5
$\begingroup$

Is the quotient of a smooth complex projective surface by an involution projective? Suppose the quotient happens to be smooth; does that change the situation?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Donu Arapura Feb 13 at 23:29
  • 4
    $\begingroup$ 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. $\endgroup$ – HYL Feb 14 at 2:02
  • $\begingroup$ @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 $\endgroup$ – Daniel Loughran Feb 14 at 10:58
  • $\begingroup$ Anyway, I agree the result should hold for surfaces. $\endgroup$ – Daniel Loughran Feb 14 at 10:59
  • 4
    $\begingroup$ @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. $\endgroup$ – Jason Starr Feb 14 at 11:07
2
$\begingroup$

The quotient of a quasi-projective variety $X$ by a finite group action $G$ is quasi-projective. As is explained in [1, Remarque V.5.1], this follows from [1, Theoreme V.4.1].

[1] M. Demazure and A. Grothendieck. Schémas en groupes I, II, III (SGA 3). Lecture Notes in Math. 151, 152, 153. Springer-Verlag, New York, 1970

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.