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?
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$\begingroup$ 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 ChowKodaira 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 counterexample for 3folds? 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 selfmap $f$ 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
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The quotient of a quasiprojective variety $X$ by a finite group action $G$ is quasiprojective. 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. SpringerVerlag, New York, 1970