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The Luna slice theorem states that if a reductive group $G$ acts on an affine space $X$ and $O$ is a closed orbit, then (in the etale topology) there exists a $G$-invariant negihborhood of $O$ with a $G$-invariant projection. It seems that the proof uses only the fact that the stabilizer of any point in $O$ is reductive. Does the theorem hold in this generality? If yes - what is the reference?

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    $\begingroup$ You might be looking for Theorem 2.1 in arxiv.org/abs/1504.06467 $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 10, 2016 at 18:07
  • $\begingroup$ Ask Michel Brion ([email protected]). He knows the answer. $\endgroup$
    – Al-Amrani
    Commented Mar 15, 2016 at 20:00
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    $\begingroup$ It would help if you made more explicit what you mean here by "affine space", as well as what kind of field you work over and what kind of source you are following for Luna slices. Aside from that, in the classical characteristic 0 setting for this theory, note that "reductive" is the same as "linearly reductive" in order to compare the passage Ariyan points to. $\endgroup$
    – Jim Humphreys
    Commented Mar 23, 2016 at 22:40
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    $\begingroup$ Thanks to all of you. Ariyan, It seems that you gave me exactly the answer I need. Could you please post it as an answer so I can accept it? $\endgroup$
    – Rami
    Commented May 2, 2016 at 18:32
  • $\begingroup$ @AriyanJavanpeykar How much the assumption of closeness of the orbit is important for the existence of the etal slice? I mean does an analog of Luna's theorem hold for open orbits? $\endgroup$
    – QGravity
    Commented Oct 4, 2022 at 19:35

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There is already a counterexample in Luna's original paper (Remark 4 on p. 98). It is the $SL(2)$-orbit of $x^2y$ in the space of binary 3-forms. Luna attributes this example to Richardson.

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    $\begingroup$ Luna's slice theorem is much stronger than just asserting "the existence of a G-invariant etale neighborhood of O with a G-invariant projection". It also asserts that this neighborhood is a pull-back from X//G. In particular, orbits are mapped isomorphically to orbits. The example above shows that this won't work for non-closed orbits. The weak version of the slice theorem is probably true. $\endgroup$
    – Friedrich Knop
    Commented Apr 6, 2016 at 19:18
  • $\begingroup$ Thank you very much. I think we need only the weaker version. $\endgroup$
    – Rami
    Commented Apr 19, 2016 at 13:43

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