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Consider a affine variety $X$ over the field of the complex numbers, and an action of a reductive group $G$ on $X$ (I will consider the case of $G$ not finite, in particular $G=\mathbb{C}^*$). Reading Hu-Keel's famous paper on Mori dream spaces, Lemma 2.1, I came across the hypothesis of having an action "with finite stabilizer". Also reading Hausen's paper "A generalization of Mumford's Geometric Invariant Theory", Lemma 5.5, I see the same notion.

Question: What is the precise definition of "action with finite stabilizer"? Does it mean for that for every point $p\in X$ the isotropy group $G_p$ has finite cardinality?

While this look the more simple solution, I'm a bit confused as having finite stabilizer implies every point is not fixed, and this for example excludes every $\mathbb{C}^*$-action on projective spaces for examples. I don't know, I think this definition is different from the one I'm thinking of; therefore I imagine it's like common knowledge. I've tried to read the proof of these statements but I didn't find any hint regarding this definition.

I apologize in advance for the simplicity of the question: if you think does not fit the criteria for this forum, I will post it on MSE. I first post it here since it is a question coming from my own research, despite being quite stupid.

Edit: the projective hypothesis is wrong: on Hu-Keel's paper the $G$-variety is only affine, I apologize! Anyway, the question still holds unfortunately.

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    $\begingroup$ If it is really for projective varieties, then almost certainly the definition allows fixed points. For example any algebraic group action of $\mathbb{C}^*$ on a normal projective variety has fixed points due to Suminhiro's theorem. $\endgroup$
    – Nick L
    Commented Aug 30, 2021 at 14:18
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    $\begingroup$ @Nick L.: No. The context is that you want to generalize results about free actions by allowing that each point has a finite stabilizer. For instance, a celebrated theorem of Keel-Mori states that if the action is proper with finite stabilizers, there exists a geometric quotient $X/G$ (as algebraic space). $\endgroup$
    – abx
    Commented Aug 30, 2021 at 17:11
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    $\begingroup$ Then the definition is never satisfied by any projective variety (when $G$ is a torus, which was the case OP was most interested in). The OP specifically required that the reductive group is acting on a projective variety, that was my point. $\endgroup$
    – Nick L
    Commented Aug 30, 2021 at 17:24

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Yes, finite sabilizers means that the isotropy group of every point is finite.

What typically happens in GIT is the following: you start with a reductive group $G$ acting on a projective variety $X$. This action is not good enough to define a naive quotient (some orbits are not closed and some points have infinite stabilizers). So what you do is remove a subvariety of ``unstable elements''. You are left with $X^{ss}$ (semistable points), which you can quotient algebraically (i.e. construct a quasi-projective variety $X^{ss}//G$ with a $G$-invariant morphism $X^{ss} \to X^{ss}//G$ satisfying some universal property).

Inside $X^{ss}$ you have the Zariski open subset $X^{s}$ of \emph{stable points}, which have closed orbit and finite stabilizer. In restriction to $X^{s}$, the GIT quotient coincides with the naive quotient (More precisely, $X^s \to X^s//G$ induces a bijection between complex points in $X^s//G$ and $G$-orbits in $X^s$), and is a smooth orbifold (orbifold points corresponding to orbits with non-trivial stabilizers).

To summarize, GIT builds algebraic quotients of actions that are not proper, but these quotients are particularly nice on the Zariski open locus where the action is proper (with finite stabilizers).

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  • $\begingroup$ Why is $X^𝑠//𝐺$ an orbifold? I.e., why does it necessarily have finite quotient singularities? A reference would be appreciated. Also are you assuming that X is smooth? $\endgroup$
    – Dr. Evil
    Commented Apr 25 at 9:41
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I am not sure if this helps. When you say ``every point is not fixed'' does that mean that the action is free, or does it mean that the stabilizer of every point of the variety is trivial? If it is the former, then I believe that I can help.

There is a result by William Haboush and others (I believe) which states that if $ G $ is a reductive group acting on a non-singular variety over an algebraically closed field field $ k $ there is an open sub-variety $ U $ for which the connected components of the stabilizers are conjugate.

What this means is that if your reductive group $ G $ is irreducible, you may replace $ G $ by $ G/G_{x} $. Because the conjugacy classes of the identity are simply the identity, we may assume that the action is free, i.e., that the only element that fixes every point of the variety is the identity.

Even if your variety is not non-singular, you may replace it by the non-singular locus which is dense in the underlying variety, and assume that the reductive group acts freely on the non-singular locus. Because the non-singular locus is dense in the underlying variety, you may assume that the action is free in general.

The result by Haboush et. al., is Deformation Theoretic Methods in the Theory of Algebraic Transformation Spaces published in the Journal of Mathematics of Kyoto University in 1974.

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