree of the field of meromorphic functions $\mathbb{C}(Y)$ is 2, i.e. $Y$ is Moishezon. Then use a theorem of Chow-Kodaira to conclude $Y$ is projective. | |
| Feb 13, 2019 at 22:54 | history | asked | Philip Engel | CC BY-SA 4.0 |