Let $G$ be algebraic group acting on a smooth algebraic variety $M$. Assume the action is proper and free. Is the orbit space $M/G$ an algebraic variety? If so, could someone point to a reference? If not, could someone point to a counterexample? Thanks.
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3 Answers
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The answer is no.
A counterexample is given in the paper by James K. Deveney and David R. Finston
A Proper $G_a$ Action on $\mathbb{C}^5$ Which is Not Locally Trivial, Proc. Amer. math. Soc. 123 (1995).
Quoting from the introduction:
An assertion which would imply that any proper, fixed point free $G_a$-action on a normal variety is locally trivial and admits a quasi-projective quotient appears in a paper of Magid and Fauntleroy [5], and the source of the error is pointed out in [4]. The example here indicates that no such result is possible.
answered May 17, 2011 at 13:58
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$\begingroup$ Free implies fixed point free but not vice versa. Do they have an example of a free action where the quotient is bad? $\endgroup$– JonathanCommented May 17, 2011 at 14:24
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3$\begingroup$ In characteristic zero, $\mathbb{G}_a$ has no nontrivial proper algebraic subgroups. So fixed point free does imply free in this case. $\endgroup$ Commented May 17, 2011 at 14:35
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$\begingroup$ Thanks! It's very helpful. How about this: Let G be complex Lie group acting on a complex manifold M. Assume the action is proper and free. Is the orbit space M/G a complex manifold? It is clear that M/G is at least an almost complex manifold. What's less clear, though I believe true, is that the almost complex structure is integrable. Does anyone has a source? Thanks! $\endgroup$– NailCommented May 17, 2011 at 15:12
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$\begingroup$ Some kind of holomorphic slice theorem would imply this (though you might need additional hypotheses). See Sjamaar's paper here: jstor.org/pss/2118628 $\endgroup$– JonathanCommented May 17, 2011 at 20:21
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See Geometric Invariant Theory by Mumford, Fogarty, and Kirwan.
If $G$ is reductive acting on a variety $X$, there will be Zariski open subsets $X^{ss}$ and $X^s$ consisting of semistable and stable points for the action, with $X^s \subseteq X^{ss} \subseteq X$. The quotients $X^{ss}/G$ and $X^s/G$ will exist as varieties. The quotient $X^{ss}/G$ is a categorical quotient, and the quotient $X^s/G$ is a geometric quotient (it satisfies many additional nice properties, like giving $X^s \to X^s/G$ the structure of a principal bundle).
If the action is scheme-theoretically proper and free, this may be enough to guarantee that $X = X^s$ (when the action is free, $X^s = X^{ss}$ is automatic). There is a lot of material in the above reference that is devoted to analysis of stability, so I am almost certain that this situation is treated in it.
When $G$ is not reductive, then extremely bad things can happen, as in Francesco's answer above.
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Harald Holmann: Quotientenräume komplexer Mannigfaltigkeiten nach komplexen Lieschen Automorphismengruppen. Math. Ann. 139 1960 383–402 (1960).
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1$\begingroup$ You probably want to add some explanation as there are already two other answers. $\endgroup$ Commented Feb 6, 2016 at 17:38